Radmila Bulajich Manfrino
José Antonio Gómez Ortega
Rogelio Valdez Delgado
Topics in
Algebra
and Analysis
Preparing for the
Mathematical Olympiad
Radmila Bulajich Manfrino
José Antonio Gómez Ortega
Rogelio Valdez Delgado
Topics in Algebra and Analysis
Preparing for the Mathematical Olympiad
Radmila Bulajich Manfrino
Facultad de Ciencias
Universidad Autónoma del Estado de Morelos
Cuernavaca, Morelos, México
José Antonio Gómez Ortega
Facultad de Ciencias
Universidad Nacional Autónoma de México
Distrito Federal, México
Rogelio Valdez Delgado
Facultad de Ciencias
Universidad Autónoma del Estado de Morelos
Cuernavaca, Morelos, México
ISBN 978-3-319-11945-8
ISBN 978-3-319-11946-5
DOI 10.1007/978-3-319-11946-5
(eBook)
Library of Congress Control Number: 2015930195
Mathematics Subject Classification (2010): 00A07, 11B25, 11B65, 11C08, 39B22, 40A05, 97Fxx
Springer Cham Heidelberg New York Dordrecht London
© Springer International Publishing Switzerland 2015
This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of
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Springer International Publishing AG Switzerland is part of Springer Science+Business Media
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Introduction
Topics in algebra and analysis have become fundamental for the mathematical
olympiad. Today, the problems in these topics that appear in the contests are
frequent, and the problems from other areas that use algebra and analysis in
their solutions are also frequent. In this book, we want to point out the principal
algebra and analysis tools that a student must assimilate and learn to use gradually
in training for mathematical contests and olympiads. Some of the topics that we
study in the book are also part of the mathematical syllabus in high school courses,
but there are other topics that are presented at the college level. That is, the book
can be used as a reference text for undergraduates in the first year of college who
will be facing algebra and analysis problems and will be interested in learning
techniques to solve them.
The book is divided in ten chapters. The first four correspond to topics from
high school and they are basic for the students that are training for the mathematical olympiad contest, at a local and national level. The next four chapters
are usually studied in the first year of college, but they have become fundamental
tools, for the students competing in an international level. The last two chapters
contain the problems and solutions of the theory studied in the book.
The first chapter covers the basic algebra, as are the numerical systems, absolute value, notable products, and factorization, among others. We expect that the
reader gain some skills for the manipulation of equations and algebraic formulae
to carry them in equivalent forms, which are easier to understand and work with
them.
In Chapter 2 the study of the finite sums of numbers is presented, for instance, the sum of the squares of the first n natural numbers. The telescopic sums,
arithmetic and geometric progressions are analyzed, as well as some of its properties.
Chapter 3 talks about the mathematical technique to prove mathematical
statements that involve natural numbers, known as the principle of mathematical
induction. Its use is exemplified with several problems. Many equivalent statements
of the principle of mathematical induction are presented.
To complete the first part of the book, in Chapter 4 the quadratic and cubic polynomials are studied, with emphasis in the study of the discriminant of a
quadratic polynomial and Vieta’s formulas for these two classes of polynomials.
v
vi
Introduction
The second part of the text begins with Chapter 5, where the complex numbers are studied, as well as its properties and some applications are given. All
these with examples related to mathematical olympiad problems. In addition, a
proof of the fundamental theorem of algebra is included.
In Chapter 6, the principal properties of functions are studied. Also, there is
an introduction to the functional equations theory, its properties and a series of
recommendations are given to solve the problems where appear functional equations.
Chapter 7 talks about the notion of sequence and series. Special sequences
are studied as bounded, periodic, monotone, recursive, among others. In addition,
the concept of convergence for sequences and series is introduced.
In Chapter 8, the study of polynomials from the first part of the book is
generalized. The theory of polynomials of arbitrary degree is presented, as well
as several techniques to analyze properties of the polynomials. At the end of the
chapter, the polynomials of several variables are studied. Most of the sections of
these first eight chapters have at the end a list of exercises for the reader, selected
and suitable to practice the topics in the corresponding sections. The difficulty of
the exercises vary from being a direct application of a result seen in the section to
being a contest problem that with the technique studied is possible to solve.
Chapter 9 is a collection of problems, each one of them close to one or more of
the topics seen in the book. These problems have a degree of difficulty greater than
the exercises. Most of the problems have appeared in some mathematical contests
around the world or olympiads. In the solution of each problem is implicit the
knowledge and skills that are need to manipulate algebraic expressions.
Finally, Chapter 10 contains the solutions to all exercises and problems presented in the book. The reader can notice that at the end of some sections there is
a ⋆ symbol, this means that the level of the section is harder than others sections.
In a first lecture, the reader can skip these sections; however, it is recommended
that the reader have them in mind for the techniques used in them.
We thank Leonardo Ignacio Martı́nez Sandoval and Rafael Martı́nez Enrı́quez
for his always-helpful comments and suggestions, which contribute to the improvement of the material presented in this book.
Contents
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
1 Preliminaries
1.1 Numbers . . . . . . . . . . . .
1.2 Absolute value . . . . . . . . .
1.3 Integer part and fractional part
1.4 Notable products . . . . . . . .
1.5 Matrices and determinants . .
1.6 Inequalities . . . . . . . . . . .
1.7 Factorization . . . . . . . . . .
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of a number
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1
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25
2 Progressions and Finite Sums
2.1 Arithmetic progressions
2.2 Geometric progressions
2.3 Other sums . . . . . . .
2.4 Telescopic sums . . . .
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31
36
38
40
3 Induction Principle
3.1 The principle of mathematical induction
3.2 Binomial coefficients . . . . . . . . . . .
3.3 Infinite descent . . . . . . . . . . . . . .
3.4 Erroneous induction proofs . . . . . . .
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43
53
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60
4 Quadratic and Cubic Polynomials
4.1 Definition and properties . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 Vieta’s formulas . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63
65
66
5 Complex Numbers
5.1 Complex numbers and their properties . . . . . .
5.2 Quadratic polynomials with complex coefficients
5.3 The fundamental theorem of algebra . . . . . . .
5.4 Roots of unity . . . . . . . . . . . . . . . . . . .
5.5 Proof of the fundamental theorem of algebra ⋆ .
75
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81
82
85
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vii
viii
Contents
6 Functions and Functional Equations
6.1 Functions . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Properties of functions . . . . . . . . . . . . . . . . .
6.2.1 Injective, surjective and bijective functions .
6.2.2 Even and odd functions . . . . . . . . . . . .
6.2.3 Periodic functions . . . . . . . . . . . . . . .
6.2.4 Increasing and decreasing functions . . . . . .
6.2.5 Bounded functions . . . . . . . . . . . . . . .
6.2.6 Continuity . . . . . . . . . . . . . . . . . . .
6.3 Functional equations of Cauchy type . . . . . . . . .
6.3.1 The Cauchy equation f (x + y) = f (x) + f (y)
6.3.2 The other Cauchy functional equations ⋆ . .
6.4 Recommendations to solve functional equations . . .
6.5 Difference equations. Iterations . . . . . . . . . . . .
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89
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102
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108
111
7 Sequences and Series
7.1 Definition of sequence . . . . . . . . . .
7.2 Properties of sequences . . . . . . . . .
7.2.1 Bounded sequences . . . . . . . .
7.2.2 Periodic sequences . . . . . . . .
7.2.3 Recursive or recurrent sequences
7.2.4 Monotone sequences . . . . . . .
7.2.5 Totally complete sequences . . .
7.2.6 Convergent sequences . . . . . .
7.2.7 Subsequences . . . . . . . . . . .
7.3 Series . . . . . . . . . . . . . . . . . . .
7.3.1 Power series . . . . . . . . . . . .
7.3.2 Abel’s summation formula . . . .
7.4 Convergence of sequences and series ⋆ .
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115
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135
8 Polynomials
8.1 Polynomials in one variable . . . . . . . . .
8.2 The division algorithm . . . . . . . . . . . .
8.3 Roots of a polynomial . . . . . . . . . . . .
8.3.1 Vieta’s formulas . . . . . . . . . . .
8.3.2 Polynomials with integer coefficients
8.3.3 Irreducible polynomials . . . . . . .
8.4 The derivative and multiple roots ⋆ . . . .
8.5 The interpolation formula . . . . . . . . . .
8.6 Other tools to find roots . . . . . . . . . . .
8.6.1 Parameters . . . . . . . . . . . . . .
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139
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145
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150
151
153
153
ix
Contents
8.7
8.8
8.6.2 Conjugate . . . . . . . . .
8.6.3 Descartes’ rule of signs ⋆ .
Polynomials that commute . . .
Polynomials of several variables .
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154
155
157
161
9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
10 Solutions to Exercises and Problems
10.1 Solutions to exercises of Chapter 1
10.2 Solutions to exercises of Chapter 2
10.3 Solutions to exercises of Chapter 3
10.4 Solutions to exercises of Chapter 4
10.5 Solutions to exercises of Chapter 5
10.6 Solutions to exercises of Chapter 6
10.7 Solutions to exercises of Chapter 7
10.8 Solutions to exercises of Chapter 8
10.9 Solutions to problems of Chapter 9
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179
190
199
214
220
229
239
248
261
Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309
Chapter 1
Preliminaries
1.1 Numbers
We will assume that the reader is familiar with the notion of the set of numbers
that we usually use to count. This set is called the set of natural numbers and it
is usually denoted by N, that is,
N = {1, 2, 3, . . . }.
In this set we have two operations, the sum and the product, that is, if we add or
multiply two numbers in the set we obtain a natural number. In some books 0 is
considered a natural number, however, in this book it is not, but we will suppose
that 0 is such that n + 0 = n, for every natural number n.
Now, suppose that we want to solve the equation x + a = 0, with a ∈ N, that
is, we want to find an x such that the equality is true. This equation does not have
a solution in the set of natural numbers N, therefore we need to define another set
which includes the set of numbers N but also the negative numbers. In other words,
we need to extend the set of numbers N in order that such equation can be solved
in the new set. This new set is called the set of integers and is denoted by Z, that is,
Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . .}.
In this set we also have two operations, the sum and the product, which satisfy
the following properties.
Properties 1.1.1.
(a) The sum and the product of integers are commutative. That is, if a, b ∈ Z,
then
a + b = b + a and ab = ba.
(b) The sum and the product of integers are associative. That is, if a, b and c ∈ Z,
then
(a + b) + c = a + (b + c) and (ab)c = a(bc).
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5_1
1
2
Chapter 1. Preliminaries
(c) There exists in Z a neutral element for the sum, the number 0. That is, if
a ∈ Z, then
a + 0 = 0 + a = a.
(d) There exists in Z a neutral element for the multiplication, the number 1. That
is, if a ∈ Z, then
a1 = 1a = a.
(e) For each a ∈ Z there exists an inverse element under the sum which is
denoted by −a. That is,
a + (−a) = (−a) + a = 0.
(f) In Z a distributive law holds, in which addition and multiplication are involved. That is, if a, b and c ∈ Z, then
a(b + c) = ab + ac.
Note that the existence of the additive inverse allows us to solve equations of
the type mentioned above, that is, x + a = b, where a and b are integers. However,
there is not necessarily an integer number x that solves the equation qx = p,
with p and q integers. Therefore the necessity arises to extend the set of integers.
Consider the set of rational numbers, which is denoted by Q, defined as
p
| p ∈ Z and q ∈ Z\{0} .
Q=
q
In general, when working with the rational number pq we ask that p and q do not
have common prime factors, that is, the numbers are relative primes, which is
denoted by (p, q) = 1. In the set of rational numbers we also have two operations,
the sum and the product, which satisfy all the properties valid in the set of integers,
but in the case of the product we have an extra property, the multiplicative inverse
element.
Property 1.1.2. If pq ∈ Q, with p = 0 and (p, q) = 1, then there exists a unique
number, pq ∈ Q, which is called the multiplicative inverse of pq , such that
p q
· = 1.
q p
Using this property we can solve equations of the form qx = p, however, there
are numbers that we cannot write as the quotient of two integers. For example, if
we want to solve the equation x2 − 2 = 0, this equation does not
√ have a solution
2 and proceed to
in the set Q.
We
write
the
solutions
of
the
equation
as
x
=
±
√
prove that 2 is not in Q.
√
Proposition 1.1.3. The number 2 is not a rational number.
√
Proof. Suppose√the contrary, that is, 2 is a rational number. Therefore, it can
be written as 2 = pq , where p and q do not have common factors. Squaring
both sides of the equation we get 2 =
p2
q2 ,
that is, 2q 2 = p2 . This means that
3
1.1 Numbers
p2 is an even number, but then p is also even. But if p is even, p is of the form
p = 2m, then 2q 2 = (2m)2 = 4m2 . Dividing by 2 both sides of the equation, give
us q 2 = 2m2 , that is, q 2 is even and therefore q is also even. Hence, p and q are
even numbers, which contradicts
the fact that p and q were assumed as not having
√
common factors. Thus, 2 is not a rational number.
We can give a geometric representation of the rational numbers as points on
a straight line, which in this case is called the number line. A straight line can
be travelled in two directions, one which we call the positive direction and the
other the negative direction. Once we have agreed about which of the directions
is to be taken as positive we talk of an oriented straight line. For example, we can
decide that the positive direction goes from left to right. If we consider two points
O and U on the straight line, we will give the same orientation to a line segment
contained on the straight line. That is, if we let the point O be 0 and U be on
the right of it, we will say that the line segment OU is traversed in the positive
direction. If U represents the point 1, we will call OU an unitary oriented straight
line segment. In this way, we place, to the right of 0, what we take to be all the
positive integers equally spaced along the line, that is, two consecutive integers
are spaced a distance equal to the length of the segment OU . To represent the
negative numbers it is sufficient to do the same, starting at O and traversing the
straight line in the opposite direction.
The rational number pq is defined as the oriented segment pq OU . This segment
is obtained when we sum p times the qth part of the segment OU . More precisely,
we do the following:
(a) Divide the segment OU into q equal parts. To do this, we make use of an
extra straight line passing through O and not perpendicular to OU , and in this
line we take q points W1 , . . . , Wq , where two consecutive points are separated a
distance OW1 . Now, we draw the line segment from Wq to U and for each Wj we
draw a parallel straight line to U Wq , the intersection points of the parallels with
OU will divide OU in q equal parts. If V is the intersection point of the parallel to
U Wq through W1 , we have that V is the point that represents the number q1 (note
that OV has the same orientation as OU ). We also consider V ′ the symmetric
point to V , with respect to O. In the following figure, we took q = 4.
Wq
W1
V′ 0
O V
1
U
2
4
Chapter 1. Preliminaries
(b) If p is a non-negative integer, take
OP = OV + OV + · · · + OV = p · OV.
p times
p
q OU .
The segment OP is, by definition,
point P , with p = 6 and q = 4.
U′
In the next figure we have marked the
0
1
O
U
2
P
(c) If p is negative, let p′ be the positive integer such that p = −p′ . Then
OP = OV ′ + OV ′ + · · · + OV ′ = p′ OV ′ = (−p′ )OV = p · OV.
p′ times
p
q OU .
The segment OP is, by definition,
point is simply denoted by pq .
Since OU is the unitary segment, this
With this representation of the rational numbers, we have that every rational
number defines a point on the number line, but there are points on the number line
that are not represented by a rational number. For example,
√ we want to determine,
on the number line, which point represents the number 2, which we proved above
is not a rational number. To do this, take the right triangle whose legs are each
equal to 1. √
Then, by the Pythagorean theorem, the hypotenuse of
√ the triangle
is equal to 2. If we take a compass and draw a circle of radius 2 and center
at 0, the point √where the circle intersects the positive part of the number line
corresponds to 2.
√
2
1
0
O
1
U
√
2
A point on the number line that does not correspond to a rational number represents an irrational number, and the set of irrational numbers is denoted by I.
The union of these two sets is called the set of real numbers and is denoted
by R, that is, R = Q ∪ I.
The set of real numbers R contains the set of natural numbers, the set of integers
and the set of rational numbers. In fact, we have the following chain of inclusions
N ⊂ Z ⊂ Q ⊂ R.
5
1.1 Numbers
Given two points on the number line that we know represent two real numbers, we can find the point which represents the sum of these two numbers in the
following way: if P and Q are two points on the straight line and O is the origin,
the sum will be the addition of the oriented segments OP and OQ, as we can see
in the following figure.
OP + OQ
OP
OQ
OP
0
O
P +Q
Q
P
We can also find the point that represents the product of two points P and
Q which are on the number line. In order to do that we consider an extra straight
line passing through the origin O and not perpendicular to the line OP . We mark
on the extra straight line the unity U and the point Q. Through Q we draw a
parallel line to U P which will intersect the real line at a point R.
OR
Since the triangles ORQ and OP U are similar it follows that OP
= OQ
OU ,
therefore OR · OU = OP · OQ, hence OR represents the product of P and Q.
Q
U
O
P
R
With this geometric representation of the numbers it is easy to find, in the
real line, the sum and the product of any two real numbers, no matter whether
they are rational or irrational numbers.
Similarly, as it happens in the set of integers, the operations in the set of real
numbers have the same properties.
Properties 1.1.4.
(a)
(b)
(c)
(d)
The sum of two real numbers is a real number.
The sum of two real numbers is commutative.
The sum is associative.
The number 0 is called the additive neutral element. That is, x + 0 = x for
all x ∈ R.
(e) Every real number x has an additive inverse. That is, there is a real number,
which we denote by −x, such that x + (−x) = 0.
6
Chapter 1. Preliminaries
(f)
(g)
(h)
(i)
The product of two real numbers is a real number.
The product of two real numbers is commutative.
The product is associative.
The number 1 is the multiplicative neutral element. That is, x · 1 = x for all
x ∈ R.
(j) Every real number x different from 0, has a multiplicative inverse. That is,
there exists a real number, which we denote by x−1 , such that x · x−1 = 1.
(k) For any three real numbers x, y, z it follows that
x (y + z) = x · y + x · z.
This property is called the distributive law.
In the set of integers there is an order. With this we want to point out that
given two integers a and b we can say which one is greater. We say that a is greater
than b if a − b is a natural number. In symbols we have that
a > b if and only if a − b ∈ N.
This is equivalent to saying that a − b > 0.
In general, the notation a > b is equivalent to b < a. The expression a ≥ b means
that a > b or a = b. Similarly, a ≤ b means that a < b or a = b.
Properties 1.1.5. If a is an integer number, one and only one of the following
relations holds:
(a) a > 0,
(b) a = 0,
(c) a < 0.
In the set of rational numbers and in the set of real numbers, we also have
the order properties. The order of the real numbers enables us to compare two
numbers and to decide which one of them is greater or whether they are equal.
Let us assume that the real number system contains a set P , which we will call the
set of positive numbers, and we will express in symbols a > 0 if a belongs to P .
In the geometric representation of the real numbers, the set P in the number
line is, of the two pieces in which O has divided the straight line, the piece which
contains U (the number 1). The following properties are satisfied.
Properties 1.1.6. Every real number x has one and only one of the following properties:
(a) x = 0.
(b) x ∈ P , that is, x > 0.
Properties 1.1.7.
(c) −x ∈ P , that is, −x > 0.
(a) If x, y ∈ P , then x + y ∈ P (in symbols x > 0, y > 0, then x + y > 0).
(b) If x, y ∈ P , then xy ∈ P (in symbols x > 0, y > 0, then xy > 0).
We will denote by R+ the set P of positive real numbers.
Now we can define the relation x is greater than y, by saying that it holds
x − y ∈ P (in symbols x > y). Similarly, x is smaller than y, if y − x ∈ P (in
7
1.1 Numbers
symbols x < y). Observe that x < y is equivalent to y > x. We can also define
that x is smaller than or equal to y, if x < y or x = y, (using symbols x ≤ y).
Example 1.1.8.
(a) If x < y and z is any number, then x + z < y + z.
(b) If x < y and z > 0, then xz < yz.
In fact, to prove (a) we see that x + z < y + z if and only if (y + z) − (x + z) > 0
if and only if y − x > 0 if and only if x < y. To prove (b), we proceed as follows:
x < y implies that y − x > 0, and since z > 0, then (y − x)z > 0, therefore
yz − xz > 0, hence xz < yz.
Exercise 1.1. Prove the following statements:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
If
If
If
If
If
If
a < 0, b < 0, then ab > 0.
a < 0, b > 0, then ab < 0.
a < b, b < c, then a < c.
a < b, c < d, then a + c < b + d.
a > 0, then a−1 > 0.
a < 0, then a−1 < 0.
Exercise 1.2. Let a, b be real numbers. Prove that, if a + b, a2 + b and a + b2 are
rational numbers, and a + b = 1, then a and b are rational numbers.
Exercise 1.3. Let a, b be real numbers such that a2 + b2 , a3 + b3 and a4 + b4 are
rational numbers. Prove that a + b, ab are also rational numbers.
Exercise 1.4.
√
(i) Prove that if p is a prime number, then p is an irrational number.
√
(ii) Prove that if m is a positive integer which is not a perfect square, then m
is an irrational number.
Exercise 1.5. Prove that there are an infinite number of pairs of irrational numbers
a, b such that a + b = ab is an integer number.
Exercise 1.6. If the coefficients of
ax2 + bx + c = 0
are odd integers, then the roots of the equation cannot be rational numbers.
Exercise 1.7. Prove that for real numbers a and b, it follows that
√
a+ b=
a+
√
a2 − b
+
2
a−
√
a2 − b
.
2
8
Chapter 1. Preliminaries
Exercise 1.8. For positive numbers a and b, find the value of:
√
√
a a a a . . ..
(ii) a b a b . . ..
(i)
Exercise 1.9 (Romania, 2001). Let x, y and z be non-zero real numbers such that
xy, yz and zx are rational numbers. Prove that:
(i) x2 + y 2 + z 2 is a rational number.
(ii) If x3 + y 3 + z 3 is a rational number different from zero, then x, y and z are
rational numbers.
Exercise√1.10 (Romania,
2011). Let a, b be different real positive numbers, such
√
that a− ab and b− ab are both rational numbers. Prove that a and b are rational
numbers.
The decimal system is a positional system in which every digit takes a value based
on its position with respect to the decimal point. That is, the digit is multiplied
by a power of 10 according to the position it occupies. For the units digit, that is,
the digit which is just to the left-hand side of the decimal point, it is multiplied
by 10n , with n = 0. Accordingly, the digit in the tens position must be multiplied
by 101 = 10. The exponent increases one by one when we move from right to left
and decreases one by one when we move in the other direction. For example,
87325.31 = 8 · 104 + 7 · 103 + 3 · 102 + 2 · 101 + 5 · 100 + 3 · 10−1 + 1 · 10−2 .
In general, every real number can be written as an infinite decimal expansion
in the following way
bm . . . b1 b0 .a1 a2 a3 . . . ,
where bi and ai belong to {0, 1, . . . , 9}, for every i. The symbol . . . means that on
the right-hand side of the number we can have an infinite number of digits, in this
way the number bm . . . b1 b0 .a1 a2 a3 . . . , represents the real number
bm · 10m + · · · + b1 · 10 + b0 · 100 + a1 · 10−1 + a2 · 10−2 + · · · .
For example,
1
3
= 0.3333 . . . ,
1
2
= 0.50000 . . . ,
3
7
√
= 0.428571428571 . . .,
2 = 1.4142135 . . . .
With this notation we can distinguish between rational and irrational numbers.
The rational numbers are those in which the decimal expansion is finite or infinite,
but there is, always, a certain number of digits which, after a certain point, repeat
34
= 0.123636 . . . is periodic of period 2
periodically, for example the number 275
after the third digit. Meanwhile, for the irrational numbers, the decimal expansion
is infinite and never becomes periodic.
9
1.1 Numbers
This manner of representing numbers is essential to solve some of the problems
that appear in the mathematical olympiad.
In the same way, as in the decimal representation of the numbers in base 10, we
can represent the integers in any base.
If m is a positive integer, we can find its representation in base b if we write
the number as a sum of powers of b, that is, m = ar br + · · · + a1 b + a0 . The
integers which appear as coefficients of the powers of b in the representation must
be smaller than b.
Observation 1.1.9. To identify a number which is not written in base 10, we will
use a subindex indicating the base, for example, 12047 means that the number 1204
is a number expressed using base 7.
Let us analyze the following example.
Example 1.1.10. In which base is the number 221 a factor of 1215?
The number 1215 in base a is written as a3 + 2a2 + a + 5 and the number 221
in base a is 2a2 + 2a + 1. Therefore, if we divide a3 + 2a2 + a + 5 by 2a2 + 2a + 1
we obtain that
1
1
a+
2
2
9
1
+ − a+
.
2
2
Therefore, since 1215a has to be a multiple of 221a, the remainder − 12 a + 29 has
to be 0 and 12 a + 21 has to be an integer number. Both conditions are satisfied if
a = 9.
a3 + 2a2 + a + 5 = (2a2 + 2a + 1)
Exercise 1.11. Write the following numbers as
integers:
(i) 0.11111 . . .
(ii) 1.14141414 . . . .
m
n,
where m and n are positive
Exercise 1.12.
(i) Prove that 121b is a perfect square in any base b ≥ 2.
(ii) Determine the smallest value of b such that 232b is a perfect square.
Exercise 1.13 (IMO, 1970). Let a, b and n be positive integers greater than 1. Let
An−1 and An be numbers expressed in the numerical system in base a and Bn−1
and Bn be two numbers in the numerical system in base b. These numbers are
related in the following way,
An = xn xn−1 . . . x0 ,
An−1 = xn−1 xn−2 . . . x0 ,
Bn = xn xn−1 . . . x0 ,
Bn−1 = xn−1 xn−2 . . . x0 ,
with xn = 0 and xn−1 = 0. Prove that a > b if and only if
Bn−1
An−1
<
.
An
Bn
10
Chapter 1. Preliminaries
1.2 Absolute value
We define the absolute value of a real number x as
|x| =
x, if
−x, if
x≥0
x < 0.
(1.1)
For a real non-negative number k, the identity |x| = k is satisfied only by
x = k and x = −k.
The inequality |x| ≤ k is equivalent to −k ≤ x ≤ k. This can be seen as
follows: If x ≥ 0, then 0 ≤ x = |x| ≤ k. On the other hand, if x ≤ 0, then
−x = |x| ≤ k, therefore, x ≥ −k. As a consequence of the previous discussion, we
observe that x ≤ |x|. In the next figure we show the values of x which satisfy the
inequality. These being the values lying between −k and k, and including the two
numbers. The set [−k, k] = {x ∈ R | − k ≤ x ≤ k} is called a closed interval, since
it contains the numbers k and −k. The numbers −k, k are called the endpoints of
the interval.
[
−k
|
O
]
k
Similarly, the inequality |x| ≥ k is equivalent to x ≥ k or −x ≥ k. In the next
figure, the values of x that satisfy the inequalities are the values falling before −k
and −k itself, or k and those values to the right of k. The set (−k, k) = {x ∈
R | − k < x < k} is called an open interval, since it does not contain either k or
−k, that is, an open interval is an interval that does not contain its endpoints.
With this definition, we see that the set of values x that satisfy |x| ≥ k are the
values x ∈
/ (−k, k).
]
−k
|
O
[
k
Example 1.2.1. Find in the Cartesian plane1 the area enclosed by the graph of the
relation |x| + |y| = 1.
For |x| + |y| = 1 we have to consider four cases:
(a) x ≥ 0 and y ≥ 0; this implies x + y = 1, that is, y = 1 − x.
(b) x ≥ 0 and y < 0; this implies x − y = 1, that is, y = x − 1.
(c) x < 0 and y ≥ 0; this implies −x + y = 1, that is, y = x + 1.
(d) x < 0 and y < 0; this implies −x − y = 1, that is, y = −x − 1.
1 The
Cartesian plane is defined as R2 = R × R = {(x, y) | x ∈ R, y ∈ R}.
11
1.2 Absolute value
We can now draw the graph of the four straight lines.
(0, 1)
(−1, 0)
(1, 0)
(0, −1)
The area enclosed by the four straight lines is formed by four isosceles right triangles, each of which has two sides of length equal to 1. Since the area of each of
1
1
these triangles is 1×1
2 = 2 , the area of the square is 4 2 = 2.
Example 1.2.2. Solve the equation |2x − 4| = |x + 5|.
We have that
|2x − 4| =
2x − 4,
−2x + 4,
if
if
x≥2
x < 2.
We also have that
|x + 5| =
x + 5,
−x − 5,
if
if
x ≥ −5
x < −5.
If x ≥ 2, then 2x−4 = x+5, that is, x = 9. If x < −5, then −2x+4 = −x−5, hence
x = 9, but this is impossible since x < −5. The last case that we have to consider
is −5 ≤ x < 2. Then, the equation that we have to solve is −2x + 4 = x + 5.
Solving for x we get x = − 31 . Therefore, the numbers which satisfy the equation
are x = 9 and x = − 13 .
Sometimes it is easier to solve these equations without using the explicit form of
the absolute value, just by observing that |a| = |b| if and only if a = ±b and
making use of the absolute value properties.
Observation 1.2.3. If x is any real number
then the relation between the square root
√
and the absolute value is given by x2 = |x|. The identity follows from |x|2 = x2
and |x| ≥ 0.
Properties 1.2.4. If x and y are real numbers the following relations hold:
(a) |xy| = |x||y|. This implies also that xy = |x|
|y| , if y = 0.
(b) |x + y| ≤ |x| + |y|, where the equality holds if and only if xy ≥ 0.
Proof. (a) The proof follows directly from |xy|2 = (xy)2 = x2 y 2 = |x|2 |y|2 , and
taking the square root on both sides gives the result.
12
Chapter 1. Preliminaries
(b) Since both sides of the inequality are positive numbers, it is enough to
2
2
verify that |x + y| ≤ (|x| + |y|) .
2
2
|x + y| = (x + y)2 = x2 + 2xy + y 2 = |x| + 2xy + |y|
2
2
2
2
2
2
≤ |x| + 2 |xy| + |y| = |x| + 2 |x| |y| + |y| = (|x| + |y|) .
In the previous chain of relations there is only one inequality, it follows trivially
from the fact that xy ≤ |xy|. Moreover, the equality holds if and only if xy = |xy|,
which is true only when xy ≥ 0.
Inequality (b) in Properties 1.2.4 can be extended in a general form as
| ± x1 ± x2 ± · · · ± xn | ≤ |x1 | + |x2 | + · · · + |xn |,
for real numbers x1 , x2 , . . . , xn . The equality holds when all the ±xi ’s have the
same sign.
This last inequality can be proved in a similar way, and also using induction2 .
Exercise 1.14. If a and b are any real numbers, prove that ||a| − |b|| ≤ |a − b|.
Exercise 1.15. Find, in each case, the numbers x that satisfy the following:
(i) |x − 1| − |x + 1| = 0.
(ii) |x − 1||x + 1| = 1.
(iii) |x − 1| + |x + 1| = 2.
Exercise 1.16. Find all the triplets (x, y, z) of real numbers satisfying
|x + y| ≥ 1,
2xy − z 2 ≥ 1,
z − |x + y| ≥ −1.
Exercise 1.17 (OMM, 2004). Find the largest number of positive integers that can
be found in such a way that any two of them, a and b (with a = b), satisfy the next
inequality
ab
.
|a − b| ≥
100
1.3 Integer part and fractional part of a number
Given any number x ∈ R, sometimes it is useful to consider the integer number
max{k ∈ Z | k ≤ x}, that is, the greatest integer less than or equal to x. This
number is denoted by ⌊x⌋ and it is known as the integer part of x.
From this definition, the following properties hold.
2 See
Section 3.1.
1.3 Integer part and fractional part of a number
13
Properties 1.3.1. Let x, y ∈ R and n ∈ N, then it follows that:
(a) x − 1 < ⌊x⌋ ≤ x < ⌊x⌋ + 1.
(b) x is an integer if and only if ⌊x⌋ = x.
(c) ⌊x + n⌋ = ⌊x⌋ + n.
= nx .
(d) ⌊x⌋
n
(e) ⌊x⌋ + ⌊y⌋ ≤ ⌊x + y⌋ ≤ ⌊x⌋ + ⌊y⌋ + 1.
Proof. The proof of the first three properties follows immediately from the definition.
(d) If we divide ⌊x⌋ by n, we have that ⌊x⌋ = an + b, for an integer number
a and for an integer number b such that
0 ≤ b < n.
= an+b
= a + nb = a. On the other
On the one hand, we have that ⌊x⌋
n
n
hand,
x =
x since
an+b+c
⌊x⌋ +c,b+cwith
0 ≤ c < 1, and b + c < n − 1 + 1 = n, we get
=
=
a
+
= a. Then the equality holds.
n
n
n
(e) Since x = ⌊x⌋ + a and y = ⌊y⌋ + b with 0 ≤ a, b < 1, then ⌊x + y⌋ =
⌊x⌋ + ⌊y⌋ + ⌊a + b⌋ by property (c). The inequalities follow if we observe that if
0 ≤ a, b < 1, then 0 ≤ ⌊a + b⌋ ≤ 1.
Example 1.3.2. For any real number x, it follows that
1
⌊x⌋ + x +
− ⌊2x⌋ = 0.
2
If we let n = ⌊x⌋, then x can be expressed as x = n + a with 0 ≤ a < 1,
hence
1
1
− ⌊2x⌋ = n + n + a +
− ⌊2(n + a)⌋
⌊x⌋ + x +
2
2
1
− 2n − ⌊2a⌋
=n+n+ a+
2
1
= a+
− ⌊2a⌋ ,
2
1
where
the second equality follows1 by property (c). Now, if 0 ≤1 a < 2 , then
1
a + 2 = ⌊2a⌋ = 0, meanwhile, if 2 ≤ a < 1, it follows that a + 2 = ⌊2a⌋ = 1.
Example 1.3.3. If n and m are positive integers without common factors, then
n 2n 3n
(m − 1)n
(m − 1)(n − 1)
.
+
+
+ ··· +
=
m
m
m
m
2
Consider, in the Cartesian plane, the straight line passing through the origin
and the point (m, n). Since m and n are relative primes, then on the segment
14
Chapter 1. Preliminaries
where points (0, 0) and (m, n) lie, there is no other point with integer coordinates.
B = (m, n)
C = (0, n)
n−1
..
.
1
A = (m, 0)
O
1
2
···
m−1
n
n
x and passes over the points (j, m
The equation of the straight line is y = m
j),
n
n
with j = 1, . . . , (m − 1), and such that m j is not an integer. The number m j is
equal to the number of points with integer coordinates lying on the straight line
n
x and y = 1 included. It follows that
x = j and between the straight lines y = m
the sum is equal to the number of points with integer coordinates that lie in the
interior of the triangle OAB, and by symmetry it is equal to half of the points
with integer coordinates inside the rectangle OABC. The number of points with
integer coordinates in the rectangle is (n − 1)(m − 1), therefore
n 2n 3n
(m − 1)n
(m − 1)(n − 1)
.
+
+
+ ··· +
=
m
m
m
m
2
Observation 1.3.4. Since the right-hand side of the last inequality is symmetric in
m and n, then
n 2n
2m
(m − 1)n
(n − 1)m
m
+
+
+ ··· +
=
+ ···+
.
m
m
m
n
n
n
For a number x ∈ R we can also consider the number {x} = x − ⌊x⌋, which
we call the fractional part of x, and for which the following properties are fulfilled.
Properties 1.3.5. Let x, y ∈ R and n ∈ Z, then it follows that:
(a)
(b)
(c)
(d)
0 ≤ {x} < 1.
x = ⌊x⌋ + {x}.
{x + y} ≤ {x} + {y} ≤ {x + y} + 1.
{x + n} = {x}.
Exercise 1.18. For any real numbers a, b > 0, prove that
⌊2a⌋ + ⌊2b⌋ ≥ ⌊a⌋ + ⌊b⌋ + ⌊a + b⌋.
Exercise 1.19. Find the values of x that satisfy the following equation:
(i) ⌊x⌊x⌋⌋ = 1.
(ii) ||x| − ⌊x⌋| = ⌊|x| − ⌊x⌋⌋.
15
1.4 Notable products
Exercise 1.20. Find the solutions of the system of equations
x + ⌊y⌋ + {z} = 1.1,
⌊x⌋ + {y} + z = 2.2,
{x} + y + ⌊z⌋ = 3.3.
Exercise 1.21 (Canada, 1987). For any natural number n, prove that
√
√
√
√
√
⌊ n + n + 1⌋ = ⌊ 4n + 1⌋ = ⌊ 4n + 2⌋ = ⌊ 4n + 3⌋.
1.4 Notable products
The area of a square is the square of the length of its side. If the length of the
sides is a + b then the area is (a + b)2 . However, the area of the square can be
divided in four rectangles as shown in the figure.
a
a2
ab
b
ab
b2
a
b
Hence, the sum of the areas of the four rectangles will be equal to the area
of the square, that is,
(a + b)2 = a2 + ab + ab + b2 = a2 + 2ab + b2 .
(1.2)
a
(a − b)2
(a − b)b
Now we give a geometric representation of the square of the difference of two
numbers a, b, where b ≤ a. The problem now is to find the area of the square of
side a − b.
(a − b)b
b2
b
In the figure we observe that the area of the square of side a is equal to the sum of
the areas of the square of sides (a−b) and b, respectively, plus the area of two equal
16
Chapter 1. Preliminaries
rectangles with sides of length b and (a − b). That is, a2 = (a − b)2 + b2 + 2b(a − b),
hence
(1.3)
(a − b)2 = a2 − 2ab + b2 .
a
(a − b)2
(a − b)b
Now, in order to find the area of the shaded region of the following figure,
(a − b)b
b2
a−b
b
a−b
b
observe that the sum of the areas of the rectangles covering the regions is a(a −
b) + b(a − b), and factorizing this sum we get
a(a − b) + b(a − b) = (a + b)(a − b),
(1.4)
which is equivalent to the area of the large square minus the area of the small
square, that is,
(1.5)
(a + b)(a − b) = a2 − b2 .
Another notable product, but now dealing with three variables, is given by
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc.
(1.6)
The geometric representation of this product is given by the equality between the
area of the square with side length a + b + c and the sum of the areas of the nine
rectangles in which the square is partitioned, that is,
(a+ b + c)2 = a2 + b2 + c2 + ab + ac+ ba+ bc+ ca+ cb = a2 + b2 + c2 + 2ab + 2ac+ 2bc.
a2
ba
ca
a
ab
b2
cb
b
ac
bc
c2
c
a
c
b
Next, we provide a series of identities, some of them very well known and some
others less known, but all of them very helpful for solving many problems.
17
1.5 Matrices and determinants
Exercise 1.22. For every x and y ∈ R, verify the following second-degree identities:
(i) x2 + y 2 = (x + y)2 − 2xy = (x − y)2 + 2xy.
(ii) (x + y)2 + (x − y)2 = 2(x2 + y 2 ).
(iii) (x + y)2 − (x − y)2 = 4xy.
x2 + y 2 + (x + y)2
.
(iv) x2 + y 2 + xy =
2
x2 + y 2 + (x − y)2
.
(v) x2 + y 2 − xy =
2
(vi) Prove that x2 + y 2 + xy ≥ 0 and x2 + y 2 − xy ≥ 0.
Exercise 1.23. For any real numbers x, y, z, verify that
(x + y)2 + (y + z)2 + (z + x)2
.
2
(x − y)2 + (y − z)2 + (z − x)2
.
(ii) x2 + y 2 + z 2 − xy − yz − zx =
2
(iii) Prove that x2 + y 2 + z 2 + xy + yz + zx ≥ 0 and x2 + y 2 + x2 − xy − yz − zx ≥ 0.
(i) x2 + y 2 + z 2 + xy + yz + zx =
Exercise 1.24. For all real numbers x, y, z, verify the following identities:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(xy + yz + zx)(x + y + z) = (x2 y + y 2 z + z 2 x) + (xy 2 + yz 2 + zx2 ) + 3xyz.
(x + y)(y + z)(z + x) = (x2 y + y 2 z + z 2 x) + (xy 2 + yz 2 + zx2 ) + 2xyz.
(xy + yz + zx)(x + y + z) = (x + y)(y + z)(z + x) + xyz.
(x − y)(y − z)(z − x) = (xy 2 + yz 2 + zx2 ) − (x2 y + y 2 z + z 2 x).
(x + y)(y + z)(z + x) − 8xyz = 2z(x − y)2 + (x + y)(x − z)(y − z).
xy 2 + yz 2 + zx2 − 3xyz = z(x − y)2 + y(x − z)(y − z).
Exercise 1.25. For any real numbers x, y, z, verify that:
(i) x2 + y 2 + z 2 + 3(xy + yz + zx) = (x + y)(y + z) + (y + z)(z + x) + (z + x)(x + y).
(ii) xy + yz + zx − x2 + y 2 + z 2 = (x − y)(y − z) + (y − z)(z − x) + (z − x)(x − y).
Exercise 1.26. For any real numbers x, y, z, verify that:
(x−y)2 +(y −z)2 +(z −x)2 = 2 [(x − y)(x − z) + (y − z)(y − x) + (z − x)(z − y)] .
1.5 Matrices and determinants
A 2 × 2 matrix is an array
a11
a21
a12
a22
18
Chapter 1. Preliminaries
where a11 , a12 , a21 and a22 are real or complex numbers3 . The determinant of the
above matrix, which is denoted by
a11 a12
a21 a22
is the real number defined by a11 a22 − a12 a21 .
A 3 × 3 matrix is an array
⎛
a11
⎜
⎝ a21
a31
a12
a22
a32
⎞
a13
⎟
a23 ⎠
a33
where, again, every aij is a number. The subindex indicates the position of the
number in the array. Then, aij is the number located in the ith row and in the
jth column. We define the determinant of a 3 × 3 matrix by the rule
a11 a12 a13
a
a
a
21 a22
21 a23
22 a23
.
+ a13
a21 a22 a23 = a11
− a12
a31 a32
a31 a33
a32 a33
a31 a32 a33
That is, we move along the first row, multiplying a1j by the determinant of a 2 × 2
matrix obtained by eliminating the first row and the jth column, and then adding
all this together, and keeping in mind to place the minus sign before a12 . The
result of the determinant is not modified if instead of choosing the first row as the
first step we chose the second or third row. In case we choose the second row, we
begin with a negative sign and if we choose the third row the first sign is positive,
that is,
a11 a12 a13
a
a
a
12 a13
11 a13
11 a12
+ a22
− a23
.
a21 a22 a23 = −a21
a32 a33
a31 a33
a31 a32
a31 a32 a33
The signs alternate according to the following diagram:
+ − +
− + − .
+ − +
There are many properties of the determinants which follow immediately from the
definitions. These properties become rules of a sort and the following are the most
frequently used.
3 Complex
numbers will be treated in Chapter 5.
19
1.5 Matrices and determinants
Properties 1.5.1.
(a) If we interchange two consecutive rows or two
of the determinant does change, for example,
a21
a11 a12 a13
a21 a22 a23 = − a11
a31
a31 a32 a33
consecutive columns, the sign
a22
a12
a32
a23
a13
a33
.
(b) We can factorize the common factor of any row or column of a matrix and
the corresponding determinants are related in the following way, for example,
a11 a12 a13
αa11 αa12 αa13
a22
a23 = α a21 a22 a23 .
a21
a31 a32 a33
a31
a32
a33
(c) If to a row (or column) we add another row (or column), the value of the
determinant does not change, as in the following example:
a11 a12 a13 a11 + a21 a12 + a22 a13 + a23
a21
a22
a23
a21 a22 a23 =
a31 a32 a33
a31
a32
a33
⎞
⎛
a11 + a12 a12 a13
⎜
⎟
= ⎝or a21 + a22 a22 a23 , ⎠ .
a31 + a32 a32 a33
(d) If a matrix has two equal rows (or two equal columns) the determinant is
zero.
Example 1.5.2. Using determinants we can establish the identity
a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca).
(1.7)
Note that
a
D= c
b
b c
a b
c a
a
= a
c
b
a
c
−b
b
b
a
c
+ c
b
a
c
= a3 − abc − abc + b3 + c3 − abc = a3 + b3 + c3 − 3abc.
On the other hand, adding to the first
a+b+c b c
D = a+b+c a b
a+b+c c a
column the other two,
1 b
= (a + b + c) 1 a
1 c
= (a + b + c)(a2 + b2 + c2 − ab − bc − ca).
we have
c
b
a
(1.8)
20
Chapter 1. Preliminaries
By properties (b) and (c), the determinants are equal.
Observe that the expression a2 + b2 + c2 − ab − bc − ca can be written as4
"
1!
(a − b)2 + (b − c)2 + (c − a)2 .
2
From this we obtain another version of identity (1.7), that is,
a3 + b3 + c3 − 3abc =
!
"
1
(a + b + c) (a − b)2 + (b − c)2 + (c − a)2 .
2
(1.9)
Observe that if the above identity satisfies the condition a + b + c = 0 or the
condition a = b = c, then the following identity holds:
a3 + b3 + c3 = 3abc.
(1.10)
Reciprocally, if identity (1.10) is satisfied, then it follows that either a + b + c = 0
or a = b = c.
√
√
3
3
Exercise 1.27. Prove that 2 + 5 + 2 − 5 is a rational number.
Exercise 1.28. Find the factors of the expression (x − y)3 + (y − z)3 + (z − x)3 .
Exercise 1.29. Find the factors of the expression (x + 2y − 3z)3 + (y + 2z − 3x)3 +
(z + 2x − 3y)3 .
Exercise 1.30. Prove that if x, y, z are different real numbers, then
√
√
√
3
x − y + 3 y − z + 3 z − x = 0.
√
1
Exercise 1.31. Let r be a real number such that 3 r − √
3 r = 1. Find the values of
r − r1 and r3 − r13 .
Exercise 1.32. Let a, b, c be digits different from zero. Prove that if the integers
(written in decimal notation) abc, bca and cab are divisible by n then also a3 +
b3 + c3 − 3abc is divisible by n.
Exercise 1.33. How many ordered pairs of integers (m, n) are there such that the
following conditions are satisfied: mn ≥ 0 and m3 + 99mn + n3 = 333 ?
Exercise 1.34. Find the locus of points (x, y) such that x3 + y 3 + 3xy = 1.
Exercise 1.35. Find the real solutions x, y, z of the equation,
x3 + y 3 + z 3 = (x + y + z)3 .
4 See
Exercise 1.23.
21
1.6 Inequalities
1.6 Inequalities
We begin this section with one of the most important inequalities. For any real
number x, we have that
x2 ≥ 0.
(1.11)
This follows from the equality x2 = |x|2 ≥ 0.
From this result, we can deduce that the sum of n squares is non-negative,
n
#
i=1
x2i ≥ 0
(1.12)
and it will be zero, if and only if all the xi ’s are zero.
If we make the substitution x = a−b, where a and b are non-negative real numbers,
in equation (1.11), we get
(a − b)2 ≥ 0.
Simplifying, the previous inequality leads to
a2 + b2 ≥ 2ab.
(1.13)
Since
a2 + b2 ≥ 2ab if and only if 2a2 + 2b2 ≥ a2 + 2ab + b2 = (a + b)2 ,
we also have the inequality
a2 + b 2
(a + b)
≥
.
2
2
(1.14)
In case both a and b are positive numbers, the inequality (1.13) guarantees that
b
a
+ ≥ 2.
b
a
(1.15)
If we take b = 1 in the previous inequality, then we have that a + a1 ≥ 2, that is,
the sum of a > 0 and its reciprocal is greater than or equal to 2, and it will be 2
if and only if a = 1.
√ √
Replacing a, b by a, b in (1.13), we obtain that
√
a+b √
a + b ≥ 2 ab if and only if
≥ ab.
2
√
Multiplying the last inequality by ab and reordering, we obtain
√
ab ≥
2ab
.
a+b
(1.16)
(1.17)
22
Chapter 1. Preliminaries
Summarizing, the inequalities (1.14), (1.16) and (1.17), which we have just proved,
imply that
√
a+b
a2 + b 2
2ab
≤ ab ≤
≤
.
(1.18)
a+b
2
2
The first expression is known as the harmonic mean (HM ), the second is the
geometric mean (GM ), the third is the arithmetic mean (AM ) and the last one
is known as the quadratic mean (QM ).
Now, we will present a geometric and a visual proof of the previous inequalities. Consider a semicircle with center in O, radius a+b
2 and right triangles ABC,
DBA and DAC, as shown in the following figure:
A
y
h
e
B
D
E
z
C
O
a
b
These triangles are similar triangles and therefore the following equality holds:
DC
AD
=
DB
DA
h
b
=
a
h
h2 = ab,
√
ab, which according to the
hence, the height h of the triangles is given by h = √
diagram is clearly smaller than the radius. Therefore, ab ≤ a+b
2 .
To prove the first inequality in (1.18), observe that the triangles DAE and OAD
are similar, hence
AO
AD
=
,
AE
AD
h2 = y(y + z),
2ab
= y,
a+b
that is,√y represents the harmonic mean. Clearly we have that y ≤ h, therefore
2ab
ab.
a+b ≤
23
1.6 Inequalities
To prove, geometrically, the last inequality in (1.18), consider the next figure.
L
A
D
We have that OD =
a+b
2 −a
=
b−a
2
DL2 = OD2 + OL2 =
that is, DL =
a2 +b2
2
O
and using the Pythagorean theorem, we obtain
$ b − a %2
2
+
$ a + b %2
2
which is clearly greater than
=
a2 + b 2
,
2
a+b
2 .
Using Example 1.5.2, we can provide a proof of the arithmetic mean and the
geometric mean inequality for three non-negative real numbers. In fact, using the
next identity
a3 + b3 + c3 − 3abc =
!
"
1
(a + b + c) (a − b)2 + (b − c)2 + (c − a)2 ,
2
it is clear that if a, b and c are non-negative numbers, then a3 + b3 + c3 − 3abc ≥ 0,
that is, a3 +b3 +c3 ≥ 3abc. Moreover, if a+b+c = 0 or (a−b)2 +(b−c)2 +(c−a)2 = 0
the equality holds, and this happens
a = b =√c. Now, if x, y and z are
√ only when
√
non-negative numbers, define a = 3 x, b = 3 y and c = 3 z, then
x+y+z
√
≥ 3 xyz
3
(1.19)
with equality if and only if x = y = z.
Example 1.6.1. For every real number x, it follows that
2
√x +2
x2 +1
≥ 2.
In fact,
x2 + 2
x2 + 1
1
√
= √
+√
=
2
2
2
x +1
x +1
x +1
1
x2 + 1 + √
≥ 2.
2
x +1
The inequality now follows if we apply inequality (1.15).
Example 1.6.2. If a, b, c are non-negative numbers, then
(a + b)(b + c)(a + c) ≥ 8abc.
24
Chapter 1. Preliminaries
As we have seen,
(a+b)
2
≥
√
ab,
a+b
2
(b+c)
2
b+c
2
≥
√
bc and
a+c
2
≥
(a+c)
2
√
≥
√
ac, then
a2 b2 c2 = abc.
Example 1.6.3. If x1 > x2 > x3 and y1 > y2 > y3 , which sum is greater?
S = x1 y1 + x2 y2 + x3 y3 ,
S ′ = x1 y2 + x2 y1 + x3 y3 .
Consider the difference,
S ′ − S = x1 y2 − x1 y1 + x2 y1 − x2 y2
= x1 (y2 − y1 ) + x2 (y1 − y2 )
= −x1 (y1 − y2 ) + x2 (y1 − y2 )
= (x2 − x1 )(y1 − y2 ) < 0,
then, S ′ < S.
In general, for any permutation {y1′ , y2′ , y3′ } of {y1 , y2 , y3 }, we have that
S ≥ x1 y1′ + x2 y2′ + x3 y3′ ,
(1.20)
which is known as the rearrangement inequality5 .
Exercise 1.36. Let a, b be real numbers with 0 ≤ a ≤ b ≤ 1, prove that:
b−a
≤ 1.
(i) 0 ≤
1 − ab
b
a
(ii) 0 ≤
+
≤ 1.
1+b 1+a
Exercise 1.37 (Nesbitt Inequality). If a, b, c ≥ 0, prove that
a
b
c
3
+
+
≥ .
b+c a+c a+b
2
Exercise 1.38. If a, b, c are the lengths of the sides of a triangle, prove that
3
a3 + b3 + c3 + 3abc
≥ max {a, b, c} .
2
Exercise 1.39. Let p and q be positive real numbers with
(i)
5 To
1
1
1
1
≤
+
≤ .
3
p(p + 1) q(q + 1)
2
(ii)
see a general version, consult Example 7.3.6.
1
p
+
1
q
= 1. Prove that:
1
1
+
≥ 1.
p(p − 1) q(q − 1)
25
1.7 Factorization
Exercise 1.40. Find the smallest positive number k such that, for any 0 < a, b < 1,
with ab = k, it follows that
b
a
b
a
+ +
+
≥ 4.
b
a 1−b 1−a
Exercise 1.41. Let a, b, c be non-negative real numbers, prove that
(a + b)(b + c)(c + a) ≥
8
(a + b + c)(ab + bc + ca).
9
Exercise 1.42. Let a, b, c be positive real numbers that satisfy the equality (a +
b)(b + c)(c + a) = 1. Prove that
ab + bc + ca ≤
3
.
4
Exercise 1.43. Let a, b, c be positive real numbers that satisfy abc = 1. Prove that
(a + b)(b + c)(c + a) ≥ 4(a + b + c − 1).
Exercise 1.44 (APMO, 2011). Let a, b, c be positive integers. Prove that it is
impossible for all three numbers a2 + b + c, b2 + c + a and c2 + a + b to be perfect
squares.
1.7 Factorization
One of the most important forms of algebraic manipulation is known as factorization. In this section we study some examples and problems whose solutions
depend on factorization formulas. Many of the problems that involve algebraic
expressions can be easily solved using algebraic transformations in which the fundamental strategy is to find appropriate factors.
We start with some elemental formulas of factorization, where x, y are real numbers:
(a) x2 − y 2 = (x + y)(x − y).
(b) x2 + 2xy + y 2 = (x + y)2 and x2 − 2xy + y 2 = (x − y)2 .
(c) x2 + y 2 + z 2 + 2xy + 2yz + 2zx = (x + y + z)2 .
These algebraic identities are cataloged as second-order identities. In fact, we
studied these four identities in the section of notable products. However, now we
would like, given an algebraic expression, to reduce it to a product of simpler
algebraic expressions.
26
Chapter 1. Preliminaries
Example 1.7.1. For real numbers a, b, x, y, with x and y different from zero, it
follows that
b2
(a + b)2
(ay − bx)2
a2
+
−
=
.
x
y
x+y
xy(x + y)
To obtain the equality, we start with the sum on the left side of the identity
and follow the ensuing equalities
a2
b2
(a + b)2
a2 y(x + y) + b2 x(x + y) − xy(a + b)2
+
−
=
x
y
x+y
xy(x + y)
a2 y 2 + b2 x2 − 2xyab
=
xy(x + y)
(ay − bx)2
.
=
xy(x + y)
An application of the previous identity helps us to prove, in a straightforward
manner, the so-called helpful inequality6 of degree 2. This inequality assures that
for real numbers a, b and real positive numbers x, y, it follows that
a2
b2
(a + b)2
+
≥
.
x
y
x+y
The following identities are known as third-degree identities, with x, y, z ∈ R:
(a) x3 − y 3 = (x − y) x2 + xy + y 2 .
(b) x3 − y 3 = (x − y)3 + 3xy(x − y).
(c) (x + y)3 − (x3 + y 3 ) = 3xy(x + y).
(d) x3 − xy 2 + x2 y − y 3 = (x + y)(x2 − y 2 ).
(e) x3 + xy 2 − x2 y − y 3 = (x − y)(x2 + y 2 ).
To prove the validity of these identities, it is enough to expand one of the sides
of the equalities or use the Newton binomial theorem, which we will study in
Section 3.2.
Another quite important identity of degree 3, previously given as (1.7), is
x3 + y 3 + z 3 − 3xyz = (x + y + z)(x2 + y 2 + z 2 − xy − yz − zx),
for all x, y, z real numbers. A proof of this identity can be obtained directly
expanding the right-hand side of the identity. Different proofs of this equality will
be given throughout this book.
An equivalent form of the above identity is
'
&
1
2
x3 + y 3 + z 3 − 3xyz = (x + y + z) (x − y) + (y − z)2 + (z − x)2 .
2
6 See
[6] or [7].
27
1.7 Factorization
The identities x2 − y 2 = (x + y)(x − y) and x3 − y 3 = (x − y) x2 + xy + y 2 are
particular cases of the nth degree identity,
xn − y n = (x − y)(xn−1 + xn−2 y + · · · + xy n−2 + y n−1 ),
(1.21)
for any x, y real numbers. If n is odd, we can replace y by −y in the last formula
to obtain the factorization for the sum of two nth odd power numbers,
xn + y n = (x + y)(xn−1 − xn−2 y + · · · − xy n−2 + y n−1 ).
(1.22)
In general, when n is even, the sum of two nth powers cannot be factored, however
there are some exceptions when it is possible to complete squares. Let us see the
following example.
Example 1.7.2 (Sophie Germain identity). For any x, y real numbers, it follows
that
x4 + 4y 4 = (x2 + 2y 2 + 2xy)(x2 + 2y 2 − 2xy).
Completing the square, we have
x4 + 4y 4 = x4 + 4x2 y 2 + 4y 4 − 4x2 y 2 = (x2 + 2y 2 )2 − (2xy)2
= (x2 + 2y 2 + 2xy)(x2 + 2y 2 − 2xy).
Another example, with even powers is the following.
Example 1.7.3. For any x, y real numbers, it follows that
x2n − y 2n = (x + y)(x2n−1 − x2n−2 y + x2n−3 y 2 − · · · + xy 2n−2 − y 2n−1 ).
To prove this we only have to divide x2n − y 2n by x + y or do the product on the
right-hand side and simplify.
Example 1.7.4. The number n4 −22n2 +9 is a composite number for any integer n.
The idea is to try to factorize the expression. We do it by completing squares,
and the common way to do it is as follows:
n4 − 22n2 + 9 = (n4 − 22n2 + 121) − 112 = (n2 − 11)2 − 112.
While doing this a problem arises: 112 is not a perfect square, therefore the factorization is not immediate. However, we can try the following strategy to complete
squares, which is less usual,
n4 − 22n2 + 9 = (n4 − 6n2 + 9) − 16n2 = (n2 − 3)2 − 16n2
= (n2 − 3)2 − (4n)2 = (n2 − 3 + 4n)(n2 − 3 − 4n)
= ((n + 2)2 − 7)((n − 2)2 − 7),
and observe that none of the factors is equal to ±1.
The following is another example of how to solve problems using basic factorizations.
Example 1.7.5. Find all the pairs (m, n) of positive integers such that |3m −2n | = 1.
28
Chapter 1. Preliminaries
When m = 1 or m = 2, it is easy to find the solutions (m, n) = (1, 1), (1, 2),
(2, 3). Now, we will prove that these are the only solutions of the equation. Suppose
that (m, n) is a solution of |3m − 2n | = 1, with m > 2 and therefore n > 3. We
analyze both cases: 3m − 2n = 1 and 3m − 2n = −1.
Suppose that 3m − 2n = −1 with n > 3, then 8 divides 3m + 1. However, if we
divide 3m by 8 we obtain as a remainder 1 or 3, depending on whether n is odd
or even, therefore in this case we do not have a solution.
Suppose now that 3m −2n = 1 with m ≥ 3, therefore n ≥ 5, since 2n +1 = 3m ≥ 27.
Then 3m − 1 is divisible by 8, hence m is even. Write m = 2k, with k > 1. Then
2n = 32k − 1 = (3k + 1)(3k − 1), therefore 3k + 1 = 2r , for some r > 3. But the
previous case tells us we know that this is impossible, therefore in this case there
are no solutions.
The following formulas are also helpful to factorize. For real numbers x, y, z we
have the following equalities:
(x + y)(y + z)(z + x) + xyz = (x + y + z)(xy + yz + zx),
3
3
3
3
(x + y + z) = x + y + z + 3(x + y)(y + z)(z + x).
(1.23)
(1.24)
To convince ourselves of the validity of these equalities, just expand both sides in
each equality. From these equalities the following observation arises.
Observation 1.7.6.
(a) If x, y, z are real numbers, with xyz = 1, then
(x + y)(y + z)(z + x) + 1 = (x + y + z)(xy + yz + zx).
(1.25)
(b) If x, y, z are real numbers with xy + yz + zx = 1, then
(x + y)(y + z)(z + x) + xyz = x + y + z.
(1.26)
Exercise 1.45. For all real numbers x, y, z, the following identities hold:
(i) (x + y + z)3 − (y + z − x)3 − (z + x − y)3 − (x + y − z)3 = 24xyz.
(ii) (x − y)3 + (y − z)3 + (z − x)3 = 3(x − y)(y − z)(z − x).
(iii) (x − y)(y + z)(z + x) + (y − z)(z + x)(x + y) + (z − x)(x + y)(y + z)
= −(x − y)(y − z)(z − x).
Exercise 1.46. For all real numbers x, y, z, prove that
(i) If f (x, y, z) = x3 + y 3 + z 3 − 3xyz, then
f (x, y, z) =
1
1
f (x + y, y + z, z + x) = f (−x + y + z, x − y + z, x + y − z).
2
4
(ii) If f (x, y, z) = x3 +y 3 +z 3 −3xyz, then f (x, y, z) ≥ 0 if and only if x+y+z ≥ 0,
and f (x, y, z) ≤ 0 if and only if x + y + z ≤ 0.
29
1.7 Factorization
Exercise 1.47. Prove that for any real numbers x, y, the following identities are
satisfied:
(i) (x + y)5 − (x5 + y 5 ) = 5xy(x + y)(x2 + xy + y 2 ).
(ii) (x + y)7 − (x7 + y 7 ) = 7xy(x + y)(x2 + xy + y 2 )2 .
Exercise 1.48. Let x, y and z be real numbers such that x = y and
x2 (y + z) = y 2 (x + z) = 2.
Find the value of z 2 (x + y).
Exercise 1.49. Find the real solutions x, y, z and w of the system of equations
x + y + z = w,
1 1
1
1
+ + = .
x y z
w
Exercise 1.50. Let x, y and z be real numbers different from zero such that x +
1
y + z = 0 and x1 + y1 + z1 = x+y+z
. Prove that for any odd integer n it follows that
1
1
1
1
+ n+ n = n
.
n
x
y
z
x + yn + z n
Chapter 2
Progressions and Finite Sums
2.1 Arithmetic progressions
In antiquity, patterns of points played an important role in the use of numbers and
cosmological conceptions. The Pythagoreans used to represent some integers as a
set of points arranged in polygonal or polyhedral forms. These integers, presented
as spatial arrays, are known as figurate numbers. In this section we will study
some of these numbers.
Suppose that we want to add the natural numbers 1 + 2 + 3 + · · · + n, where n
is any natural number. Call tn the sum of these numbers, for example, t1 = 1,
t2 = 1 + 2 = 3, t3 = 1 + 2 + 3 = 6, t4 = 1 + 2 + 3 + 4 = 10. We will represent them
as patterns of points in the following form:
t1
t2 = 3
t3 = 6
t4 = 10
We can obtain the sum tn from the following figure, where the geometrical arrangement shows that 2tn = n(n + 1).
tn
···
···
tn
n-points
···
···
n + 1-points
These numbers are known as triangular numbers.
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5_2
31
32
Chapter 2. Progressions and Finite Sums
This sum is also known as the Gauss sum, named after Carl Friedrich Gauss who,
when he was only a child, calculated this sum using the following trick. Let tn be
the sum of the n numbers that we want to add. Since the sum is commutative,
we can arrange the numbers starting from the higher value and ending with the
lower. If now we sum, term by term, both arrangements, we get
and
tn
tn
2tn
=
1
+
2
+ ··· +
=
n
+ n− 1 + ··· +
= (n + 1) + (n + 1) + · · · +
therefore 2tn = n(n + 1), that is, tn =
n
1
(n + 1),
(2.1)
n(n+1)
.
2
Remember that even numbers can be represented as 2n, where n = 1, 2, . . . and
odd numbers can be written as 2n − 1, where n = 1, 2, . . . .
If we want to sum the first n odd numbers, that is, the ones that run from 1 to
2n − 1, and we call cn the sum of these numbers, then we have, following Gauss,
that
1
+
3
+ · · · + 2n − 1
cn =
1
and cn = 2n − 1 + 2n − 3 + · · · +
,
(2.2)
2cn =
2n
+
2n
+ ··· +
2n
therefore, 2cn = n · 2n = 2n2 , where we have multiplied by n because we have
exactly n odd numbers. Therefore, cn = n2 . We can represent these sums using
patterns of points in the following way:
c1
c2 = 4
c3 = 9
c4 = 16
Then the sum of the first n odd natural numbers corresponds to the representation
of the perfect square n2 , that is,
···
n-points
..
.
···
n-points
Every corridor represents the corresponding odd number we are adding, therefore,
the sum of the first n odd numbers is the number of points in the square, that is,
n · n = n2 . These numbers are known as the square numbers.
33
2.1 Arithmetic progressions
An arithmetic progression is a collection or sequence of numbers such that each
term of the sequence can be obtained from the preceding number adding a fixed
quantity. That is, a collection {a0 , a2 , . . . } is an arithmetic progression if, for each
n ≥ 0, an+1 = an + d, where d is a constant. This common constant is called the
difference of the progression and the collection or sequence will be represented
by {an }.
Proposition 2.1.1. If {an } is an arithmetic progression with difference d, it follows
that:
(a) The term an is equal to a0 + nd, for n = 0, 1, 2, . . . .
n
(n + 1) = 2a02+nd (n + 1), for n = 0, 1, 2, . . . .
(b) a0 + a1 + · · · + an = a0 +a
2
an−1 +an+1
(c) an =
, for n = 1, 2, 3, . . . , that is, each term is the arithmetic mean
2
of its two neighbours.
Proof. (a) an −a0 = (an −an−1 )+(an−1 −an−2 )+· · ·+(a1 −a0 ) = d+d+· · ·+d = nd.
(b) Similarly as we did with the Gauss sum, if S = a0 + a1 + · · · + an = an +
an−1 + · · · + a0 , then
2S = (a0 + an ) + (a1 + an−1 ) + · · · + (an + a0 ) = (a0 + an )(n + 1),
a result of the identity
aj + an−j = aj+1 + an−j−1 ,
since aj+1 − aj = an−j − an−j−1 = d.
(c) It follows from an − an−1 = an+1 − an .
Example 2.1.2. The following figures are formed with toothpicks and they are composed by equilateral triangles.
Figure 1
Figure 2
Figure 3
How many toothpicks are needed to construct the figure with n triangles?
To build the first figures we need 3, 5 and 7 toothpicks, respectively. The fourth
figure will have 4 triangles, that is, one more than the third figure; but we only
need 2 more toothpicks to make an additional triangle. In general, this happens
always as we go from figure j to figure j + 1, that is, a new triangle is constructed,
but only 2 additional toothpicks are needed. Therefore, the difference of toothpicks
going from one figure to the next is 2, that is, if aj and aj+1 are the number of
toothpicks necessary to construct the jth figure and the next one, respectively, we
have that aj+1 − aj = 2. Then,
an = an−1 + 2 = an−2 + 2 · 2 = · · · = a1 + 2 · (n − 1) = 3 + 2n − 2 = 2n + 1.
34
Chapter 2. Progressions and Finite Sums
Proposition 2.1.3. The sequence {an } is an arithmetic progression if and only if
there exist real numbers m and b such that an = mn + b, for every n ≥ 0.
Proof. If {an } is an arithmetic progression with difference d, part (a) of the previous
proposition, an = a0 + nd, leads to m = d and b = a0 been the numbers we were
searching for.
Reciprocally, if an = mn + b, then an+1 − an = m is constant and therefore {an }
is an arithmetic progression, with difference m.
An harmonic
progression is a sequence {an } which satisfies that the sequence
1
1
is
an
arithmetic
progression. That is, an+1
− a1n is constant, for every natural
an
number n.
(
)
For example, the sequence 1, 21 , 13 , 41 , . . . , n1 , . . . is an harmonic progression since
{1, 2, 3, 4, . . . } is an arithmetic progression.
Exercise 2.1. Calculate the sum of the first n even numbers. Can you exhibit a
geometric representation of them?
Exercise 2.2.
(i) If {an } and {bn } are arithmetic progressions, then {an + bn } and {an − bn }
are arithmetic progressions, this can be shortened saying that {an ± bn } are
arithmetic progressions.
(
)
(ii) If {an } is an arithmetic progression, then bn = a2n+1 − a2n is an arithmetic
progression.
Exercise 2.3. If {an } is an arithmetic progression with aj = 0, for every j =
0, 1, 2, . . . , then
1
1
1
n
+
+ ···+
=
.
a0 a1
a1 a2
an−1 an
a0 an
Exercise 2.4. Prove that a sequence {an } is an arithmetic progression if and only if
there exist real numbers A and B such that Sn = a0 + a1 + · · · + an−1 = An2 + Bn,
for every n ≥ 0.
Exercise 2.5. A sequence {an} is an arithmetic progression of order 2 if {an+1 − an}
is an arithmetic progression.
Prove that {an } is an arithmetic progression of order 2 if and only if there exists
a degree 2 polynomial P (x) such that P (n) = an , for every n ≥ 0.
Exercise 2.6. If {an } ⊂ R+ is an arithmetic progression, then
√
√
a1 + an
.
a1 an ≤ n a1 a2 · · · an ≤
2
35
2.1 Arithmetic progressions
Exercise 2.7. Prove that there are 5 prime numbers which are in arithmetic progression with difference 6. Is the progression unique?
Exercise 2.8. For any natural number n, let Sn be the sum of the integers m, with
2n < m < 2n+1 . Prove that Sn is a multiple of 3, for all n.
Exercise 2.9. If a < b < c are real numbers which are in harmonic progression,
then
4
1
1 1
1
+
+
= − .
b−c c−a a−b
c a
Exercise 2.10. If a, b, c and d are in harmonic progression, then a + d > b + c.
Exercise 2.11. If a, b and c are real numbers, prove that b + c, c + a and a + b are
in harmonic progression if and only if a2 , b2 and c2 are in arithmetic progression.
Exercise 2.12. An increasing arithmetic progression satisfies that the product of
any two terms of the progression is also an element of the progression. Prove that
each term of the progression is an integer number.
Exercise 2.13. In the following arrangement, all the odd numbers were placed in
such a way that in the jth row there are j consecutive odd numbers,
1
3 5
7 9 11
13 15 17 19
..
..
.
.
(i) Which is the first number (on the left ) in the 100th row?
(ii) Which is the sum of the numbers in the 100th row?
Exercise 2.14. Consider the following array, in which the numbers from 1 to 9 are
placed as indicated.
1
2
3
4
5
6
7
8
9
Observe that the sum of the integers in the two main diagonals is 15. If we construct
a similar array with the numbers from 1 to 10000, what is the value of the sum of
the numbers in the two main diagonals?
36
Chapter 2. Progressions and Finite Sums
Exercise 2.15. Fill the following board in such a way that the numbers in the rows
and columns are in arithmetic progression.
74
186
103
0
2.2 Geometric progressions
A geometric progression is a sequence of numbers related in such a way that
each number can be obtained by multiplying the previous one by a fixed constant,
different from zero, which we call the common ratio or the ratio of the progression.
That is, the sequence {an } is a geometric progression if the ratio an+1
an is constant.
Let this ratio be denoted by r. Note that a0 = 0 and r = 0 are not included.
Proposition 2.2.1. If {an } is a geometric progression with ratio r, then:
(a) The nth term is an = a0 rn , for n = 0, 1, 2, . . . .
1 − rn+1
(b) The sum a0 + a1 + · · · + an = a0
, for n = 0, 1, 2, . . . .
1 −√r
(c) If an are positive numbers then an = an−1 an+1 , for n = 1, 2, 3, . . . .
Proof. (a) Since
an−1
an
an−1 · an−2
(b) Using part (a) we get
·· · ·· aa01 = rn we have, after simplifying, that an = a0 rn .
S = a0 + a1 + · · · + an = a0 + a0 r + · · · + a0 r n
= a0 (1 + r + · · · + rn ) =
a0 (1 − rn+1 )
.
1−r
The last equality follows from identity (1.21).
√
an
(c) Since an+1
an−1 an+1 .
an = an−1 , it follows that an =
Example 2.2.2. If {an } and {bn } are geometric progressions, then {an · bn } is a
geometric progression.
If {an } and {bn } are geometric progressions, then an = a0 rn and bn = b0 sn , for
some real numbers r and s. Then, an bn = (a0 b0 )(rs)n is a geometric progression
with ratio rs.
Exercise 2.16.
If {an } and {bn } are geometric progressions, with bn = 0 for all n,
an
prove that bn is a geometric progression.
37
2.2 Geometric progressions
Exercise 2.17. Find the geometric progressions {an } satisfying that an+2 = an+1 +
an , for all n ≥ 0.
Exercise 2.18. If {an } is a geometric progression with ratio r, and if the product
of a0 , a1 , . . . , an−1 is Pn , prove that:
(i) Pn = an0 rn(n−1)/2 .
(ii) (Pn )2 = (a0 an−1 )n .
Exercise 2.19. If {an } ⊂ R+ is a geometric progression with ratio r, prove that
{bn = log an } is an arithmetic progression with difference log r.
Exercise 2.20. If a, b and c are in geometric progression, then
a3 b3 + b3 c3 + c3 a3 = abc(a3 + b3 + c3 ).
Exercise 2.21. Prove that it is possible to eliminate terms of an arithmetic progression of positive integers in such a way that the remaining terms form a geometric
progression.
Exercise 2.22. The lengths of the sides of a right triangle, given by a < b < c, are
in geometric progression. Find the ratio of the progression.
Exercise 2.23 (Slovenia, 2009). Let {an } be a non-constant arithmetic progression
with initial term a1 = 1. The terms a2 , a5 , a11 form a geometric progression. Find
the sum of the first 2009 terms.
Exercise 2.24. In the next figure the polygonal line has been constructed between
the sides of the angle ABC as follows: the first segment AC of length b is perpendicular to BC, the second segment, of length a, starts where the previous segment
ended and it is perpendicular to AB; proceeding in the same manner, the following
segments start where the previous one ended, and keeping the perpendicularity to
BC and to AB alternately.
A
b
a
B
C
(i) What is the length of the nth segment?
(ii) What is the length of the polygonal line of n sides?
(iii) What is the length of the polygonal line if it has an infinite number of sides?
38
Chapter 2. Progressions and Finite Sums
2.3 Other sums
Now, we proceed to study more complicated sums. For example, let us see how to
sum the squares of the first natural numbers,
1 2 + 2 2 + 3 2 + · · · + n2 .
Consider the following array of numbers:
1
2
3
4
5
... k
... n
1
2
3
4
5
... k
... n
1
..
.
2
..
.
3
..
.
4
..
.
5
..
.
1
2
3
4
5
... k
..
..
.
.
... k
... n
..
..
.
.
... n
n-rows
We will calculate the sum of the numbers on the array in two different ways.
First, add the numbers in each of the rows of the array which, according to the
Gauss sum formula, is a triangular number, that is,
Rn = 1 + 2 + 3 + · · · + n =
n(n + 1)
.
2
But since we have n rows, then the sum of all the numbers in the array is
ST = n
n(n + 1)
.
2
On the other hand, if we add each of the corridors marked, we have that the sum
of the first corridor is only 1. The sum of the second corridor is 1 + 2 · 2 = 1 + 22 .
The sum of the third corridor is 1 + 2 + 3 · 3 = 1 + 2 + 32 . Continuing in the same
fashion, the sum of the kth corridor is
Ck = 1 + 2 + · · · + (k − 1) + k · k =
3
k
k(k − 1)
+ k2 = k2 − .
2
2
2
Now, if we sum all the corridors, we get the sum of all the numbers in the array
3 2
(1 + 22 + · · · + n2 ) −
2
3
= (12 + 22 + · · · + n2 ) −
2
ST = C1 + C2 + · · · + Cn =
1
(1 + 2 + · · · + n)
2
1 n(n + 1)
·
.
2
2
Equating both sums ST , we get
n
3
1 n(n + 1)
n(n + 1)
= (12 + 22 + · · · + n2 ) − ·
.
2
2
2
2
39
2.3 Other sums
Then
n(n + 1)
1 n(n + 1)
3 2
(1 + 22 + · · · + n2 ) = ·
+n
,
2
2
2
2
therefore
1 2 + 2 2 + · · · + n2 =
n(n + 1)(2n + 1)
.
6
Now, we will study the sum of the cubes of the first n natural numbers,
1 3 + 2 3 + · · · + n3 .
To that purpose, consider the following array of numbers:
6
7
...
1
2
3
4
5
2
4
6
8
10 12 14 . . .
3
6
9
12 15 18 21 . . .
4
8
12 16 20 24 28 . . .
5
..
.
10 15 20 25 30 35 . . .
..
.
If we add the numbers in each of the squares, we obtain
S1 = 1
S2 = 1 + 2 + 2(1 + 2)
S3 = 1 + 2 + 3 + 2(1 + 2 + 3) + 3(1 + 2 + 3)
..
..
.
.
Sn = (1 + 2 + · · · + n) + 2(1 + 2 + · · · + n) + · · · + n(1 + 2 + · · · + n)
=
n(n + 1)
2
2
.
If we now sum all the corridors
C1 = 1
C2 = 2 · 2 + 2 · 2 = 22 + 22 = 22 (1 + 1) = 23
C3 = 2(3 + 2 · 3) + 32 = 2 · 3(1 + 2) + 32 = 2 · 32 + 32 = 32 (2 + 1) = 33
..
..
.
.
Cn = 2(n + 2n + 3n + · · · + (n − 1)n) + n2
= 2n
(n − 1)n
2
+ n2 = n3 − n2 + n2 = n3 .
40
Chapter 2. Progressions and Finite Sums
But, since C1 + C2 + C3 + · · · + Cn = Sn , we get
1 3 + 2 3 + 3 3 + · · · + n3 =
n(n + 1)
2
2
.
The above sum can also be written as
13 + 23 + 33 + · · · + n3 = (1 + 2 + 3 + · · · + n)2 .
Exercise 2.25. Using the corridor technique in the following array
1 2 2 2 3 2 4 2 5 2 . . . k 2 . . . n2
1 2 2 2 3 2 4 2 5 2 . . . k 2 . . . n2
1 2 2 2 3 2 4 2 5 2 . . . k 2 . . . n2
..
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.
.
.
2
2
2
2
2 ...
2 ...
1 2 3 4 5
k
n2
n-rows
prove the following equality,
n(n + 1)
2
2
= 1 3 + 2 3 + 3 3 + · · · + n3 .
2.4 Telescopic sums
When we develop a sum of the form
n
#
k=1
[f (k + 1) − f (k)] ,
we have that
n
#
k=1
[f (k + 1) − f (k)] = f (2) − f (1) + f (3) − f (2) + · · · + f (n + 1) − f (n),
then the terms f (k), for k between 2 and n, cancel out and we obtain that the
sum is equal to f (n + 1) − f (1). This type of sums are called telescopic sums. Let
us see some examples.
Example 2.4.1. Evaluate the sum
n
#
k=1
1
.
k(k + 1)
41
2.4 Telescopic sums
Observe that
n
#
k=1
1
k(k+1)
n
#
1
=
k(k + 1)
k=1
=
1
k
−
1
k+1 ,
therefore
1
1
−
k k+1
1 1
1
−
+
2
2 3
n
1
=
.
=1−
n+1
n+1
1−
=
+
1 1
−
3 4
Example 2.4.2 (Canada, 1969). Calculate the sum
1
1
−
n n+1
+ ··· +
*n
k=1
k! · k.
Observe that k! · k = k!(k + 1 − 1) = (k + 1)! − k!, then we have that
n
#
k=1
k! · k =
n
#
k=1
((k + 1)! − k!)
= (2! − 1!) + (3! − 2!) + · · · + ((n + 1)! − n!)
= ((n + 1)! − n!) + (n! − (n − 1)!) + · · · + (2! − 1!)
= (n + 1)! − 1.
Example 2.4.3. Evaluate the product
1−
1
22
1−
1
32
··· 1 −
1
20112
.
We have that
1−
1
22
=
=
=
=
1
20112
1
1
1
1
1
1+
1−
1+
1−
··· 1 +
2
2
3
3
2011
1
1
1
1
1
1+
1+
··· 1+
1−
1−
2
3
2011
2
3
1 2 3
2012
2010
3 4 5
· · ·····
· · ·····
2 3 4
2011
2 3 4
2011
2012
1
1006
.
=
2
2011
2011
1−
1
32
··· 1 −
Exercise 2.26. Find the sum
1
1
1
1
+
+
+ ···+
.
1 · 4 4 · 7 7 · 10
2998 · 3001
1−
1
2011
··· 1 −
1
2011
42
Chapter 2. Progressions and Finite Sums
Exercise 2.27. Calculate the following sums:
n
n
#
#
2k + 1
1
(ii)
.
(i)
k(k + 2)
k 2 (k + 1)2
k=1
k=1
Exercise 2.28. Calculate the following sums:
n
n
#
#
k+1
k
(ii)
.
(i)
(k + 1)!
(k − 1)! + k! + (k + 1)!
k=1
k=1
Exercise 2.29. Find the sum
1+
1
1
+ 2+
2
1
2
1+
1
1
+ 2 + ···+
2
2
3
1+
1
1
+
.
2
2011
20122
Exercise 2.30. Find the following product
1+
1
2
1+
1
22
··· 1 +
1
22n
.
Chapter 3
Induction Principle
3.1 The principle of mathematical induction
In Chapter 2 we deduced several formulas for finite sums of numbers. Thus we
learned that the sum of the first n natural numbers is given by the identity
1 + 2 + ···+ n =
n(n + 1)
.
2
(3.1)
In fact, this formula is a collection of statements, P(n), which we have proved
using algebra, that are valid for every positive integer n.
The validity of these series of statements can be proved using what is known as
the principle of mathematical induction, which will be developed in this chapter.
The principle of mathematical induction claims that a sequence of propositions
P(1), P(2), P(3), . . . are valid if:
1. The statement P(1) is true.
2. The statement “P(k) implies P(k + 1)” is true.
We can guarantee that this last statement is true assuming that P(k) is true and
proving the validity of P(k + 1)7 .
Observation 3.1.1. Statement 1 of the principle of mathematical induction is called
the induction basis and statement 2 is known as the inductive step.
We will prove using this principle the validity of the identity (3.1) for every natural
number.
Example 3.1.2. The identity 1 + 2 + 3 + · · ·+ n =
integer n.
7 To
n(n+1)
2
is valid for every positive
prove 2. is equivalent to prove that: “not P (k + 1) implies not P (k)”, is true.
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5_3
43
44
Chapter 3. Induction Principle
If n = 1 the left-hand side of the identity has a unique term, that is, 1. The right, since this last term is also 1. We have that the formula is valid
hand side is 1(1+1)
2
for n = 1.
Now, suppose that the statement P(k) is true, that is, the equality
1 + 2 + 3 + ···+ k =
k(k + 1)
2
(3.2)
is valid. We will prove the corresponding identity for k + 1.
According to the formula, the terms on the left-hand side, for k + 1, are 1 + 2 +
3 + · · · + k + (k + 1), and if we use (3.2), we have that
k(k + 1)
+ (k + 1) = (k + 1)
2
(k + 1)(k + 2)
=
,
2
1 + 2 + 3 + · · · + k + (k + 1) =
k
+1
2
which is the same as the right-hand side of the formula for k + 1. This proves the
validity of P(k + 1). Therefore, by the principle of mathematical induction the
identity is valid for every positive integer n
Example 3.1.3. For every natural number n, the number n3 − n is a multiple of 6.
If n = 1, then n3 − n = 0 is a multiple of 6, since 0 = 6 · 0.
Suppose that k 3 − k is a multiple of 6, that is, k 3 − k = 6 r, for some integer
number r. We will prove that the statement is true for k + 1, that is, we will show
that (k + 1)3 − (k + 1) is a multiple of 6.
Now, we have that
(k + 1)3 − (k + 1) = k 3 + 3 k 2 + 3 k + 1 − k − 1
= (k 3 − k) + 3(k 2 + k) = 6 r + 3(k 2 + k).
Here we have used the induction hypothesis that P(k) holds, that is, k 3 − k = 6r.
Since k 2 +k = k(k+1) is the product of consecutive numbers, one of them has to be
even, then k 2 + k is even. Then, 3(k 2 + k) is a multiple of 6. Thus, (k + 1)3 − (k + 1)
is a sum of multiples of 6. Therefore, by the principle of mathematical induction
the result is valid for every positive integer n.
Example 3.1.4 (Hanoi Towers). A beautiful legend of the creation of the world tells
us that Brahma placed on the Earth three bars of diamond and 64 golden discs.
The discs had all different sizes and at the beginning they were located in the first
diamond bar following a decreasing order of the diameters from bottom to top.
Also Brahma created a monastery where the monks had the fixed task of moving
all the discs from the first bar to the third, following some rules. The allowed rules
were to move one disc from one bar to any other bar but under the condition that
a disc with a greater diameter could never be placed on top of a smaller disc. The
45
3.1 The principle of mathematical induction
legend also tells us that when the monks would finish their duty, the world will end.
What is the smallest number of necessary moves that the monks have to make to
accomplish the work?
1◦ bar
2◦ bar
3◦ bar
Let xn be the minimum number of necessary moves to move n discs. If n = 1,
there is only one disc and therefore just one move is needed, that is, x1 = 1. With
two discs in the first bar, we move the small disc to the second bar, then the big
one from the first bar to the third one, and finally the small one from the second
bar to the third. Therefore, x2 = 3. With three discs the problem starts to become
interesting.
Expressing as i → j the move of the top disc from bar i to bar j, the discs can
move in the following sequence of moves 1 → 3, 1 → 2, 3 → 2, 1 → 3, 2 → 1,
2 → 3, 1 → 3, that is, x3 = 7. Observing the obtained sequence 1, 3, 7, we can
note that if we sum 1 to each term we obtain consecutive powers of 2. Therefore,
we conjecture that xn = 2n − 1. To prove this conjecture by induction, we only
need to prove the inductive step. Suppose that xn = 2n − 1 moves and let us see
what happens with n + 1 discs. By the inductive hypothesis, the n superior discs
can be moved to the second bar in 2n − 1 moves. After this, the largest disc moves
from the first to the third bar in one move, and finally the n discs in the second
bar can be moved to the third bar in 2n − 1 moves. Therefore, in total we made
(2n − 1) + 1 + (2n − 1) = 2n+1 − 1 moves.
Therefore, 2n − 1 moves are enough, but is this the minimum? We will prove by
induction that in fact this is the minimum. For n = 1, this is obvious. Suppose
that it is true for n. Now, we want to move n + 1 discs from the first to the third
bar, that is, in some moment we have to move the largest disc to the third bar.
At this moment the remaining n discs have to be all together in another bar and,
by the induction hypothesis, they could not have arrived there in less than 2n − 1
moves. In the same way, when the largest disc moves to its final position in the
third bar, the n remaining discs have to be all in another bar, and to move them
to the third bar we need at least 2n − 1 moves.
Adding, we see that we cannot move n+1 discs in less than (2n −1)+1+(2n −1) =
2n+1 − 1 moves, which completes the proof.
With 64 discs, as in the legend, it is necessary to do 264 − 1 moves, and if we
suppose that the monks can do one move every second, to finish their task they
will need 500 thousand millions of years.
46
Chapter 3. Induction Principle
Observation 3.1.5. In many cases, the basis of induction is not for n = 1 but for
n = k, for some natural number k. Then, if we prove the induction step, we can
conclude that P(n) is true for all natural numbers n ≥ k.
Example 3.1.6. For x, y real numbers and for n ≥ 2, it follows that
xn − y n = (x − y)(xn−1 + xn−2 y + · · · + xy n−2 + y n−1 ).
If n = 2 the result follows because x2 − y 2 = (x − y)(x + y).
Suppose that xk − y k can be written as
xk − y k = (x − y)(xk−1 + xk−2 y + · · · + xy k−2 + y k−1 ).
We will prove the result for k + 1. We start from the following identity,
xk+1 − y k+1 = xk+1 − xy k + xy k − y k+1
= x(xk − y k ) + y k (x − y).
Using the induction hypothesis, we have that
x(xk − y k ) + y k (x − y) = x(x − y)(xk−1 + xk−2 y + · · · + y k−1 ) + y k (x − y)
= (x − y)(x(xk−1 + xk−2 y + · · · + y k−1 ) + y k )
= (x − y)(xk + xk−1 y + · · · + xy k−1 + y k ).
Therefore, xk+1 − y k+1 = (x − y)(xk + xk−1 y + · · · + xy k−1 + y k ), as we wanted
to prove.
Example 3.1.7. The sum of the interior angles of an n-sided convex polygon is
180◦(n − 2).
The statement makes sense for n ≥ 3, that is, when we have a triangle. For
n = 3, the statement holds since the sum of the interior angles of a triangle is
180◦ = 180◦ (3 − 2).
Suppose the statement is valid for any convex n-gon. Given a convex (n + 1)-gon
with vertices A1 , . . . , An+1 , the diagonal An A1 divides the polygon in a convex
n-gon A1 A2 . . . An and a triangle An An+1 A1 . The sum of the interior angles of the
(n+ 1)-gon will be the sum of the angles of the n-gon A1 . . . An plus the sum of the
interior angles of the triangle An An+1 A1 , that is, 180◦(n−2)+180◦ = 180◦ (n−1).
To verify a series of statements P(n), in some cases it is better to work with
more general propositions P ′ (n), that is, propositions such that the validity of
P ′ (n) will guarantee the validity of P(n). Let us see some examples.
Example 3.1.8. For any positive integer n ≥ 2, it follows that
1
1
3
1
+ 2 + ···+ 2 < .
22
3
n
4
3.1 The principle of mathematical induction
47
To use the principle of mathematical induction to prove the inequality can
be complicated, since if we suppose that 212 + 312 + · · · + n12 < 43 , we would have
1
3
to prove that 212 + 312 + · · · + n12 + (n+1)
2 < 4 , but the margin of manoeuvre is
limited. Then we will try to work with the following stronger result
Sn ≤
3
− an , for all n ≥ 2,
4
where Sn is the sum on the left-hand side of the inequality that we want to prove,
and {an } are positive numbers that we have to discover. In order to produce a
proof by mathematical induction we need to show that the basis of induction is
valid, that is
3
1
1
≤ − a2 or a2 ≤ .
4
4
2
The inductive step consists in showing that the condition Sn ≤ 43 −an implies
1
that Sn+1 ≤ 34 − an+1 . Since Sn+1 = Sn + (n+1)
2 , the above will be true if an and
an+1 satisfy
1
− an + an+1 ≤ 0,
(n + 1)2
which is equivalent to
an − an+1 ≥
Now, the numbers an =
1
n
1
, for all n ≥ 2.
(n + 1)2
satisfy the above inequality, since
1
1
1
1
−
=
≥
.
n n+1
n(n + 1)
(n + 1)2
Also, a2 = 12 satisfies the condition of the basis of induction. Therefore, the numbers an = n1 are good candidates to make possible the induction. In fact we have
proved that, for every n ≥ 2, it follows that Sn ≤ 34 − n1 , thus
1
1
1
3
+ 2 + ···+ 2 < .
22
3
n
4
A similar example is the following.
Example 3.1.9. For any positive integer n, it follows that
1
1
1
√ < 2.
√ + √ + ···+
(n + 1) n
2 1 3 2
As in the previous example, it is sufficient to prove that
1
2
1
1
√ <2− √
√ + √ + ···+
.
(n + 1) n
n+1
2 1 3 2
For n = 1, we have
1
√
2 1
< 2 − √22 , which is true, that is, the induction basis holds.
48
Chapter 3. Induction Principle
For the inductive step, it suffices to prove that
2
2
1
√
< √
−√
,
(n + 2) n + 1
n+1
n+2
which can be reduced to prove that
√
√
√
1
2
√
< 2( n + 2 − n + 1) = √
,
n+2
n+2+ n+1
but this last inequality is obviously true.
Another frequently used version of the principle of mathematical induction is the
following.
Strong induction principle. The set of propositions P(1), P(2), P(3), . . . , P(n),
. . . are true if:
1. The statement P(1) is true.
2. The statement “P(1), . . . , P(n) implies that P(n + 1)” is true.
Observations 3.1.10. (a) In the strong induction principle, the inductive step is
valid if each time that P(k) is valid for k ≤ n then, starting with this hypothesis,
we prove that P(n + 1) is valid.
(b) It is clear that the strong induction principle implies the simple principle of induction. But in fact both are equivalent, since strong induction is a
consequence of simple induction. To see this, it is enough to consider the logical
conjunction8 Q(n) of the propositions P(1), . . . , P(n). If P(1) is true also Q(1)
is true (since they are exactly the same). If Q(n) is valid so are P(1), P(2), . . . ,
P(n), and by the strong induction hypothesis also P(n + 1) is valid, this implies
that Q(n + 1) is true. Then, by simple induction Q(n), is true for any natural
number n and the same is true for P(n).
Although strong induction and the simple one are logically equivalent, in some
cases it is easier to use one rather than the other.
The strong induction is implicit in the definition of sequences by recurrence
relations.
Example 3.1.11. If a is a real number such that a +
is an integer for all n ≥ 1.
1
a
is an integer, then an +
1
an
For n = 1, the statement is true by the hypothesis that a + a1 is an integer number.
Let us see how to solve for n = 2; sometimes this case gives us ideas about how
to justify the inductive step.
2
2
Note that a + a1 = a2 + a12 + 2; then a2 + a12 = a + a1 − 2 is an integer number
and then our statement for n = 2 is true.
8 The logical conjunction of P(1), . . . , P(n) means that all the propositions are true at the same
time.
49
3.1 The principle of mathematical induction
Let us analyze again our formula
2
1
1 1
1
1
+ a · + · a = a2 + 2 + a0 + 0 ,
2
a
a a
a
a
then we get, a2 + a12 = a + a1 · a + a1 − (a0 + a10 ), but this gives us another
idea, the statement for n = 2 depends on the statements for n = 1 and for n = 0
(which by the way are also valid).
To obtain the statement for n = 3, we will work following the previous idea,
a+
1
a
=
a+
1
a
a+
a2 +
1
a2
1
a
= a2 +
· a+
1
a
= a3 +
1
1
+a+ .
a3
a
It follows that
a3 +
1
=
a3
a2 +
1
a2
· a+
1
a
− a+
1
a
,
the right-hand side is an integer number if a + a1 and also a2 + a12 are integers,
but this is already known. It is now clear how we can prove the inductive step.
Suppose that the statement is valid for integers less than or equal to n, then from
the identity
an+1 +
1
=
an+1
an +
1
an
· a+
1
a
− an−1 +
1
an−1
,
(3.3)
it follows that the statement for n + 1 is also valid.
There are other ways to prove statements inductively.
Cauchy’s induction principle. The set of propositions P(1), P(2), . . . , P(n), . . .
are all valid if:
1. The statement P(2) is true.
2. The statement “P(n) implies P(n − 1)” is true.
3. The statement “P(n) implies P(2n)” is true.
Observation 3.1.12. Let us see why Cauchy’s induction principle implies the principle of mathematical induction. First, note that 1 and 2 guarantee that P(1) is
true, that is, the induction basis holds. Since 3 holds, the validity of P(n) implies
that P(2n) is true. Now, applying n−1 times the condition 2, we get that P(2n−1),
P(2n− 2), . . . , P(n+ 1) are all true. In particular, P(n+ 1) is true. Thus, we have
proved the inductive step of the principle of mathematical induction. Therefore all
P(n) are true.
We apply this inductive process to prove the inequality between the geometric
mean and the arithmetic mean.
50
Chapter 3. Induction Principle
Example 3.1.13. For real non-negative numbers x1 , . . . , xn the following inequality
holds for all n:
√
x1 + x2 + · · · + xn
≥ n x1 x2 . . . xn .
(3.4)
n
√
n
Denote by A = x1 +x2 +···+x
and G = n x1 x2 . . . xn , the arithmetic mean and the
n
geometric mean of the numbers x1 , . . . , xn , respectively. The proof will be done
by induction over n, using the inductive principle just described.
1. For n = 1 we have the equality, and the case n = 2 was proved in Section 1.6.
√
2. Let x1 , x2 , . . . , xn be non-negative numbers and let g = n−1 x1 · · · xn−1 . If
we add this number to the numbers x1 , . . . , xn−1 , we obtain n numbers to
which we apply P(n) and obtain
x1 + · · · + xn−1 + g
√
≥ n x1 x2 · · · xn−1 g =
n
n
g n−1 · g = g.
Then, x1 + · · · + xn−1 + g ≥ ng, which in turn leads to
therefore P(n − 1) is valid.
3. Let x1 , x2 , . . . , x2n be non-negative numbers, then
x1 +···+xn−1
n−1
≥ g, and
x1 + x2 + · · · + x2n = (x1 + x2 ) + (x3 + x4 ) + · · · + (x2n−1 + x2n )
√
√
√
≥ 2 x1 x2 + x3 x4 + · · · + x2n−1 x2n
√
1
1
√
√
≥ 2n x1 x2 x3 x4 · · · x2n−1 x2n n = 2n (x1 x2 · · · x2n ) 2n .
In the previous sequence we applied several times the statement P(2), which
we know is true, and after that the same was done with statement P(n) to the
√
√
√
numbers x1 x2 , x3 x4 , . . . , x2n−1 x2n .
Example 3.1.14. Let us see another way to prove the previous inequality, but in this
case using another variant of the induction principle. Observe first that if some
xi = 0, then the inequality is clear.
Suppose then that every xi > 0, and that x1 x2 . . . xn = 1. Prove the statements
P(n): x1 + x2 + · · · + xn ≥ n.
Clearly the basis of induction is true, that is, P(1) : x1 ≥ 1 is true, in fact x1 = 1.
Suppose that P(n) is valid for any n positive numbers whose product is 1,
and let x1 , . . . , xn , xn+1 be positive numbers whose product is 1. Then, there will
be some xi ≥ 1 and some xj ≤ 1. Without loss of generality, we can suppose that
x1 ≥ 1 and x2 ≤ 1. Then, by the previous statement, (x1 − 1)(x2 − 1) ≤ 0, then
x1 x2 +1 ≤ x1 +x2 . Therefore, x1 +x2 +· · ·+xn +xn+1 ≥ 1+x1 x2 +x3 +· · ·+xn+1 .
Now apply the induction hypothesis to the n numbers x1 x2 , x3 , . . . , xn+1 to show
that x1 + x2 + · · · + xn + xn+1 ≥ 1 + n, that is, the statement P(n + 1) is true.
√
For the general case, if a1 , . . . , an are positive and letting G = n a1 a2 . . . an , if we
an
a1
consider the identities x1 = G , . . . , xn = G , we have that x1 . . . xn = 1. Then,
a1 + · · · + an
√
x1 + · · · + xn ≥ n, which is equivalent to
≥ n a1 . . . an .
n
51
3.1 The principle of mathematical induction
Example 3.1.15. Let x1 , x2 , . . . , xn and y1 , y2 , . . . , yn be natural numbers. Suppose
that x1 + x2 + · · · + xn = y1 + y2 + · · · + ym < mn. Then it is possible to cancel
out some terms (not all of them) from both sides of the above equality while always
preserving the equality.
We use induction over k = m + n. Since n ≤ x1 + x2 + · · · + xn < mn, then
m > 1, and similarly n > 1, then m, n ≥ 2 and k ≥ 4. For m + n = 4, we have
that m = n = 2, and the only possible cases are 1 + 1 = 1 + 1 and 1 + 2 = 1 + 2
(maybe in a different order) and the result is immediate.
Suppose that k = m + n > 4 and consider
s = x1 + x2 + · · · + xn = y1 + y2 + · · · + ym < mn.
Without loss of generality, we can suppose that x1 is the largest term of the set
of xi ’s, with i = 1, 2, . . . , n, and y1 is the largest term of the set of yj ’s, with
j = 1, 2, . . . , m. We can also assume that x1 > y1 , because if x1 = y1 the problem
is solved. Then we have
(x1 − y1 ) + x2 + · · · + xn = y2 + · · · + ym .
We need to prove that the sum s′ = y2 + · · · + ym satisfies the required condition,
s
, hence
that is, s′ < n(m − 1). Since y1 ≥ y2 ≥ · · · ≥ ym , it follows that y1 ≥ m
s
m−1
m−1
=s
< mn
= n(m − 1),
m
m
m
and now we can apply the principle of mathematical induction to reach the desired
conclusion.
s′ = s − y 1 ≤ s −
Exercise 3.1.
(i) Prove by induction that for q = 1,
1 + q + · · · + q n−1 =
1 − qn
.
1−q
(ii) Prove that 1 + 21 + 22 + · · · + 2n = 2n+1 − 1.
Exercise 3.2. For the Fibonacci sequence, defined by a1 = 1, a2 = 1 and, for
n ≥ 3, an = an−1 + an−2 , prove that
an+2 = 1 + a1 + a2 + · · · + an .
n
Exercise 3.3. Prove that 3n+1 divides 23 + 1, for any integer number n ≥ 0.
Exercise 3.4. There are 3n coins with identical aspect, but one of the coins is false
and its weight is less than the weight of the real coins. Prove how, with a plate
weighing scale9 , in n weighings we can identify the false coin.
9 A plate weighing scale is a balance with two plates that will be at the same level if the weight
of the objects placed in each one of the plates is the same.
52
Chapter 3. Induction Principle
Exercise 3.5.
(i) For which integers n does it follow that 7 divides 2n − 1?
(ii) For which positive integers n does it follow that 7 divides 2n + 1?
Exercise 3.6. Find the values of an , if a1 = 1 and for each n ≥ 2, it follows that
a1 + a2 + · · · + an = n2 .
Exercise 3.7. For n ≥ 1, prove that
n n n + 1
1
2
3
=
+
+
+ ···+
.
2
2
2
2
2
2
Exercise 3.8. Find the values of an , if a1 = 1 and for each n ≥ 2
√
n an+1
√
√
√
.
a1 + a2 + · · · + an =
2
Exercise 3.9. Prove the next inequality relating the geometric mean and the arithmetic mean,
√
x1 + x2 + · · · + xn
≥ n x1 x2 · · · xn ,
n
by following the indicated steps:
(i) Use induction to prove that
xn+1 − (n + 1)x + n = (x − 1)2 (xn−1 + 2xn−2 + 3xn−3 + · · · + n).
Therefore, for x > 0, it follows that xn+1 − (n + 1)x + n ≥ 0.
n+1
and b =
(ii) Apply the previous inequality to x = ab , where a = x1 +···+x
n+1
x1 +···+xn
, and conclude that
n
x1 + · · · + xn+1
n+1
n+1
≥ xn+1
x1 + · · · + xn
n
n
= xn+1 bn .
(iii) Now use induction again in order to finish the proof.
*n
Exercise 3.10.
that i=1 ei ai
0 < a1 < a2 < · · · < an and ei = ±1. Prove
n+1Let
has at least 2 different values when the ei vary over the 2n possible elections
of the signs.
Exercise 3.11. A sequence a1 , a2 , . . . , a2n of numbers 0 or 1 is said to be evenbalanced if a1 +a3 +· · ·+a2n−1 = a2 +a4 +· · ·+a2n . Prove that an arbitrary sequence
of numbers 0 or 1, with 2n + 1 elements, has a subsequence10 even-balanced with
2n elements.
10 See
Subsection 7.2.7, for the definition of subsequence.
53
3.2 Binomial coefficients
Exercise 3.12. Prove that the only infinite sequence {an } of positive numbers such
that for each positive integer number n, the equality
a31 + a32 + · · · + a3n = (a1 + a2 + · · · + an )2
holds, is the sequence given by an = n, for n = 1, 2, . . . .
Exercise 3.13. If a1 < a2 < · · · < an are positive integers, then
a31 + a32 + · · · + a3n ≥ (a1 + a2 + · · · + an )2 ,
with equality if and only if ak = k, for each k = 1, 2, . . . , n.
Exercise 3.14.
(i) Prove that, for n ≥ 1, it follows that
1+
1
2
1+
1
22
··· 1 +
1
2n
<
5
.
2
(ii) Prove that, for n ≥ 1, it follows that
1+
1
13
1+
1
23
··· 1 +
1
n3
< 3.
Exercise 3.15. Let a1 , a2 , . . . , an be real numbers greater than or equal to 1. Prove
that
n
#
1
n
≥
.
√
n
1
+
a
1
+
a
i
1 · · · an
i=1
3.2 Binomial coefficients
The factorial of an integer number n ≥ 0, denoted by n!, can be defined by
induction as follows:
(a) 0! = 1,
(b) n! = n(n − 1)!, for n ≥ 1.
Observation 3.2.1. If n ≥ 1, then
n! = n(n − 1) · · · 2 · 1.
11
all integers n and m, with 0 ≤ m ≤ n, we define the binomial coefficient
For
n
m as
n
n!
=
.
(3.5)
m!(n − m)!
m
11 For
a combinatorial meaning of the binomial coefficients and the factorial, see [19].
54
Chapter 3. Induction Principle
Properties 3.2.2. For 0 ≤ m ≤ n, it follows that:
(a)
n
0
= 1 and
(b)
n
m
=
n
n
= 1.
n
.
n−m
(c) For each m = 1, 2, . . . , n − 1, it follows that
n
m
=
n−1
n−1
+
.
m−1
m
(3.6)
This identity is known as Pascal’s formula.
(d)
n
m
is a positive integer.
Proof. To prove (a) and (b) we just apply the definition of binomial coefficient.
(c) To prove this property, observe that
n−1
n−1
+
m−1
m
(n − 1)!
(n − 1)!
+
(m − 1)!(n − 1 − m + 1)! m!(n − 1 − m)!
m(n − 1)! + (n − m)(n − 1)!
n!
=
=
=
m!(n − m)!
m!(n − m)!
=
n
.
m
(d) This statement
can be justified
by induction over n. For n = 0 and n = 1,
it is clear since 00 = 1 and 10 = 11 = 1.
n
Suppose that for n − 1 the assumption is true,
us prove that m
is
let
nand
n
=
1
are
integers.
By
=
an integer, for m = 0, . . . , n. By (a), we have
that
n
0
n
is a sum of two integers and therefore
(c), for m = 1, . . . , n − 1, we have that m
it is an integer.
Theorem 3.2.3 (Binomial theorem). Let a and
b be real numbers and let n and m
n
are the binomial coefficients in the
be integers with 0 ≤ m ≤ n. The numbers m
binomial development (a + b)n , that is,
(a + b)n =
n n
n n−1
n n−i i
n n
a +
a
b + ···+
a b + ···+
b .
0
1
i
n
Using the sum notation we can write the previous equality as
(a + b)n =
n
#
n n−i i
a b.
i
i=0
This identity is known as Newton’s binomial formula.
(3.7)
55
3.2 Binomial coefficients
Proof. Induction over n yields aprove
of this identity.
If n = 0, then (a + b)0 = 1 and 00 a0 b0 = 1.
Suppose that n > 0 and that the identity is valid for n − 1, that is,
(a + b)n−1 =
n−1
#
i=0
n − 1 n−1−i i
a
b
i
is true, then
(a+b)n = (a + b)(a + b)n−1 = (a + b)
n−1
#
i=0
=
=
n−1
#
n − 1 n−1−i i
b
a
i
n−1
n−1 0 n
n − 1 n−i i # n − 1 n−i i
a b
a b +
a b +
i
−
1
n
−1
i
i=1
i=1
,
n−1 +
n−1
n n # n−1
n−1 n
a +
b
+
an−i bi +
0
n
−1
i
i
−
1
i=1
n−1 n
a +
0
n
n−1
=
n n # n n−i i
n n # n n−i i
a b.
a b +
b =
a +
n
0
i
i
i=0
i=1
In the last step we have used Pascal’s formula.
Example 3.2.4. If n is a positive integer, then
n
#
i=0
(−1)i
n
i
= 0.
Apply the binomial theorem with a = 1 and b = −1, to obtain
n
n
#
#
n
n
i
0 = (1 − 1) =
.
(−1)i
(−1) =
i
i
i=0
i=0
n
Exercise 3.16. Prove the following equalities:
n
#
n
(i)
j
j=0
n
=2 .
n
#
n j
(ii)
2 = 3n .
j
j=0
Exercise 3.17. Prove the following equalities:
n
m
n
n−r
(i)
=
.
m
r
r
m−r
(ii)
n
m
=
n n−1
.
m m−1
56
Chapter 3. Induction Principle
Exercise 3.18. Prove the following equalities:
n
2
#
2n
n
=
.
(i)
n
j
j=0
(ii)
r
#
n
k
k=0
(iii)
m
r−k
n
#
m+k
k
=
m+n+1
.
n
=
k=0
(iv)
n
#
k=m
k
m
=
n+m
.
r
n+1
.
m+1
Exercise 3.19. Calculate the following sums:
n
#
n
(i)
.
j
j
j=1
(ii)
n
#
j=0
n
1
.
j+1 j
Exercise 3.20. Prove the following equalities:
(i)
(ii)
(iii)
n
#
(−1)j+1 n
j
j
j=1
n
#
(−1)j+1 n
j(j + 1) j
j=1
n
#
(−1)j n
(j + 1)2 j
j=0
=1+
1
1 1
+ +···+ .
2 3
n
=
1
1 1
+ + ··· + .
2 3
n
=
1
n+1
1+
1
1
+ ··· +
.
2
n
Exercise 3.21. Prove that the following relations hold:
n
n
#
#
n
n
j
= 0.
(−1)j j 2
(−1) j
= 0.
(ii)
(i)
j
j
j=1
j=1
Exercise 3.22. Prove that the number of odd integers in the following list
n
n
n
,
,...,
0
1
n
is a number which is a power of 2.
Exercise 3.23. For each prime number p ≥ 3, prove that the number
divisible by p2 .
2p−1
p−1
− 1 is
57
3.3 Infinite descent
3.3 Infinite descent
The method of infinite descent was frequently used by Pierre Fermat (1601–1665),
therefore it is also known as Fermat’s method. In general, it is used to prove that
something does not happen. For instance, Fermat used it to prove that there are
no integer solutions of the equation x4 + y 4 = z 2 , with xyz = 0.
The theoretical basis of his method rests on the fact that there is no such thing
as an infinite decreasing collection of positive integers. In other words, we cannot
find an infinite collection of positive integers such that n1 > n2 > n3 > · · · .
There are two ways to use this idea in order to prove a statement. The first is to
start with a statement P(n1 ) which we suppose is true. If from this statement, we
can find a positive integer number n2 < n1 such that P(n2 ) is valid and if from
this last statement we can find a positive integer number n3 < n2 such that in
turn P(n3 ) is valid, and so on and so forth, then an infinite number of positive
integers is generated satisfying n1 > n2 > n3 > · · · , but this is not possible, so
P(n1 ) is not true. Let us see an example in order to illustrate this method.
√
Example 3.3.1. The number 2 is not a rational number.
√
√
1
Suppose that 2 is a rational number, then 2 = m
n1 , with m1 and n1 positive
√
1
, we have that
integers. Since 2 + 1 = √2−1
√
2+1=
m1
n1
n1
1
,
=
m1 − n 1
−1
therefore
√
2=
2n1 − m1
n1
−1=
.
m1 − n 1
m1 − n 1
√
√
1
Since 1 < 2 < 2, substituting the suppose rational value 2 leads to 1 < m
n1 < 2,
hence n1 < m1 < 2n1 . From here we have that, 2n1 − m1 > 0 and m1 − n1 > 0.
Then, if we define m2 = 2n1 − m1 and n2 = m1 − n1 , we have√that m2 < m1
m2
1
and n2 < n1 , since n1 < m1 and m1 < 2n1 , respectively. Then, 2 = m
n1 = n2 ,
with m2 < m1 and n2 < n1 . Continuing this process we can generate an infinite
number of positive integers mi and ni such that
√
m1
m2
m3
2=
=
=
= ··· ,
n1
n2
n3
with m1 >√m2 > m3 > · · · and n1 > n2 > n3 > · · · , but this is not possible.
Therefore, 2 is not a rational number.
Example 3.3.2. Find all the pairs of positive integers a, b that satisfy the equation
a2 − 2b2 = 0.
Suppose that there exist positive integers a1 , b1 such that
a21 − 2b21 = 0.
(3.8)
58
Chapter 3. Induction Principle
This implies that a1 is an even number, that is, a1 = 2a2 for some positive integer
a2 . Then, since
(2a2 )2 − 2b21 = 0,
it follows that
2a22 − b21 = 0.
Therefore b1 is even, that is, b1 = 2b2 with b2 a positive integer. Substitution in
the last equation leads to
2a22 − (2b2 )2 = 0,
or a22 − 2b22 = 0.
(3.9)
This implies that a2 and b2 is another pair of positive integers satisfying equation
(3.8). Since a1 = 2a2 , we have that a1 > a2 . Moreover, the above equations imply
that a1 > b1 > a2 > b2 . Equation (3.9) proves that a2 is even, that is, a2 = 2a3
for some positive integer a3 .
Repeating the above arguments we obtain an infinite sequence of natural
numbers in which each term is smaller than the previous one, that is,
a1 > b 1 > a2 > b 2 > a3 > b 3 > · · · .
However, this sequence cannot exist. Therefore, there is no pair of natural√numbers
that satisfy the equation (3.8). Note that this example proves also that 2 is not
a rational number.
The other way to use the infinite descent method has a more positive character.
It can be used to show that a series of propositions P(a) are valid, where a is an
element of a set A ⊂ N. To do this we use the following argument: suppose that
P(a) is not valid for some a ∈ A and define the set B = {a ∈ A | P(a) is not true}.
Since B = ∅, in B there is a first element, say b. Now, using the hypothesis of the
problem, we can find a positive integer number c < b such that P(c) is not valid.
This leads to a contradiction since b was the minimum element of B. Then, P(a)
has to be true for all a ∈ A.
Example 3.3.3 (Putnam, 1973). Let a1 , a2 , . . . , a2n+1 be integers such that if we
take one out, then the remaining numbers can be divided in two sets of n integers
which have the same sum. Prove that all numbers are equal.
We can suppose that a1 ≤ a2 ≤ · · · ≤ a2n+1 . If we subtract the smallest number from all these numbers, the new numbers also satisfy the inequality and the
conditions of the problem. Then, without loss of generality, we can suppose that
a1 = 0.
The sum of the 2n remaining numbers different from a1 satisfies the condition of
being congruent to 0 modulo 2. Now, let us see that if we choose any two numbers
the pair will have the same parity. Let ai and aj be any two such numbers and
S = a1 + · · · + a2n+1 . Since S − ai ≡ S − aj ≡ 0 mod 2, we have that ai ≡ aj
mod 2.
59
3.3 Infinite descent
If we divide by 2 all the numbers, the new collection has the same properties.
Using the same arguments we can conclude that
a1 ≡ a2 ≡ · · · ≡ a2n+1 ≡ 0
mod 22 .
We can continue this argument to conclude that
a1 ≡ a2 ≡ · · · ≡ a2n+1 ≡ 0
mod 2k ,
for each k ≥ 1, but this is possible only in the case in which all the numbers are
equal to zero, and therefore the original numbers are equal.
Example 3.3.4. Let ABC be an acute triangle. Let A1 be the foot of the altitude
from A, A2 the foot of the altitude from A1 over CA, A3 the foot of the altitude
from A2 over AB, A4 the foot of the altitude from A3 over CB, and so on and so
forth. Prove that all the Ai ’s are different.
A
A3
B
A4 A1
A5
A2
C
Observe that each An is a point in one of the sides of the triangle, that is, it
cannot be in the extension of the side and it cannot be a vertex of the triangle.
This follows from the fact that if An is in the extension of one side of the triangle,
then the triangle is obtuse, and if it is a vertex, then the triangle is either obtuse
or a right triangle.
Note also that An and An+1 do not coincide, because they belong to different
sides of the triangle and they are not vertices.
Suppose now that An coincides with Am , with n < m, and suppose also
that n is the smallest index with the property that An coincides with some Am .
We now see that n = 1, since otherwise we would have that An−1 coincides with
Am−1 , and therefore n does not satisfy the property of being the smallest index n
such that An = Am , for some m > n.
Now if A1 coincides with Am , with m ≥ 3, then Am−1 has to be the vertex A,
but we already saw that no Am is vertex of a triangle. Then, A1 cannot coincide
with any Am .
60
Chapter 3. Induction Principle
Exercise 3.24 (Hungary, 2000). Find the prime numbers p for which it is impossible
to find integers a, b and n, with n positive, such that pn = a3 + b3 .
Exercise 3.25. The equation x2 + y 2 + z 2 = 2xyz has no integer solutions except
when x = y = z = 0.
Exercise 3.26. Find all the pairs of positive integers (a, b) such that ab + a + b
divides a2 + b2 + 1.
Exercise 3.27. Prove that for n = 4, it does not exist a regular polygon with n
sides such that the vertices are points with integer coordinates.
3.4 Erroneous induction proofs
In this section we present some examples showing the necessity of verifying all
the steps required in a proof which uses the induction mathematical principle.
Sometimes, if we miss proving one detail this can lead us to some absurd situations,
as we will see.
Example 3.4.1. All non-negative powers of 2 are equal to 1.
The induction basis is n = 0. The statement is true because 20 = 1.
Suppose this is true for all k ≤ n, that is, 20 = 21 = · · · = 2n = 1. Let us now
verify the statement for 2n+1 . Follow the identities and see that
2n+1 =
22n
2n · 2n
1·1
= 1.
=
=
2n−1
2n−1
1
Then the proof is complete.
Error. The inductive step is not valid for n = 0, that is, P (0) ⇒ P (1) is a false
statement.
Example 3.4.2. All integers greater than or equal to 2 are even.
The induction base for n = 2 is clearly valid, since 2 is even.
Suppose that for each integer number k with 2 ≤ k ≤ n, the statement is true,
that is, such numbers k are even. Let us see now that n + 1 is also an even number.
We write n + 1 as n + 1 = k1 + k2 , with k1 , k2 ≤ n. By the induction hypothesis
k1 and k2 are even numbers, then the sum n + 1 is also even. This completes the
proof.
Error. The inductive step is erroneous. Numbers k1 and k2 have to be greater
than or equal to 2. But this is not always true, for example, for n = 3. Also, we
can justify that in the inductive step we require two previous numbers satisfying
the statement. But it turns out that the statement is not valid for the first two
numbers, 2 and 3, since 3 is not an even number.
61
3.4 Erroneous induction proofs
Example 3.4.3. Consider the statement
In : n(n + 1)
is odd for all positive integers n.
Where is the mistake in the following proof ?
Suppose that the statement In is valid for n and we will show that it is true for
n + 1. We start from the identity
(n + 1)(n + 2) = n(n + 1) + 2(n + 1).
On the right-hand side of the identity we have, by the induction hypothesis, that
n(n + 1) is odd and, if we add to this last number the even number 2(n + 1), we
have that n(n + 1) + 2(n + 1) is also odd, and therefore the statement In+1 is also
valid.
Error. We have not verified the induction basis.
Example 3.4.4. Consider the following statement:
Rn : if there are n straight lines, not two of them parallel,
then all the lines have a common point.
Where is the error in the following proof ?
The statement R1 is true. Also R2 is true, since two non-parallel straight lines
meet in one point.
Suppose the statement valid for n − 1 straight lines, and consider now n straight
lines l1 , l2 , . . . , ln , where no two of them are parallel. By the induction hypothesis
the n − 1 straight lines l1 , . . . , ln−1 have a common point, call it P . Now, instead
of taking out the straight line ln , take out the line ln−1 . Then, by the induction
hypothesis, the n − 1 straight lines l1 , . . . , ln−2 , ln have a common point, call it Q.
But l1 and l2 have just one common point, then P = Q. Then the n straight lines
l1 , . . . , ln have P as a common point.
Error. The induction basis has to be proved for n = 3, but the statement R3 is
false. Also note that the proof of the inductive step from R2 to R3 does not hold
since l2 is eliminated in the second part.
Exercise 3.28. Consider the statement: Every function defined on a finite set is
constant.
Find the error in the following proof.
Let f : A → B be a function defined on a finite set A. We will perform the
proof by induction over n, the number of elements of the set A. If n = 1, it is clear
that f is constant. Suppose that the statement is valid for all the functions defined
on sets with n elements and let f be a function defined on a set A = {a1 , . . . , an+1 },
with n + 1 elements. Consider C = A {an+1 }, which is a set with n elements,
by the induction hypothesis, the function f restricted to C is constant. If we now
62
Chapter 3. Induction Principle
consider D = A {an } , which also has n elements, the function f restricted to
D is constant, but these two constants coincide with f (a1 ). Then the function f
is constant in all the set A and therefore the proof is complete.
Exercise 3.29. Consider the statement: For any n ≥ 2, if there are n coins, one of
which is false and lighter than the other coins, then the false coin can be identified
in at most 4 weighing of the coins in a balance with two plates.
Find the error in the following proof.
Of course, if we just have 2 coins, weighing the coins just once is enough to
identify the false coin, we just put one in each plate of the balance and the plate
that raises up has the false coin.
Suppose that the result is valid for k coins and consider k + 1 coins, where
just one of them is lighter than the others. Now leave one coin out, if in the rest
k coins we cannot identify the false coin weighing 4 times we are done; the false
coin is the one we took out, otherwise weighing 4 times we found the false coin
among the k coins.
Exercise 3.30. Consider the statement: Every non-negative integer is equal to 0.
Find the error in the following proof.
For n ≥ 0, let P(n) be the statement: “n = 0”.
Of course P(0) is true. Suppose that for k ≥ 0 the statements P(0), P(1), . . . ,
P(k) are true. The veracity of P(k + 1) follows from the fact that (k + 1) = k + 1,
because if P(k) and P(1) are true, “k = 0”, “1 = 0”, then k + 1 = 0 + 0 = 0
and therefore P(k + 1) is true. By the strong induction principle “n = 0” for all
n ≥ 0.
Chapter 4
Quadratic and Cubic Polynomials
4.1 Definition and properties
Consider an expression of the form
P (x) = a3 x3 + a2 x2 + a1 x + a0 ,
where a0 , a1 , a2 and a3 are constant numbers. We say that P (x) is a cubic polynomial or a polynomial of degree 3 in the variable x if a3 = 0; if a3 = 0 and
a2 = 0 we say that P (x) is a quadratic polynomial or a polynomial of degree 2;
in case that a3 = a2 = 0 and a1 = 0, we say that P (x) is a linear polynomial or a
polynomial of degree 1; finally, if a3 = a2 = a1 = 0 and a0 = 0, we say that P (x)
is a constant polynomial and its degree is 0. The degree of the polynomial P (x) is
denoted by deg(P ). The constants a3 , a2 , a1 and a0 are called the coefficients of
the polynomial. A polynomial with all the coefficients equal to zero is known as
the zero polynomial12 . If in a polynomial, the coefficient of the highest power of
x is 1, we say that the polynomial is a monic polynomial.
We say that a3 x3 + a2 x2 + a1 x + a0 and b3 x3 + b2 x2 + b1 x + b0 , are two equal
polynomials if ai = bi , for i = 0, 1, 2, 3.
We can evaluate the polynomials by replacing the variable x by a number t, the
value of the polynomial P (x), in x = t, is P (t).
A zero of a polynomial P (x) is a number r such that P (r) = 0. We also say that
r is a root of the polynomial or a solution of the equation P (x) = 0.
If the coefficients of a polynomial P (x) are integers, we say that P (x) is a polynomial over the integers or a polynomial with integer coefficients; similarly, if the
coefficients are rational numbers, we say that the polynomial is a polynomial over
the rationals, etc.
12 In
this book, the zero polynomial has no degree.
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5_4
63
64
Chapter 4. Quadratic and Cubic Polynomials
In many aspects, the polynomials are like the integers, they can be added, subtracted, multiplied and divided. To see how this is done consider the polynomials
P (x) = a0 + a1 x + a2 x2 + a3 x3 ,
Q(x) = b0 + b1 x + b2 x2 + b3 x3 .
We define the sum of polynomials as
(P + Q)(x) = P (x) + Q(x) = (a0 + b0 ) + (a1 + b1 )x + (a2 + b2 )x2 + (a3 + b3 )x3 ;
the subtraction as
(P − Q)(x) = P (x) − Q(x) = (a0 − b0 ) + (a1 − b1 )x + (a2 − b2 )x2 + (a3 − b3 )x3 ,
and the product of a polynomial by a constant c as
(cP )(x) = cP (x) = ca0 + ca1 x + ca2 x2 + ca3 x3 .
The product of two polynomials is defined as
(P Q)(x) = (P (x))(Q(x)) = a0 b0 + (a0 b1 + a1 b0 )x + (a0 b2 + a1 b1 + a2 b0 )x2
+ (a0 b3 + a1 b2 + a2 b1 + a3 b0 )x3 + (a1 b3 + a2 b2 + a3 b1 )x4
+ (a2 b3 + a3 b2 )x5 + a3 b3 x6 .
Finally, we define the polynomial division. Given the above two polynomials, with
the following restrictions
P (x) = a3 x3 + a2 x2 + a1 x + a0
3
2
Q(x) = b3 x + b2 x + b1 x + b0 ,
with
a3 = 0,
with some coefficient different from zero,
there are always S(x) and R(x) such that
P (x) = S(x)Q(x) + R(x),
with deg(R) < deg(Q) or R(x) = 0.
We call the polynomial S(x) the quotient and R(x) the remainder of the
division of P (x) by Q(x).
If R(x) = 0, we say that Q(x) divides (exactly) P (x) and we write Q(x)|P (x).
In the next section, we will see that a number a is a zero of a polynomial P (x) if
and only if x − a divides P (x).
A polynomial H(x) is the greatest common divisor of P (x) and Q(x) if and only if
1. H(x) divides both P (x) and Q(x),
2. If K(x) is any other polynomial which divides both P (x) and Q(x) then K(x)
also divides H(x).
It can be proved that H(x) is unique up to a multiplication by a constant number.
65
4.1 Definition and properties
4.1.1 Vieta’s formulas
(a) If a monic polynomial P (x) = x2 + px + q has roots a and b, then
x2 + px + q = (x − a)(x − b) = x2 − (a + b)x + ab,
comparing the coefficients, it follows that
p = −(a + b) and q = ab.
(4.1)
(b) If a monic polynomial P (x) = x3 + px2 + qx + r has roots a, b and c, then
x3 + px2 + qx + r = (x − a)(x − b)(x − c) = x3 − (a + b + c)x2 + (ab + bc + ca)x − abc
and comparing the coefficients we have that
p = −(a + b + c),
q = ab + bc + ca,
r = −abc.
(4.2)
Formulas (4.1) and (4.2) are known as Vieta’s formulas.
Let us see the following examples.
Example 4.1.1 (USSR, 1986). The roots of the polynomial x2 + ax + b + 1 = 0 are
natural numbers. Prove that a2 + b2 is not a prime number.
If r and s are roots of the polynomial, formula (4.1), assures us that r+s = −a
and rs = b + 1. Then, a and b are integers and a2 + b2 = (r + s)2 + (rs − 1)2 =
(r2 + 1)(s2 + 1) is the product of two numbers greater than 1.
Example 4.1.2 (Germany, 1970). Let p and q be real numbers, with p = 0, and let
3
2
a, b, c be roots of the polynomial
1 1px 1− px + qx + q.
Prove that (a + b + c) a + b + c = −1.
Since px3 − px2 + qx + q = p(x3 − x2 + pq x + pq ) = p(x − a)(x − b)(x − c), by
formula (4.2), we have that
(a + b + c)
1 1 1
+ +
a
b
c
= (a + b + c)
ab + bc + ca
abc
=
q
p
− pq
= −1.
The generalization of Vieta’s formulas for polynomials of degree greater than
3, will be presented in Chapter 8, dedicated to the theory of polynomials.
Exercise 4.1. Find all solutions of m2 − 3m + 1 = n2 + n − 1, where m and n are
positive integers.
Exercise 4.2. Let P (x) = ax2 + bx + c be a quadratic polynomial with real coefficients. Suppose that P (−1), P (0) and P (1) are integers. Prove that P (n) is an
integer number for every integer number n.
66
Chapter 4. Quadratic and Cubic Polynomials
Exercise 4.3. If a, b, c, p and q are integers, with q = 0, (p, q) = 1 and
of the equation ax2 + bx + c = 0, prove that p divides c and q divides a.
p
q
a root
Exercise 4.4. Let a, b, c be real numbers, with a and c non-zero. Let α and β be the
roots of the polynomial ax2 + bx + c and let α′ and β ′ be the roots of the polynomial
cx2 +bx+a. Prove that if α, β, α′ , β ′ are positive numbers then (α+β)(α′ +β ′ ) ≥ 4.
Exercise 4.5. Find all real numbers a such that the sum of the squares of the roots
of P (x) = x2 − (a − 2)x − a − 1 is a minimum.
Exercise 4.6. If p, q and r are solutions of the equation x3 − 7x2 + 3x + 1 = 0,
find the value of 1p + q1 + r1 .
Exercise 4.7. The solutions of the equation x3 + bx2 + cx + d = 0 are p, q and r.
Find a quadratic equation with roots p2 + q 2 + r2 and p + q + r, in terms of b, c
and d.
Exercise 4.8. For which positive real numbers m, the roots x1 and x2 of the
equation
2m − 1
m2 − 3
= 0,
x+
x2 −
2
2
satisfy the condition x1 = x2 − 21 ?
Exercise 4.9 (Kangaroo, 2003). Let P (x) be a polynomial such that P (x2 + 1) =
x4 + 4x2 . Find P (x2 − 1).
Exercise 4.10. The natural numbers a, b, c and d satisfy that a3 + b3 = c3 + d3
and a + b = c + d. Prove that two of these numbers are equal.
4.2 Roots
If we divide a degree 3 polynomial P (x) by x − a we get
P (x) = (x − a)Q(x) + r,
with
r∈R
and deg(Q) = 2.
Let x = a, then it follows that P (a) = r, therefore
P (x) = (x − a)Q(x) + P (a).
(4.3)
It follows, from equation (4.3), that
P (a) = 0
if and only if P (x) = (x − a)Q(x),
for some polynomial Q(x). This result is known as the factor theorem.
(4.4)
67
4.2 Roots
If a1 and a2 are two different zeros of P (x), then, by equation (4.4), it follows
that P (x) = (x − a1 )Q(x). Since, P (a2 ) = (a2 − a1 )Q(a2 ) = 0 and a2 = a1 , then
Q(a2 ) = 0, hence Q(x) = (x − a2 )Q1 (x), for some polynomial Q1 (x). Then,
P (x) = (x − a1 )(x − a2 )Q1 (x)
with deg(Q1 ) = 1.
If deg(P ) = 3 and P (ai ) = 0, for a1 , a2 , a3 , then
P (x) = c(x − a1 )(x − a2 )(x − a3 ) with c ∈ R.
If there exists m ∈ N and a polynomial Q(x) such that
P (x) = (x − a)m Q(x)
with Q(a) = 0,
(4.5)
we say that the multiplicity of the root a is m.
Example 4.2.1. If P (x) = ax2 + bx + c is a quadratic polynomial, then its roots are
√
√
−b − b2 − 4ac
−b + b2 − 4ac
and
.
(4.6)
2a
2a
Since a = 0, we can rewrite the polynomial P (x) as
b
c
P (x) = ax2 + bx + c = a x2 + x +
a
a
2
2
b
2bx
b
4c
+ 2− 2+
= a x2 +
2a
4a
4a
4a
.
2
1 2
b
− 2 b − 4ac
=a x+
2a
4a
⎡
2 ⎤
√
2
2 − 4ac
b
b
⎦
= a⎣ x +
−
2a
2a
..
√
√
b
b2 − 4ac
b2 − 4ac
b
x+
=a x+
−
+
2a
2a
2a
2a
.
.
√
√
−b + b2 − 4ac
−b − b2 − 4ac
=a x−
x−
.
2a
2a
Then,
−b +
√
b2 − 4ac
2a
and
−b −
√
b2 − 4ac
2a
are its roots.
The number ∆ = b2 − 4ac is called the discriminant of the quadratic polynomial P (x) = ax2 + bx + c.
68
Chapter 4. Quadratic and Cubic Polynomials
Observation 4.2.2. If the discriminant, b2 −4ac of the quadratic polynomial P (x) =
ax2 + bx + c is zero, the polynomial P (x) can be written as a constant multiplied
by the square of a linear polynomial. In this case the polynomial P (x) has only
b
.
one real root, which is equal to − 2a
In the case where the discriminant is greater than zero, that is, b2 − 4ac > 0
then the polynomial P (x) has two different real roots.
Finally, when the discriminant is negative, there are two distinct complex
roots. This case will be analyzed in Chapter 5.
Summarizing, if r and s are the roots of P (x), then the discriminant is zero
if and only if r = s. In case the discriminant is different from zero, then r =
s,
and P (x) = a(x − r)(x − s).
Now, we shall see what is the geometric meaning of the discriminant of a
quadratic polynomial. Remember that P (x) = ax2 + bx + c can be written as
P (x) = a
-
b
x+
2a
2
.
1 2
b
− 2 b − 4ac = a x +
4a
2a
2
−
∆
.
4a
To construct the graph of the previous equation, that is, to locate the set
of pairs of points (x, y) = (x, P (x)) in the Cartesian plane, we let y = P (x) and
obtain the equation
y+
∆
b
=a x+
4a
2a
2
,
(4.7)
b
∆
which represents the parabola with vertex at the point − 2a
, − 4a
, where the sign
of the coefficient a determines if the parabola opens up (a > 0), or down (a < 0).
In fact, the equation (4.7) tells us much more about the quadratic polynomial
P (x). Suppose that a > 0, that is, the parabola opens up. The second coordinate
∆
, is positive if and only if −∆ > 0, that is, if the discriminant of
of the vertex, − 4a
P (x) is negative, which means that the graph of the parabola does not intersect
the X-axis. Then P (x) has no real roots.
∆
is negative, the graph of the parabola intersects the X-axis in two
If − 4a
points x1 and x2 , which are the roots of P (x). Observe that in this case, we
have that ∆ is positive, which agrees with the fact that P (x) has two real roots.
b
,
Moreover, the polynomial P (x) reaches the minimum value at the point x = − 2a
∆
and − 4a is the minimum value of P (x).
Making a similar analysis if a < 0, we can conclude that when the parabola
opens down it will intersect the X-axis if and only if ∆ ≥ 0. Here, the polynomial
b
where the maximum value
P (x) reaches its maximum value at the point x = − 2a
∆
of P (x) is − 4a .
69
4.2 Roots
∆>0
a>0
x1
−b
∆
2a , − 4a
x2
−b
∆
2a , − 4a
∆<0
a<0
When the graph of the parabola is tangent to the X-axis, we are in the case
in which the discriminant is zero, that is, when both roots are equal.
As an application for the quadratic polynomial theory we can prove the following
inequality.
Example 4.2.3 (Cauchy–Schwarz inequality). If a1 , . . . , an and b1 , . . . , bn are real
numbers, it follows that
n
n
1/2
1/2 n
#
#
#
2
2
ai b i ≤
.
bi
ai
i=1
i=1
i=1
*n
The expression P (x) = i=1 (ai x + bi )2 is a quadratic polynomial in x. Since
P (x) ≥ 0, we have that P (x) cannot have two different real roots, therefore the
discriminant cannot be positive. Now, in order to calculate the discriminant of
this polynomial, we expand each term of the sum (ai x + bi )2 = a2i x2 + 2ai bi x + b2i ,
from where the polynomial takes the form
n
n
n
#
#
#
2
2
2
P (x) =
bi ,
ai b i x +
ai x + 2
i=1
i=1
i=1
therefore, the discriminant is
4
n
#
i=1
ai b i
2
−4
n
#
i=1
a2i
n
#
i=1
b2i
≤ 0.
Rewriting the above expression and taking the square root, we obtain the desired
result, that is,
n
n
1/2
1/2 n
#
#
#
2
2
ai b i ≤
bi
ai
.
i=1
i=1
i=1
70
Chapter 4. Quadratic and Cubic Polynomials
Note that the equality holds if the discriminant is zero, that is, when the polynoi
mial has just one real root. Observe that this holds if x = −b
ai for all i = 1, 2, . . . , n,
−bi
and this means ai is constant.
It is also possible to prove the inequality between the geometric mean and the
arithmetic mean of two numbers, by analyzing the discriminant of a certain quadratic polynomial.
√
Example 4.2.4. For a, b ≥ 0, it follows that a+b
ab and the equality holds if
2 ≥
and only if a = b.
√
√
Consider the quadratic polynomial P (x) = (x − a)(x − b), which has
two real roots, so$ its discriminant
positive
or zero. But the discriminant of
$is
√ %
√ %
√
√
√
2
P (x) = (x − a) x − b = x −
a + b x + ab is
$√
√
√ %2
√
a + b − 4 ab = a + b − 2 ab.
√
Since the discriminant
is
positive
or
zero,
then
a
+
b
≥
2
ab. Also, the equality
√
√
holds if and only if a = b and then a = b.
Example 4.2.5. The following construction dates from the Greek period and it is the
procedure to find geometrically what in modern terms are the roots of a quadratic
polynomial x2 + bx − c2 , with b and c positive numbers.
Construct first an isosceles triangle OP Q with base P Q of length b, and the
altitude from O of length c. We draw a circumference through the vertices P and
Q of the triangle and with center O.
O
B
A
c
C
P
b
Q
D
Let AB be the diameter of the circumference parallel to P Q and construct
the rectangle ABCD. The positive root of x2 + bx − c2 is equal to the length of the
segment CP and the other root (the negative root) is equal to the negative length
of the segment CQ.
Let us see how to prove this statement. By Vieta’s formulas (4.1) it follows
that if α and β are the roots of the polynomial, α + β = −b and αβ = −c2 ,
then we necessarily have a negative root. Let us consider the right triangles BCP
71
4.2 Roots
and QCB. Since ∠P BC = ∠BQC, these triangles are similar. Then we have that
CP · CQ = BC 2 = c2 , therefore
CP (−CQ) = −c2 .
On the other hand, since CQ = CP + b, we have that CP − CQ = −b. Then, CP
and −CQ fulfill Vieta’s relations, therefore these last numbers are the roots of the
equation.
Example 4.2.6. Factorize a3 + b3 + c3 − 3abc.
Let us consider the polynomial
P (x) = (x − a)(x − b)(x − c) = x3 − (a + b + c)x2 + (ab + bc + ca)x − abc,
with roots a, b and c. That is,
a3 − (a + b + c)a2 + (ab + bc + ca)a − abc = 0,
b3 − (a + b + c)b2 + (ab + bc + ca)b − abc = 0,
c3 − (a + b + c)c2 + (ab + bc + ca)c − abc = 0.
Adding these three equalities and factoring a + b + c, we get
a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca).
(4.8)
Remember that the expression a2 + b2 + c2 − ab − bc − ca can be written as
"
1!
(a − b)2 + (b − c)2 + (c − a)2 .
2
From this we get the following factorization,
a3 + b3 + c3 − 3abc =
"
!
1
(a + b + c) (a − b)2 + (b − c)2 + (c − a)2 .
2
(4.9)
Example 4.2.7 (Czechoslovakia, 1969). Let a, b and c be real numbers such that
a + b + c > 0, ab + bc + ca > 0 and abc > 0. Prove that a, b, c are positive.
Let us consider the cubic and monic polynomial with roots a, b and c, that is,
P (x) = (x − a)(x − b)(x − c) = x3 − (a + b + c)x2 + (ab + bc + ca)x − abc.
For x ≤ 0, we have that P (x) < 0, then we can guarantee that the roots are
positive.
Exercise 4.11. For what values λ the polynomial λx2 + 2x + 1 −
equal roots?
1
λ
does have two
72
Chapter 4. Quadratic and Cubic Polynomials
Exercise 4.12. Let a, b and c be positive real numbers. Is it possible, for each of the
following polynomials P (x) = ax2 +bx+c, Q(x) = bx2 +cx+a, R(x) = cx2 +ax+b,
to have both real roots?
Exercise 4.13. For what integer values of k are the solutions of the equation
kx2 − (1 − 2k)x + k − 2 = 0
rational numbers?
Exercise 4.14 (Czech-Slovakia, 2006). Find all pairs of integers (a, b) such that
a + b is a root of the polynomial x2 + ax + b.
Exercise 4.15. Find all integer values of x for which the polynomial x2 − 5x − 1
is a perfect square.
Exercise 4.16. Solve, using geometry, the equation x2 + bx − c2 , with b and c
positive numbers, using the following construction due to R. Descartes. Draw a
circumference with center O and radius 2b . Draw QR, a tangent to the circumference through Q, with QR = c. Let S and T be the points where the straight line
through R and O cuts the circumference.
R
S
c
O
b
2
Q
T
The quadratic polynomial has one positive root and one negative root. In the
figure, the length of the segment RS is equal to the positive root and the negative
root is equal to the negative length of the segment RT .
Exercise 4.17. Let P (x) = ax2 +bx+c be a quadratic polynomial such that P (x) = x
does not have real solutions. Prove that P (P (x)) = x has no real solutions either.
Exercise 4.18 (Poland, 2007). Let P (x) be a polynomial with integer coefficients.
Prove that if P (x) and P (P (P (x))) have a common root, then they have a common
integer root.
73
4.2 Roots
Exercise 4.19 (Russia, 2009). Let a, b and c be non-zero real numbers such that
ax2 + bx + c > cx, for all real numbers x. Prove that cx2 − bx + a > cx − b, for all
real numbers x.
Exercise 4.20 (Russia, 2009). Two different real numbers a and b are such that
the equation (x2 + 20ax + 10b)(x2 + 20bx + 10a) = 0 has no real solutions. Prove
that 20(b − a) cannot be an integer number.
Exercise 4.21 (Russia, 2011). Let P (x) be a monic quadratic polynomial such that
P (x) and P (P (P (x))) have a common root. Prove that P (0)P (1) = 0.
Exercise 4.22. Let a, b, c, d, e and f be positive integers such that they satisfy the
relation ab + ac + bc = de + df + ef , and let N = a + b + c + d + e + f . Prove that
if N divides abc + def then N is a composite number.
Exercise 4.23. Let P (x) and Q(x) be two quadratic polynomials with integer coefficients. If both polynomials have an irrational number as a common zero, prove
that one of them is a multiple of the other.
Exercise 4.24. Determine if there exist polynomials x2 − b1 x + c1 = 0 and x2 −
b2 x + c2 = 0, with b1 , c1 , b2 and c2 different, such that the four roots are b1 , c1 ,
b2 and c2 .
Exercise 4.25. Let a, b and c be real numbers. Prove that at least one of the
following equations has a real solution:
x2 + (a − b)x + (b − c) = 0,
x2 + (b − c)x + (c − a) = 0,
x2 + (c − a)x + (a − b) = 0.
Exercise 4.26. Let a, b and c be real numbers such that a + b + c = 0. Prove that
a5 + b 5 + c5
=
5
a2 + b 2 + c2
2
a3 + b 3 + c3
3
.
Exercise 4.27. Let a, b and c be real numbers such that a+b+c = 3, a2 +b2 +c2 = 5,
a3 + b3 + c3 = 7. Find a4 + b4 + c4 .
Exercise 4.28 (OMCC, 2001). Let a, b and c be real numbers such that the equation
ax2 +bx+c = 0 has two different real solutions p1 , p2 and the equation cx2 +bx+a =
0 has two different real solutions q1 , q2 . Also the numbers p1 , q1 , p2 , q2 , in this
order, form an arithmetic progression. Prove that a + c = 0.
74
Chapter 4. Quadratic and Cubic Polynomials
Exercise 4.29. Let a, b and c be real numbers different from zero, with a+ b + c = 0
and a3 + b3 + c3 = a5 + b5 + c5 . Prove that a2 + b2 + c2 = 65 .
Exercise 4.30 (Russia, 2010). Let P (x) be a cubic polynomial with integer coefficients such that there exist different integers a, b, and c such that P (a) = P (b) =
P (c) = 2. Prove that no integer number d satisfying P (d) = 3 exists.
Chapter 5
Complex Numbers
5.1 Complex numbers and their properties
The roots of a quadratic equation are not always real numbers. For instance, the
roots of the equation x2 + 2x + 10 = 0, which can be calculated using directly
equation (4.6) of Section 4.2, are
√
√
−2 + −36
−2 − −36
and
.
2
2
√
For this reason it is necessary to consider “imaginary” numbers, like −36.
A complex number z is an expression of the form x + iy, where x and y are real
numbers, and i2 = −1. At this moment we will not worry about the meaning of i,
for the time being we will be interested only in the fact that its square is −1. The
real part of z, which will be denoted by Re z, is the number x, and the imaginary
part of z, denoted by Im z, is the number y.
The set C of all complex numbers x + iy can be identified with the set of points
(x, y) in the Cartesian plane, and when this is done, due to this representation it
is called the complex plane. The X-axis is called the real axis and the Y -axis is
known as the imaginary axis.
In order to work with complex numbers, we need the following three definitions:
the complex conjugate of z, denoted by z̄, is the complex number x − iy; the
module or norm of z, denoted by |z|, is the real number x2 + y 2 , which is the
distance from the origin to the point (x, y) representing z. Finally, the argument
of z = 0 is the angle between the positive real axis and the line through 0 and
z, taken in the counterclockwise direction. The argument of z is denoted by arg z
and generally it is assigned a value between 0 and 2π.
There is another form to write a complex number z, which is known as the polar
form of the complex number z. Let r be the norm of z, r = |z|, and θ = arg z,
then z = r(cos θ + i sin θ).
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5_5
75
76
Chapter 5. Complex Numbers
Y -axis
|z |
z = x + iy
r=
y
θ
X-axis
x
z̄ = x − iy
The set of complex numbers is similar to the set of real numbers, in the sense
that there are two operations that can be applied to its elements, the sum and the
product of complex numbers.
Let z = x + iy and w = u + iv be two complex numbers. Then the sum of these
complex numbers is given by
z + w = (x + iy) + (u + iv) = (x + u) + i(y + v),
and the product by
z · w = (x + iy)(u + iv) = xu + xiv + iyu + i2 yv
= (xu − yv) + i(xv + yu).
The set of real numbers can be seen as a subset of the complex numbers, if
we identify each real number x with the complex number (x, 0).
Observe that to number i corresponds the number (0, 1) and that
(x, y) = (x, 0) + (0, y) = (x, 0) + (0, 1) · (y, 0),
and this is also represented as x + iy, that is, to (x, y) corresponds the complex
number x + iy.
The operations of sum and product of complex numbers satisfy the same
properties that these operations satisfy in the set of real numbers, as being commutative, associative and share also the existence of neutral elements for both
operations, these being 0 = (0, 0) and 1 = (1, 0), respectively. The sum of complex numbers is exactly the same operation as the sum of vectors in the Cartesian
plane; the operation that can be a novelty is the product of complex numbers. Let
us see how to find the inverse of a complex number. For that purpose we need an
77
5.1 Complex numbers and their properties
important relation that exists between the norm of a complex number z and its
conjugate, that is
(5.1)
z z̄ = |z|2 .
In order to see that, note that if z = x + iy, then z z̄ = (x + iy)(x − iy) = x2 + y 2 =
|z|2 . From identity (5.1), if z = 0, tells us that its multiplicative inverse, z1 , is equal
to |z|z̄ 2 .
Exercise 5.1. Let z, w be two complex numbers. Prove that:
(i) z + w = z̄ + w̄, zw = z̄ w̄,
z̄ = z.
1
(ii) |z̄| = |z|, |zw| = |z| |w|, 1z = |z|
, if z = 0.
(iii)
(iv)
(v)
(vi)
Re z = 12 (z + z̄) ≤ |z|.
1
(z − z̄) ≤ |z|.
Im z = 2i
|z + w| ≤ |z| + |w|.
|z + w|2 + |z − w|2 = 2(|z|2 + |w|2 ).
Exercise 5.2. Find the complex numbers z such that Im z + z1 = 0.
Exercise 5.3. If z and w are complex numbers with |z + w| = |z − w| and w = 0,
iz
then
is a real number.
w
Exercise 5.4. If z and w are complex numbers, prove that:
(i) |1 − z̄w|2 − |z − w|2 = (1 + |zw|)2 − (|z| + |w|)2 .
2
2
2
2
(ii) |1 + z̄w| − |z + w| = (1 − |z| )(1 − |w| ).
2
2
Exercise 5.5. If z and w are complex numbers such that (1 + |w| )z = (1 + |z| )w,
prove that z = w or z̄w = 1.
Exercise 5.6. Let z1 , z2 , z3 be complex numbers such that
z1 + z2 + z3 = 0
Prove that z12 + z22 + z32 = 0.
and
|z1 | = |z2 | = |z3 | = 1.
Exercise 5.7. Prove that if z1 , z2 are complex numbers with |z1 | = |z2 | = 1 and
z1 +z2
z1 z2 = −1, then 1+z
is a real number.
1 z2
Exercise 5.8. Let a, b, c and d be complex numbers with the same norm, and such
that a + b + c = d. Prove that d is equal to a, b or c.
Now, we present the geometric meaning of the complex product. Let us see two
examples that will help us to understand such meaning. First, let us consider the
transformation z → iz, that is, x + iy → i(x + iy) = ix − y, which in vectorial
form means that the vector (x, y) becomes the vector (−y, x), and note that both
vectors are perpendicular. Then, the transformation z → iz corresponds to the
78
Chapter 5. Complex Numbers
rotation of the complex plane in the counterclockwise direction, around zero, with
angle π2 .
Y
iz
z
X
i2 z = −z
If we apply the previous transformation twice, we obtain z → iz → i(iz) = i2 z =
−z, which is a rotation with angle π.
The previous examples show that complex multiplication implicitly carries a rotation of the Cartesian plane. If instead of taking the product of a complex number
z with i, we take the product with another complex number w, certain rotation
appears in a natural way. We will see this now.
Let z = x+iy = r(cos θ +i sin θ) and w = u+iv = t(cos φ+i sin φ) be two complex
numbers written in polar form. Taking its product, we obtain
z · w = r(cos θ + i sin θ) · t(cos φ + i sin φ)
= r t(cos θ cos φ + i cos θ sin φ + i sin θ cos φ + i2 sin θ sin φ)
= r t (cos θ cos φ − sin θ sin φ + i(cos θ sin φ + sin θ cos φ))
(5.2)
= r t (cos(θ + φ) + i(sin θ + φ)) ,
where we used the sine and cosine formula for the sum of two angles.
Thus, arg(z · w) = arg z + arg w, modulo 2π.
By identity (5.2), we conclude that geometrically the product of two complex
numbers, z and w, is the complex number whose norm is the product of the norms
of z and w, and its argument is the sum of the arguments of the same two numbers.
Y
zw
w
z
θ
θ
1
X
79
5.2 Quadratic polynomials, complex coefficients
Using formula (5.2) repeatedly, for z = cos θ + i sin θ, we obtain the so-called de
Moivre’s formula, where for every integer n we have
(cos θ + i sin θ)n = cos nθ + i sin nθ.
(5.3)
Exercise 5.9. Prove that, for complex numbers a, b and c, the following are equivalent:
(i) The points a, b and c are collinear.
c−a
∈ R.
(ii)
b−a
(iii) cb̄ − cā − ab̄ ∈ R.
1 a ā
(iv) 1 b b̄ = 0.
1 c c̄
Conclude that the equation of the line through b and c is Im( z−c
z−b ) = 0.
Exercise 5.10. Find the complex numbers z such that z, i, iz are collinear.
Exercise 5.11. Let z, w be two vertices of a square, find the other two vertices in
terms of z and w.
Exercise 5.12. Prove by induction de Moivre’s formula (5.3).
Exercise 5.13. Prove that if z +
integer n ≥ 1.
1
z
= 2 cos θ, then z n +
1
zn
= 2 cos nθ, for every
5.2 Quadratic polynomials with complex coefficients
Now, let us see that it is possible to obtain explicitly the roots of every polynomial
of degree 2 with complex coefficients. In order to do that, first we need to be able
to solve the following type of equations
(x + iy)2 = a + ib,
(5.4)
where a and b are real numbers. That is, we need to find the values of x and y,
which means to find the square root of a + ib.
First, observe that since (x + iy)2 = x2 − y 2 + i2xy, equation (5.4) is equivalent to
a = x2 − y 2
and b = 2xy,
(5.5)
80
Chapter 5. Complex Numbers
the second equation implies that the sign of b determines if the signs of x and y
are equal or different. Now, taking the norm on both sides of equation (5.4), it
follows that
(5.6)
x2 + y 2 = a2 + b2 .
Adding this last equation to the first equality of (5.5), we obtain 2x2 = a +
√
a2 + b2 , that is,
x2 =
1$
a+
2
a2 + b 2
%
or x = ±
1$
a+
2
%
a2 + b 2 .
√
Similarly, we obtain y = ± 12 −a + a2 + b2 . Note that x, y are well defined
√
√
since a + a2 + b2 ≥ a + |a| ≥ 0 and −a + a2 + b2 ≥ −a + |a| ≥ 0.
Once we know how to calculate the square roots of complex numbers, we can calculate the zeros of every quadratic polynomial. In fact, if we consider the quadratic
polynomial P (z) = az 2 + bz + c, with a, b, c ∈ C, then, by using formula (4.6)
we can always find both zeros of the polynomial. For instance, if the discriminant
b2 − 4ac is a complex number, we can obtain its square root and then calculate
the two roots, which are now complex numbers too.
As a consequence, every quadratic polynomial with complex coefficients has at
least one complex root, and therefore can be written as the product of two linear
factors with complex coefficients.
Observe also that if P (z) is a polynomial with real coefficients, then for every complex number w we have that P (w̄) = P (w), since the conjugate of each coefficient
is the same number, just because it is a real number. Thus, if w is a zero of P (z),
then w̄ is also a zero. Hence, for a polynomial with real coefficients, the complex
roots, if they exist, appear in conjugate pairs.
Example 5.2.1. Solve the equation z 8 + 4z 6 − 10z 4 + 4z 2 + 1 = 0.
Dividing the expression by z 4 , we obtain that
z4 +
1
z4
+ 4 z2 +
1
z2
− 10 =
z+
1
z
4
− 6 − 10 = 0.
(5.7)
Making the change of variable u = z + 1z we get equation u4 = 16, which has
solutions u1 = 2, u2 = −2, u3 = 2i, u4 = −2i.
From equation u = z + 1z we have that z = u2 ± u2 /4 − 1. By substituting the
√
four values of u√we obtain the eight solutions: z1,2 = 1, z3,4 = −1, z5,6 = i(1 ± 2),
z7,8 = −i(1 ± 2).
Example 5.2.2 (Romania, 1999). Let p and q be complex numbers with q = 0.
Prove that if the roots of the quadratic polynomial x2 + px + q 2 = 0 have the same
norm, then pq is a real number.
81
5.3 The fundamental theorem of algebra
Let x1 , x2 be the roots of the given equation and let r = |x1 | = |x2 |. Then
(x1 + x2 )2
x1
x2
x1 x̄2
x2 x̄1
2
p2
=
=
+
+ 2 = 2 + 2 + 2 = 2 + 2 Re(x1 x̄2 )
q2
x1 x2
x2
x1
r
r
r
is a real number. Moreover, Re(x1 x̄2 ) ≥ −|x1 x̄2 | = −r2 and then
Thus, pq is a real number.
p2
q2
≥ 0.
Exercise 5.14. Find all the complex numbers z such that |z| = 1 and |z 2 + z̄ 2 | = 1.
Exercise 5.15. Let a, b, c be complex numbers with |a| = |b| = |c| = 0.
(i) Prove that if a root of the equation az 2 + bz + c = 0 has norm 1, then
b2 − ac = 0.
(ii) If each of the equations az 2 + bz + c = 0 and bz 2 + cz + a = 0 have a root of
norm 1, then |a − b| = |b − c| = |c − a|.
Exercise 5.16 (Romania, 2003). If the complex numbers z1 , z2 , z3 , z4 , z5 all have
*5
*5
2
norm 1 and satisfy
i=1 zi =
i=1 zi = 0, prove that these numbers are the
vertices of a regular pentagon.
Exercise 5.17 (Romania, 2007). Let a, b, c be complex numbers of norm 1. Prove
that there exist numbers α, β, γ ∈ {−1, 1}, such that |αa + βb + γc| ≤ 1.
Exercise 5.18 (Romania, 2008). Let a, b, c be complex numbers
that satisfy a |bc|+
√
b |ca| + c |ab| = 0. Prove that |(a − b)(b − c)(c − a)| ≥ 3 3 |abc|.
Exercise 5.19 (Romania, 2009). Find the complex numbers a, b, c, all with the
same norm, such that a + b + c = abc = 1.
5.3 The fundamental theorem of algebra
One of the goals of polynomial theory is to find the roots or the factors in which
a polynomial can be decomposed. In this spirit we have the result known as the
fundamental theorem of algebra.
Theorem 5.3.1 (The fundamental theorem of algebra). Every polynomial P (z) =
an z n + an−1 z n−1 + · · · + a1 z + a0 , where n ≥ 1, ai ∈ C and an = 0, has at least
one root in C.
For the proof of this theorem, see Section 5.5.
Let us first remember the factor theorem.
82
Chapter 5. Complex Numbers
Theorem 5.3.2 (Factor theorem). If a is a zero of a polynomial P (z), then z − a
is a factor of P (z).
This theorem has the following interpretation: to know the zeros of the polynomial
is to know the polynomial. This is made precise in the following result.
Corollary 5.3.3. Every polynomial P (z) of degree n with coefficients in C, can be
written in the form
P (z) = c(z − z1 )(z − z2 ) · · · (z − zn ),
zi ∈ C, c ∈ C.
That is, the polynomial has exactly n complex roots. The numbers z1 , z2 , . . ., zn
are not necessarily distinct.
If it is difficult to find the roots of the polynomial, then it is a good idea to find
another polynomial for which it could be easier to find the roots. Let us see an
example.
Example 5.3.4 (USA, 1975). Let P (x) be a polynomial of degree n such that P (k) =
k/(k + 1), for k = 0, 1, 2, . . . , n. Find P (n + 1).
The condition P (k) = k/(k + 1) does not say anything about the roots of P (x).
Then, we can consider the polynomial of degree n + 1,
Q(x) = (x + 1)P (x) − x.
Clearly, the roots of Q(x) are 0, 1, 2, . . . , n, therefore we can write
(x + 1)P (x) − x = Cx(x − 1)(x − 2) · · · (x − n),
where C is a constant that is going to be determined. Evaluating in x = −1, we
n+1
obtain that 1 = C(−1)(−2)(−3) · · · (−n)(−(n + 1)), from where C = (−1)
(n+1)! .
Finally, if x = n + 1, we get (n + 2)P (n + 1) − n − 1 = (−1)n+1 , hence P (n + 1) =
n+1+(−1)n+1
.
n+2
5.4 Roots of unity
Using de Moivre’s formula (5.3) it follows immediately that the polynomial z n −1 =
2π
0 has roots 1, w, w2 , . . . , wn−1 , where w = cos 2π
n +i sin n . These roots are known
as the nth roots of unity and in the complex plane they can be identified as the
vertices of a regular n-sided polygon inscribed in the unit circle with center at the
origin13 . By the factor Theorem 5.3.2, we have the decomposition
z n − 1 = (z − 1)(z − w)(z − w2 ) · · · (z − wn−1 ).
13 More
generally, the equation z n − a = 0, for any complex number
a = 0, has n different
complex solutions called the nth roots of a. These solutions are n |a|w j , for j = 0, 1, . . . , n − 1
+ i sin 2π
.
and w = cos 2π
n
n
83
5.4 Roots of unity
For instance,
for the case n = 3, the roots
of z 3 − 1 = (z − 1)(z 2 + z + 1) = 0 are 1,
√
√
−1−i 3
−1+i 3
1
2
and w = w = w̄ =
, and they are known as the cubic roots
w=
2
2
of unity. Note that w satisfies w3 = 1 and 1 + w + w2 = 0.
2π
We have seen that w = cos 2π
n +i sin n generates all the nth roots of unity, that is,
2
n−1
n
, w = 1}. We say that one element u ∈ Un is a primitive
Un = {w, w , . . . , w
root of unity if um = 1 for all positive integers m < n. Now, we can state the
following result.
Theorem 5.4.1.
(a) If n divides q, then any root of z n − 1 = 0 is a root of z q − 1 = 0.
(b) The common roots of z m − 1 = 0 and z n − 1 = 0 are the roots of z d − 1 = 0,
where d is the greatest common divisor of m and n, which is denoted by
d = (m, n).
(c) The primitive roots of z n − 1 = 0 are wk , where 0 ≤ k ≤ n and (k, n) = 1.
Proof. (a) If q = pn and w is a root of z n − 1, it follows that wq − 1 = wpn − 1 =
(wn )p − 1 = (1)p − 1 = 0, then w is root of z q − 1.
(b) If w is a root of z m − 1 and z n − 1, and d = (m, n), we have that
d = am+bn for some integers a and b. Hence, wd −1 = wam+bn −1 = wam wbn −1 =
(wm )a (wn )b − 1 = 1a 1b − 1 = 0, therefore, w is a root of z d − 1.
Conversely, since d divides m and n, by property (a), if w is a root of z d − 1 then
it is root of z m − 1 and z n − 1.
(c) To prove this part, let us see the following lemma.
2π
Lemma 5.4.2. Let w = cos 2π
n + i sin n . The smallest positive integer m such that
n
k m
(w ) = 1 is m = (k,n) , where (k, n) is the greatest common divisor of k and n.
As a consequence of this lemma part (c) follows and wk is a primitive root if
and only if (k, n) = 1.
Proof of the lemma. First, note that ws = 1 if and only if s = an. Let m be the
smallest positive integer such that (wk )m = 1, then km = an. If d = (k, n), then
k1 dm
k1 m
k = k1 d and n = n1 d, with (k1 , n1 ) = 1, hence a = km
n = n1 d = n1 is an
integer. Thus n1 |m, and since wkn1 = wk1 dn1 = wk1 n = 1, it follows that m = n1 .
n
Finally, we have that n = n1 d = md, therefore m = nd = (k,n)
.
Example 5.4.3. Prove the identity
sin
2π
(n − 1)π
n
π
· sin
· · · · · sin
= n−1 .
n
n
n
2
Consider the polynomial P (z) = (1−z)n −1, which can be written as wn −1, where
2πk
w = 1 − z. The roots of wn = 1 are the nth roots of unity, wk = cos 2πk
n + i sin n ,
for k = 0, 1, . . . , n − 1. Then, the roots of P (z) are zk = 1 − wk .
84
Chapter 5. Complex Numbers
Looking at the polynomial P (z), we observe that it can be written as P (z) =
z(−n+Q(z)), where Q(z) is a polynomial of degree n−1. Hence, if we let P (z) = 0
we see that the roots should satisfy, by Vieta’s formula (8.4) Section 8.3, that
3n−1
3n−1
(−1)n n = k=1 zk , and from this n = k=1 |zk |.
Now, we need to calculate |zi |.
|zk | = |1 − wk | =
1 − cos
2
2πk
n
=
1 − 2 cos
2πk
n
+ cos2
=
2 − 2 cos
2πk
n
=
2πk
n
4 sin2
+ sin
2πk
n
+ sin2
2πk
n
πk
n
= 2 sin
2
πk
,
n
where we used that cos2 x + sin2 x = 1 and 1 − cos(2x) = 2 sin2 x. The identity
now follows.
Example 5.4.4 (AMC, 2002). Find the number of ordered pairs of real numbers
(a, b) such that (a + ib)2002 = a − ib.
√
Let z = a + ib, z̄ = a − ib, and |z| = a2 + b2 . The relation given above is
equivalent to z 2002 = z̄. Note that |z|2002 = |z 2002 | = |z̄| = |z|, thus
|z|(|z|2001 − 1) = 0.
Hence, |z| = 0 and therefore (a, b) = (0, 0) or |z| = 1. In the latter case we have
that z 2002 = z̄, which is equivalent to z 2003 = z̄ · z = |z|2 = 1. Since the equation
z 2003 = 1 has 2003 different solutions, then there are 1 + 2003 = 2004 ordered
pairs that satisfy the equation.
Exercise 5.20. Solve the equation
x4 + x3 + x2 + x + 1 = 0.
Exercise 5.21. Find the solutions of the equation
x6 + 2x5 + 2x4 + 2x3 + 2x2 + 2x + 1 = 0.
Exercise 5.22 (Romania, 2007). Let n be a positive integer. Prove that there exists
a complex number with norm 1 that is a solution of the equation z n + z + 1 = 0 if
and only if n = 3m + 2, for some positive integer m.
Exercise 5.23. If w = 1 is an nth root of unity, prove that:
(i) 1 + w + w2 + · · · + wn−1 = 0.
(ii) 1 + 2w + 3w2 + · · · + nwn−1 =
n
w−1 .
5.5 Proof of the fundamental theorem of algebra
85
Exercise 5.24. If w = 1 is a nth root of unity:
(i) Prove that (1 − w)(1 − w2 ) . . . (1 − wn−1 ) = n.
1
1
1
+ 1−w
(ii) Find the value of 1−w
2 + · · · + 1−w n−1 .
Exercise 5.25.
(i) Prove that if ω = 1 is a cubic root of unity (that is, ω 3 = 1), then for a, b,
c ∈ C, it follows that
a2 + b2 + c2 − ab − bc − ca = (a + bω + cω 2 )(a + bω 2 + cω).
(ii) Use (i) and equation (4.8) to obtain the following identity:
a3 + b3 + c3 − 3abc = (a + b + c)(a + bω + cω 2 )(a + bω 2 + cω).
(5.8)
Exercise 5.26. Two regular polygons are inscribed in the same circle. The first
polygon has 1982 sides and the second one has 2973 sides. If the polygons have
some vertices in common, how many vertices in common do they have?
Exercise 5.27. Find the positive integers n for which x2 +x+1 divides x2n +xn +1.
Exercise 5.28. Let S be the set of integers x that can be written in the form
x = a3 + b3 + c3 − 3abc, for some integers a, b, c. Prove that if x, y ∈ S , then
xy ∈ S .
Exercise 5.29 (USA, 1976). If P (x), Q(x), R(x), S(x) are polynomials such that
P (x5 ) + xQ(x5 ) + x2 R(x5 ) = (x4 + x3 + x2 + x + 1)S(x),
prove that x − 1 is a factor of P (x).
Exercise 5.30. Find all the polynomials P (z) of degree at most 2 with coefficients
in C, that satisfy P (z)P (−z) = P (z 2 ).
5.5 Proof of the fundamental theorem of algebra ⋆
As we have seen, the fundamental theorem of algebra states that every nonconstant polynomial with complex coefficients has at least one complex root. This
theorem includes polynomials with real coefficients.
The fundamental theorem of algebra can also be stated saying that every polynomial of degree n, with complex coefficients, has exactly n roots, counting the
multiple roots as many times as they appear. The equivalence of both theorems
can be proved using repeatedly the factor theorem.
86
Chapter 5. Complex Numbers
In spite of the name of the theorem, there are no purely algebraic elementary
proofs, since these proofs use the fact that the real numbers are complete and
this is not an algebraic concept. This theorem was considered fundamental for
algebra when the study of this discipline was concentrated in finding the roots of
the polynomial equations with real or complex coefficients.
Some proofs of the theorem only show that every non-constant polynomial, with
real coefficients, has a complex root. This is sufficient to establish the theorem
in the general case, since given a non-constant polynomial P (z) with complex
coefficients, the polynomial Q(z) = P (z)P (z̄) has real coefficients and if z is a
root of Q(z), then z or its conjugate z̄ is also a root of P (z).
The proof that we present here uses a result known as the growth lemma.
Lemma 5.5.1 (Growth lemma). Given a polynomial P (z) = an z n + an−1 z n−1 +
· · · + a1 z + a0 of degree n ≥ 1, with complex coefficients, there is a real number
R > 0 such that if |z| ≥ R then
1
|an ||z n | ≤ |P (z)| ≤ 2|an ||z n |.
2
*n−1
k
Proof. Let r(z) =
k=0 |ak ||z| . By the triangle inequality, it follows that for
every complex number z,
|an ||z|n − r(z) ≤ |P (z)| ≤ |an ||z|n + r(z).
Now, if |z| ≥ 1 and m < n, it follows that |z|m ≤ |z|n−1 , therefore r(z) ≤
*n−1
M |z|
, where M = k=0 |ak |. Taking R =max {1, 2M |an|−1 }, for |z| ≥ R, it
follows that
n−1
|P (z)| ≤ |an ||z|n + r(z) ≤ |an ||z|n + M |z|n−1
= |z|n−1 (|an ||z| + M ) ≤ |z|n−1 (|an ||z| + |an ||z|)
= 2|an ||z|n ,
where the last inequality follows from the fact that |z| ≥ R ≥
Again, for |z| ≥ R, we have that
2M
|an |
>
M
|an | .
|P (z)| ≥ |an ||z|n − r(z) ≥ |an ||z|n − M |z|n−1
1
≥ |an ||z|n − |an ||z|n
2
1
= |an ||z|n ,
2
where the last inequality holds if and only if 12 |an ||z|n ≥ M |z|n−1 , which is true if
2M
, and the proof is complete.
and only if |z| ≥ |a
n|
The proof of the fundamental theorem of algebra that we present here is
based only on advanced calculus. Calculus provide a very useful result presented
as the following lemma, which can be consulted in [17].
5.5 Proof of the fundamental theorem of algebra
87
Lemma 5.5.2. If f : D → R is a continuous function on D, a closed and bounded
subset of R2 , then f attains its minimum and maximum values in points of D.
Theorem 5.5.3 (The fundamental theorem of algebra). Every polynomial
P (z) = an z n + an−1 z n−1 + · · · + a1 z + a0 ,
where n ≥ 1, ai ∈ C and an = 0, has at least one root in C.
The proof is based on the following two lemmas.
Lemma 5.5.4. Let P (z) be a polynomial with complex coefficients. Then |P (z)|
attains its minimum value in some point z0 ∈ C.
Proof. Let s = |P (0)| = |a0 | and R1 = max R, n 2s|an |−1 , where the number
R is given by the growth lemma. If |z| > R1 , then it follows that
|P (z)| ≥
1
1
1
2s
|an ||z n | > |an ||R1n | ≥ |an |
= s.
2
2
2
|an |
Thus, for all z such that |z| > R1 , it follows that |P (z)| > |P (0)|. In particular, if R2 > R1 , then for every z such that |z| ≥ R2 , the same inequality
holds.
In this way we have found a closed disk D of radius R2 with center at 0 such
that |P (z)| > |P (0)|, for all |z| > R2 . Since |P (z)| is a continuous function with real
values, it follows by Lemma 5.5.2 that |P (z)| attains its minimum value in D.
Lemma 5.5.5. Let P (z) be a non-constant polynomial with complex coefficients. If
P (z0 ) = 0, then |P (z0 )| is not the minimum value of |P (z)|.
Proof. Let P (z) be a non-constant polynomial with complex coefficients, and let
z0 be a point such that P (z0 ) = 0. Making the change of variable z + z0 for z
moves z0 to the origin, and then we can assume that P (0) = 0. Now, we multiply
P (z) by P (0)−1 so that we may assume that P (0) = 1. Then, we must show that
1 is not the minimum value of |P (z)|.
Let k be the smallest non-zero power of z in P (z). Then we can assume that
P (z) has the form P (z) = 1 + az k + g(z), with g(z) a polynomial of degree greater
than k.
Let α be a kth root of −a−1 . Then, making one last change of variable, αz
by z, we obtain that P (z) has the form
P (z) = 1 − z k + z k+1 g(z), for some polynomial g(z).
For real positive values of z we obtain, using the triangle inequality, that
|P (z)| ≤ |1 − z k | + z k+1 |g(z)|.
88
Chapter 5. Complex Numbers
Since z k < 1, for |z| < 1, then
|P (z)| ≤ 1 − z k + z k+1 |g(z)| = 1 − z k (1 − z|g(z)|).
For z small, z|g(z)| is also small, then we can choose z1 so that z1 |g(z1 )| < 1. It
follows that z1k (1 − z1 |g(z1 )|) > 0, and then |P (z1 )| < 1 = |P (0)|, and this finishes
the proof.
Using these two lemmas we obtain the proof of the fundamental theorem of
algebra.
Proof. Let P (z) be a non-constant polynomial with complex coefficients. By Lemma 5.5.4, |P (z)| has a minimum value in some point z0 ∈ C. Then, by Lemma 5.5.5,
it follows that |P (z0 )| = 0. Therefore, P (z) has a complex root.
Chapter 6
Functions and Functional Equations
6.1 Functions
The concept of function is one of the most important in mathematics. A function is
a relation between elements of two sets X and Y , which we denote by f : X → Y ,
that satisfies:
(a) Every element x ∈ X is related to some element y ∈ Y , and we write y =
f (x).
(b) Every element x ∈ X is related to one and only one element of Y , that is, if
f (x) = y1 and f (x) = y2 , then necessarily y1 = y2 .
The set X is called the domain, the set Y the codomain and the relation
y = f (x) the correspondence rule. We define the image or the range of a function
f : X → Y as
Img f = {y ∈ Y | there exists x ∈ X with f (x) = y}.
We say that f and g are equal functions if they have the same domain X, the
same correspondence rule (f (x) = g(x), for all x ∈ X) and the same codomain.
We define the graph of a function f : X → Y as
Γ(f ) = {(x, y) ∈ X × Y | y = f (x)},
where X × Y = {(x, y) | x ∈ X, y ∈ Y } is the Cartesian product of X and Y .
Two simple but important functions are the constant function and the identity function, which are defined as follows: if f : X → Y is such that f (x) = b
for all x ∈ X, with b ∈ Y fixed, then f is called the constant function equal to b,
while the image is the set with only one element {b}; the identity function has the
same domain and codomain, that is, Id : X → X, and it is defined as Id(x) = x
for all x ∈ X.
For functions f : R → R, the geometric representation of the graph in the
Cartesian plane is useful. For instance, the geometric representation of the constant
function b and the identity function IdR are the following:
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5_6
89
90
Chapter 6. Functions and Functional Equations
y
y
IdR
b
x
x
Some of the problems that appear in the mathematical olympiad contests,
which make reference to functions, ask to find all the functions that satisfy a given
property, or to find a specific value of some function. Often, these are difficult tasks,
therefore it is important to understand the general behavior of the function before,
so as to be able to decide correctly which functions satisfy or not the property. In
this first section we will offer several examples of the kind of problems that can
appear, and we will point out a series of facts that we can ask about a function.
Example 6.1.1. Let f : N → N be given by f (n) = n(n + 1). Let us find the values
of m and n such that 4f (n) = f (m), where m and n are natural numbers.
Suppose that 4f (n) = f (m), then 4n2 + 4n = m2 + m. If we complete the
square on the left side of the equation, we obtain
4n2 + 4n + 1 = m2 + m + 1,
(2n + 1)2 = m2 + m + 1,
but m2 + m + 1 cannot be the square of an integer because m2 < m2 + m + 1 <
(m+1)2 . Therefore, there are no natural numbers m and n satisfying the condition.
Example 6.1.2. Let f : N → N be a function such that f (3n) = n + f (3n − 3),
for every positive integer n greater than 1 and such that f (3) = 1. Find the value
of f (12).
It is natural to use the fact that 12 = 3 · 4 in order to find f (12) = f (3 · 4).
Using the formula or relation of the hypothesis, we have that f (12) = f (3 · 4) =
4 + f (3 · 3). We can repeatedly apply these substitutions to get
f (12) = 4 + f (3 · 3) = 4 + 3 + f (3 · 2)
= 4 + 3 + 2 + f (3 · 1) = 4 + 3 + 2 + 1 = 10.
Observe that we can find f (3n) for all n, in the following way:
f (3n) − f (3(n − 1)) = n
f (3(n − 1)) − f (3(n − 2)) = n − 1
..
..
.
.
91
6.1 Functions
..
..
.
.
f (6) − f (3) = 2
f (3) = 1.
Adding these equations results in f (3n) = 1 + 2 + 3 + · · · + n =
n(n+1)
.
2
Functions can be combined to form new functions. For instance, if we have
functions with the same domain and codomain, we can add, subtract or multiply
them to obtain new functions. For two functions f , g : X → Y , with Y ⊂ C, we
define the sum and the difference of the functions f and g as
(f ± g)(x) = f (x) ± g(x), for all x ∈ X.
We define the product of the functions f and g as
(f · g)(x) = f (x) · g(x), for all x ∈ X.
Finally, if g(x) = 0 for all x, we define the quotient of the functions f and g as
f
g
(x) =
f (x)
, for all x ∈ X.
g(x)
Example 6.1.3. Find all the functions f : R → R that satisfy
f (y + x) − f (y − x) = 4yx, for all x, y ∈ R.
(6.1)
If we let y = x, we get that f (2y)−f (0) = 4y 2 and taking f (0) = c, we obtain
f (2y) = 4y 2 + c. Now, if we let 2y = x, we get that the solution of the functional
equation is of the form f (x) = x2 + c. It is easy to see that these functions satisfy
equation (6.1).
Example 6.1.4. Find all the functions f : R+ → R+ satisfying
f (xy) = f
x
y
, for all x, y ∈ R+ .
In this case, if we let y = x we have that f (x2 ) = f (1). This makes sense
since the function is defined on the positive real numbers. On the other hand,
if we let y = x2 , then f (y) = f (1) for all y ∈ R+ . Hence, the functions solving
the functional equation are the constant functions. It is easy to see that these
functions satisfy the functional equation.
In the last two examples we made the substitution y = x, which directly gave
us the solution of the given functional equations. Next we present an example
where, after substituting the variables for numerical values, we get information
about the function.
92
Chapter 6. Functions and Functional Equations
Example 6.1.5. Find all the functions f : R → R such that
f (x + y) + f (x − y) = f (x) + 6xy 2 + x3 , for all x, y ∈ R.
Let y = 0 to see that 2f (x) = f (x) + x3 . Then f (x) = x3 is the function we
are looking for. We can directly check that f (x) = x3 satisfies the equation.
Example 6.1.6. Find all the functions f : N → R such that f (1) = 3, f (2) = 2 and
f (n + 2) +
1
= 2,
f (n)
for all
n ∈ N.
(6.2)
Observe that the original equation gives us
f (3) = 2 −
1
1
5
=2− =
f (1)
3
3
and f (4) = 2 −
1
1
3
6
=2− = = .
f (2)
2
2
4
Hence, we can conjecture that f (n) = n+2
n is true, for all natural numbers n. It is
true by hypothesis for the cases n = 1 and n = 2, and we already verified the result
for n = 3 and n = 4. We will finish the proof using induction, that is, assuming the
result holds for n, and then proving it for n + 2. Suppose that f (n) = n+2
n , then
1
n
1
= 2 − n+2 = 2 −
f (n)
n
+
2
n
n+4
2n + 4 − n
=
.
=
n+2
n+2
f (n + 2) = 2 −
Hence, we have the result for all n ∈ N and, in fact, the function satisfies the
condition.
Example 6.1.7 (India, 2010). Find all functions f : R → R satisfying,
f (x + y) + xy = f (x)f (y),
for all
x, y ∈ R.
(6.3)
Let x = y = 0 in equation (6.3), then f (0) = f (0)2 , hence f (0) = 0 or
f (0) = 1.
If f (0) = 0, we let y = 0 in equation (6.3) to get f (x) = 0, for all x ∈ R. But
the function f (x) = 0, for all x ∈ R, (the constant function zero) does not satisfy
condition (6.3) when xy = 0.
Suppose that f (0) = 1. If we let x = 1, y = −1, we get that f (1)f (−1) =
f (1 − 1) − 1 = f (0) − 1 = 0, then f (1) = 0 or f (−1) = 0.
If f (−1) = 0, letting y = −1 we get f (x − 1) − x = 0, then f (x − 1) = x, and
using y = x − 1 in this last equality, we get f (y) = y + 1.
If f (1) = 0, letting y = 1 we obtain f (x + 1) + x = f (x)f (1) = 0 so that
f (x + 1) = −x. Finally, if we take y = x + 1 we obtain f (y) = 1 − y.
In this way, the only solutions are f (x) = x + 1 and f (x) = 1 − x. It is easy to
check that these functions satisfy the functional equation (6.3).
93
6.1 Functions
Another way to create a new function given two functions f : X → Y and
g : Y → Z is the composition of f and g, which is the function g ◦ f : X → Z
defined for all x ∈ X by
(g ◦ f )(x) = g(f (x)).
Observe that the composition g ◦f is defined only if the codomain of f is contained
in the domain of g.
Example 6.1.8 (IMO, 1977). Consider f : N → N such that f (n + 1) > f (f (n)),
for every positive integer n. Prove that f (n) = n, for every n ∈ N.
Let A = {f (1), f (2), . . . } be the image of f . By hypothesis, note that for
every n ≥ 2, it follows that f (n) > f (f (n − 1)). Hence for n ≥ 2, f (n) cannot be
the minimum of the image of f . It follows that the minimum of A is f (1) and that
f (n) > f (1) for n ≥ 2.
Observe that if m ≥ p, then f (m) ≥ p. For p = 2, the result follows from
the discussion above. Suppose the result true for p ≥ 2 and let us show it holds
for p + 1. Let m ≥ p + 1, then m − 1 ≥ p and f (m − 1) ≥ p; now, by hypothesis,
f (m) > f (f (m − 1)) ≥ p, hence f (m) ≥ p + 1.
Now, let Ap = {f (p), f (p + 1), f (p + 2), . . . }. For every n ≥ p + 1, it follows
that f (n) > f (f (n−1)). Since the observation guarantees that f (f (n−1)) belongs
to the set Ap , then f (n) cannot be the minimum of Ap , hence the minimum must
be f (p). Therefore, f (n) > f (p) for all n ≥ p + 1.
Finally, from the last paragraph, it follows that
f (1) < f (2) < · · · < f (p) < f (p + 1) < . . . .
(6.4)
Now, let us show that f (n) ≥ n. We have that f (1) ≥ 1 and f (2) > f (1), hence
f (2) ≥ 2. Similarly, from f (2) ≥ 2 and f (3) > f (2), it follows that f (3) ≥ 3, and
using induction we can prove that f (n) ≥ n. Finally, if for some n we have that
f (n) > n, then f (n) ≥ n + 1 and, by (6.4), f (f (n)) ≥ f (n + 1), contradicting the
hypothesis. Thus, f (n) = n, for every n ∈ N.
Exercise 6.1. Find all functions f : R\{0, 1} → R satisfying the functional equation
1
= x, for all x = 0, 1.
f (x) + f
1−x
Exercise 6.2 (Canada, 1969). Let f : N → N be a function with the following
properties:
(i) f (2) = 2.
(ii) f (mn) = f (m)f (n), for all m and n.
(iii) f (m) > f (n), for m > n.
Prove that f (n) = n.
94
Chapter 6. Functions and Functional Equations
Exercise 6.3. Find all functions f : R\{0} → R such that
xf (x) + 2xf (−x) = −1, for x = 0.
Exercise 6.4. Find all functions f : R\{0} → R that satisfy the equation
1
f (−x) + f
x
1
x
= x, for x = 0.
Exercise 6.5 (Ireland, 1995). Find all functions f : R → R for which
xf (x) − yf (y) = (x − y)f (x + y), for all x, y ∈ R.
Exercise 6.6 (Ukraine, 1997). Find all functions f : Q+ ∪ {0} → Q+ ∪ {0}, that
satisfy the following conditions:
(a) f (x + 1) = f (x) + 1, for all x ∈ Q+ ∪ {0}.
(b) f (x2 ) = f (x)2 , for all x ∈ Q+ ∪ {0}.
Exercise 6.7. Find all functions f : R → R such that
xf (y) + yf (z) + zf (x) = yf (x) + zf (y) + xf (z), for x, y, z real numbers.
6.2 Properties of functions
In this section, we will study important properties that a function may or may
not have. A good knowledge of these properties can help us to detect what kind
of function we have.
6.2.1 Injective, surjective and bijective functions
We say that a function f : X → Y is injective (also known as one-to-one) if for
any x1 , x2 ∈ X, with x1 = x2 , it follows that f (x1 ) = f (x2 ). The condition is
equivalent to saying that if f (x1 ) = f (x2 ), then x1 = x2 .
We say that a function f : X → Y is surjective (also known as onto) if Img f = Y ,
that is, if for every y ∈ Y there exists x ∈ X such that f (x) = y.
Finally, we say that a function is bijective if it is injective and surjective.
For functions from the real numbers to the real numbers knowing the graph
of the function can be very useful. The graph can help us to determine if the
function is injective, surjective or both.
More precisely, a function is injective if any parallel line to the x-axis intersects the graph of the function in at most one point. A function is surjective, if
any horizontal line y = y0 , with y0 in the codomain of the function, intersects
the graph in at least one point. The function shown in the following graph is
95
6.2 Properties of functions
not injective, since any parallel line above the x-axis, intersects the graph in two
points. Moreover, it is not surjective since a horizontal line below the x-axis never
intersects the graph of the function.
y = f (x)
Example 6.2.1. A function f : N → N that satisfies f (f (m) + f (n)) = m + n for
all m, n ∈ N, is injective.
The function is injective, because if f (m) = f (n), then f (m)+ f (n) = f (n)+ f (n),
and from here it follows that m + n = f (f (m) + f (n)) = f (f (n) + f (n)) = n + n,
thus m = n.
Example 6.2.2. The functions f : R+ → R+ that satisfy the condition
f (xf (y)) + f (yf (x)) = 2xy, for all x, y ∈ R+
are injective.
Letting x = y, we have that f (xf (x)) = x2 , in particular f (f (1)) = 1. Letting
x = f (1) in the last equation, we get that
f (1)2 = f (f (1)f (f (1))) = f (f (1)) = 1,
hence f (1) = 1. If one takes y = 1 in the original equation, we obtain that
f (x) + f (f (x)) = 2x.
With this last equality we can show that f is injective. If f (x) = f (y), then
2x = f (x) + f (f (x)) = f (y) + f (f (y)) = 2y, hence x = y.
Example 6.2.3. A function f : R → R that satisfies
f (f (x) + y) = 2x + f (f (y) − x), for all x, y ∈ R
is surjective.
If we take y = −f (x), it follows that f (f (x) − f (x)) = 2x + f (f (−f (x)) − x),
that is, f (f (−f (x)) − x) = f (0) − 2x. Now, if y ∈ R we need to find x0 such
96
Chapter 6. Functions and Functional Equations
that f (0) − 2x0 = y, but x0 =
surjective.
f (0)−y
2
satisfies f (f (−f (x0 )) − x0 ) = y, thus f is
Example 6.2.4. A function f : Q+ → Q+ that satisfies
f (xf (y)) =
f (x)
, for all x, y ∈ Q+
y
is bijective.
If x = 1, then f (f (y)) =
f (1)
y ,
this will help us to show that f is injective.
If f (y1 ) = f (y2 ), then f (f (y1 )) = f (f (y2 )), that is, fy(1)
= fy(1)
. Hence y1 = y2 ,
1
2
+
and therefore f is injective. It is left to show that f is surjective. Let m
n ∈ Q ,
.
If
this
happens,
then
we want to show that there is x ∈ Q+ such that f (x) = m
n
m
f (1)
f (f (x)) = f m
,
hence
,
and
solving
for
x
we
get x = ff (1)
=
f
, which
n
x
n
(m
n )
is a rational number. That is, f is surjective, and therefore f is bijective.
Observation 6.2.5. In the previous examples of this section, we were faced with a
situation where
f (g(x)) = h(x).
Notice that if h is injective then g is injective. Also, if h is surjective then f is
surjective.
6.2.2 Even and odd functions
The functions that satisfy f (x) = f (−x) are called even functions and the functions such that f (x) = −f (−x) are called odd functions. A function f : R → R
is even if its graph is symmetric with respect the y-axis, whereas the graph of an
odd function is symmetric with respect to the origin.
y = f (x)
Graph of an even function.
y = f (x)
Graph of an odd function.
97
6.2 Properties of functions
Example 6.2.6. Find all the functions f : Q → Q, that satisfy
f (x + y) + f (x − y) = 2f (x) + 2f (y), for all x, y ∈ Q.
Letting x = y = 0, we have f (0) = 0. If x = y, then f (2x) = 4f (x), and by
induction f (nx) = n2 f (x) for every positive integer n. With x = 0, we have that
f (y) + f (−y) = 2f (y), then f (y) = f (−y), hence f is even and f (nx) = n2 f (x)
for every integer number n and every rational number x. If r = pq is a rational
$ %
$ %
$ %
number, with p ≥ 1, then p2 f (1) = f (p) = f q pq = q 2 f pq , and then f pq =
$ %2
p
f (1). Hence f (r) = cr2 , for all r ∈ Q with c = f (1). It is easy to check that
q
these functions satisfy the given condition.
6.2.3 Periodic functions
Periodicity plays an important role in mathematics and for this reason we include
some examples about this topic.
We say that a function f : R → R is periodic if there exists a = 0 ∈ R such
that
f (x + a) = f (x), for all x ∈ R.
The number a is called a period of f . It is clear that for all n = 0, the number na
is also a period of f .
a
−a
Example 6.2.7. A function f : R → R is periodic, if for some a ∈ R and every
x ∈ R, it is true that
1 + f (x)
.
f (x + a) =
1 − f (x)
From the equation f (x + a) =
1+f (x)
1−f (x) ,
evaluating in x − a, we obtain that
1+f (x−a)
f (x) = 1−f
(x−a) . After solving for f (x) from the original equation, we get that
f (x+a)−1
(x−a)
f (x) = f (x+a)+1 . Equating the last two equations, we get that 1+f
1−f (x−a) =
f (x+a)−1
−1
f (x+a)+1 and after simplifying f (x + a) = f (x−a) . Evaluating this last equation in
x + a, we get that f (x + 2a) = f−1
(x) , and then
f (x + 4a) =
−1
−1
= −1 = f (x),
f (x + 2a)
f (x)
that is, f is periodic with period 4a.
98
Chapter 6. Functions and Functional Equations
Example 6.2.8 (Belarus, 2005).
(a) Consider a function f : N → N that satisfies f (n) = f (n + f (n)) for all
n ∈ N. Prove that, if the image of f is finite, then f is periodic.
(b) Find a non-periodic function f : N → N such that f (n) = f (n + f (n)), for
all n ∈ N.
To prove (a) we do as follows. Since f (n) = f (n + f (n)), then f (n + f (n) +
f (n)) = f (n + f (n) + f (n + f (n))) = f (n + f (n)) = f (n). Therefore, f (n) =
f (n + 2f (n)) and, by induction, f (n) = f (n + kf (n)) for all k ∈ N.
Let A = f (N) = {a1 , . . . , at } and T = a1 · · · · · at . Let us see that T is a
period of f . Since f (n) = f (n + kf (n)), for all k, it follows that
f (n) = f
n+
T
f (n)
f (n)
= f (n + T ),
T
where k = f (n)
. That is, T is a period of f .
(b) We would like to find a non-periodic function f that satisfies the equation.
For n = 2k m with k ∈ N ∪ {0} and m odd, we define f (n) = 2k+1 .
Now, let us see that the function satisfies the equation, f (n + f (n)) = f (2k m +
2k+1 ) = f (2k (m + 2)) = 2k+1 = f (n) and moreover, f is not periodic, otherwise
its range would be finite, however, its image is the set of powers of 2.
6.2.4 Increasing and decreasing functions
We say that a function is increasing if for x < y it follows that f (x) < f (y). We say
that a function is non-decreasing if for x < y it follows that f (x) ≤ f (y). Similarly,
we say that a function is decreasing if for x < y it follows that f (x) > f (y), and
it is non-increasing if for x < y it happens that f (x) ≥ f (y).
Another way to guarantee the injectivity of a real function, whose domain is
the real numbers, is to know if the function is increasing or decreasing.
Example 6.2.9 (IMO, 1992). Find all functions f : R → R that satisfy
f (x2 + f (y)) = y + f (x)2 , for all x, y ∈ R.
Let a = f (0). With x = 0, we have that f (f (y)) = a2 + y. With x = y = 0,
then f (a) = a2 , hence f (x2 + a) + a2 = f (x)2 + f (a). Applying f to both sides
of the equation it follows that f (f (x2 + a) + a2 ) = f (f (x)2 + f (a)). The left-hand
side of the equality is
f (f (x2 + a) + a2 ) = x2 + a + f (a)2
= x2 + a + a4 ,
(6.5)
and the right-hand side of the equality is
f (f (x)2 + f (a)) = a + f (f (x))2
= a + (a2 + x)2 = a + a4 + 2a2 x + x2 .
(6.6)
6.2 Properties of functions
99
Comparing both equations (6.5) and (6.6), we need to have that 2a2 x = 0, hence
a = 0. From here we conclude that f (f (y)) = y, for all y ∈ R and f (x2 ) = f (x)2 ,
for all x ∈ R.
The last equation guarantees that if x ≥ 0, then f (x) ≥ 0. Since f (f (y)) = y
we have that f is injective, hence f (x) = 0 if and only if x = 0. In this way when
x > 0, then f (x) > 0.
Since f (f (x)2 + f (y)) = f (f (x))2 + y = x2 + y, it follows that f (x2 + y) =
f (f (f (x)2 + f (y))) = f (x)2 + f (y) = f (x2 ) + f (y).
√
If y < x, then x − y > 0 and x = x − y + y. Applying the last equality to x − y
and y, it follows that f (x) = f (x − y + y) = f (x − y) + f (y) > f (y), that is, f is
increasing. But f non-decreasing and f (f (x)) = x, guarantee that f (x) = x. In fact
if f (x) > x, then x = f (f (x)) > f (x) and if f (x) < x, then x = f (f (x)) < f (x).
Therefore, the only function that satisfies the functional equation is f (x) = x.
6.2.5 Bounded functions
We say that a function f : A ⊂ R → R is bounded above in A, if there exists
M ∈ R such that f (a) ≤ M , for all a ∈ A.
We say that a function f : A ⊂ R → R is bounded below in A, if there exists
N ∈ R such that f (a) ≥ N , for all a ∈ A.
We say that a function f : A ⊂ R → R is bounded in A, if there exists M > 0
such that |f (x)| ≤ M , for all x ∈ A, or equivalently −M ≤ f (x) ≤ M , for all
x ∈ A.
Example 6.2.10. Let m ≥ 2 be an integer number. Find all the bounded functions
f : [0, 1] → R such that, for x ∈ [0, 1], it follows that
4
$x%
2x
(m − 1) x
1
+f
+ ···+ f
.
f (x) = 2 f (0) + f
m
m
m
m
If |f (x)| ≤ M for x ∈ [0, 1], then, using the triangle inequality, it follows that
m−1
mM
j
M
1 #
f
x ≤
.
=
|f (x)| ≤ 2
2
m j=0
m
m
m
Hence |f (x)| ≤ M
m , for x ∈ [0, 1]. With this new bound, we can repeat the last
M
argument to show that |f (x)| ≤ m
2 , for x ∈ [0, 1]. An inductive argument guaranM
tees us that for all n ∈ N, it follows that |f (x)| ≤ m
n , for x ∈ [0, 1], and when we
let n go to infinity, we have that f (x) = 0, for all x ∈ [0, 1], and this is the only
function that satisfies the equation.
6.2.6 Continuity
When the values of a function f (x) get closer to b as x tends to a, we say that b is
the limit of f (x) as x tends to a. In mathematical language this is usually written
100
Chapter 6. Functions and Functional Equations
as follows:
∀ ǫ > 0, ∃ δ > 0 such that 0 < |x − a| < δ ⇒ |f (x) − b| < ǫ
and we write limx→a f (x) = b.
Also, we can consider limx→∞ f (x) = b, which is defined as: for all ǫ > 0
there exists M > 0 such that if x > M then |f (x) − b| < ǫ.
One characterization of the limit concept when dealing with sequences is the
following theorem.
Theorem 6.2.11. Let f be a function, then limx→a f (x) = b if and only if for every
sequence {an } with limn→∞ an = a, it follows that limn→∞ f (an ) = b.
The proof of this theorem will be given in Chapter 7.
Observation 6.2.12. The previous theorem also states that limx→a f (x) is not b, if
there exists a sequence {an } such that limn→∞ an = a and limn→∞ f (an ) is not b.
Example 6.2.13 (IMO, 1983). Find all functions f : R+ → R+ that satisfy:
(a) f (xf (y)) = yf (x), for all positive real numbers x, y.
(b) limx→∞ f (x) = 0.
If x = 1, then f (f (y)) = yf (1), hence the function is bijective. In fact, if
f (x) = f (y), then xf (1) = f (f (x)) = f (f (y)) = yf (1), and since f (1) = 0 we have
+
that x = y, therefore f is injective; let us
%%that f is surjective, let c ∈ R
$ see
$ now
c
c
= f (f (y)) = yf (1) = c, hence
, then it follows that f f f (1)
and take y = f (1)
f is surjective. Thus, f is bijective.
In particular, there is a y0 such that f (y0 ) = 1. Since f (xf (y0 )) = y0 f (x) for
all x > 0, taking x = 1, we have that f (1) = y0 f (1), and then y0 = 1, hence x = 1
is a fixed point.
If x = y, then f (xf (x)) = xf (x), and xf (x) is also a fixed point of f . If we show
1
that the only fixed point
1 can
conclude that f (x) = x .
1 of f is 1, we
From the equation f a f (a) =
af
a and the fact that f is injective, we have
that f (a) = a if and only if f a1 = a1 . If there is a fixed point different from 1,
hence there is a fixed point a greater than 1. But f (a) = a implies, by induction,
that f (an ) = an , (for instance, f (a2 ) = f (af (a)) = af (a) = a2 ). Since an → ∞
and f (an ) = an → ∞, this contradicts the fact that limx→∞ f (x) = 0. Therefore,
the only fixed point is 1.
The function f (x) = x1 satisfies the conditions of the problem.
We say that a function f is continuous at some point a if when we let x tend to a,
f (x) tends to f (a), that is, limx→a f (x) = f (a). Also, we say that f is continuous
on a set A if f is continuous at every point a ∈ A.
One characterization of the continuity property is given by the following
result.
6.2 Properties of functions
101
Theorem 6.2.14. A function f is continuous at a if and only if for every sequence
{an } with limn→∞ an = a, it follows that limn→∞ f (an ) = f (a).
A set D ⊂ R is dense in the set of real numbers if every open interval in R
has points in D.
Theorem 6.2.15. The set of rational numbers is dense in the set of real numbers.
Theorem 6.2.16. If a function f is continuous on R and f is zero in a dense subset
of the real numbers, then f is identically zero in R.
As a consequence, if the functions f and g are continuous and coincide on a
dense subset of R, then they coincide on all R.
The proofs of these three theorems will be given in Section 7.4.
Example 6.2.17 (Nordic, 1998). Find all the continuous functions f : R → R that
satisfy the equation,
f (x + y) + f (x − y) = 2(f (x) + f (y)), for all x, y ∈ R.
We have seen in Example 6.2.6 that f (x) = f (1)x2 , for x ∈ Q and, by Theorem
6.2.16, f (x) = f (1)x2 for all x ∈ R.
Exercise 6.8. Let f , g : R → R be two functions that satisfy f (g(x)) = g(f (x)) =
−x, for any real number x. Prove that f and g are odd functions.
Exercise 6.9. Find all surjective functions f : R → R that satisfy
f (f (x − y)) = f (x) − f (y), for all x, y ∈ R.
Exercise 6.10. Find all continuous functions f : R → R that satisfy
f (xf (y)) = xy, for all x, y ∈ R.
Exercise 6.11 (Belarus, 2005). Find all functions f : N → N that satisfy
f (m − n + f (n)) = f (m) + f (n), for all m, n ∈ N.
Exercise 6.12 (IMO, 1990). Find a function f : Q+ → Q+ that satisfies the
equation
f (x)
, for all x, y ∈ Q+ .
f (xf (y)) =
y
Exercise 6.13. Find all continuous functions f : R → R that satisfy
xf (y) + yf (x) = (x + y)f (x)f (y), for all x, y ∈ R.
102
Chapter 6. Functions and Functional Equations
Exercise 6.14 (IMO, 1968). Let f : R → R be a function with the property
f (x + a) =
1
+
2
f (x) − f (x)2 ,
for all x ∈ R and a a fixed number.
(i) Prove that f is periodic.
(ii) In case that a = 1, give an example of a function of this type.
Exercise 6.15. Let a, b > 0, find the values of m such that the equation
|x − a| + |x − b| + |x + a| + |x + b| = m(a + b),
has at least one real solution.
6.3 Functional equations of Cauchy type
The functional equations of Cauchy type are:
(C1 )
f (x + y) = f (x) + f (y).
(C2 )
(C3 )
f (x · y) = f (x) + f (y).
f (x + y) = f (x) · f (y).
(C4 )
f (x · y) = f (x) · f (y).
In order to establish what functions satisfy certain functional equations we should
take into account the domain and the codomain where we want to solve the equation. For instance, if we take equation (C2 ), and we want to solve it in all R,
considering y = 0, we obtain that the solution of the equation is f (x) = 0, for
all x, which means the equation sought for was very simple. Therefore, it is more
adequate for this functional equation to consider the set of real positive numbers
as its domain.
6.3.1 The Cauchy equation f (x + y) = f (x) + f (y)
The first of the equations of Cauchy type is the most important. With this equation
we will illustrate how some functional equations are solved. First, we will see how
to determine some of the values that the functions take, and this will allow us to
find, in a natural way, other values until we learn how the functions behave in the
set of rational numbers.
Letting x = y = 0, we have that f (0) = 2f (0), then f (0) = 0. If y = −x,
we have that 0 = f (0) = f (x + (−x)) = f (x) + f (−x), and then f (−x) = −f (x),
which tells us that the function f should be odd.
103
6.3 Equation Cauchy type
With x = y, we have that f (2x) = 2f (x). Now, using induction, we can
conclude that f (nx) = nf (x), for any positive integer n. In fact, f ((n + 1)x) =
f (nx + x) = f (nx) + f (x) = nf (x) + f (x) = (n + 1)f (x).
Recalling that f (−x) = −f (x), we get f (nx) = nf (x) for all n ∈ Z. Now,
x
x
x
1
since f (x) = f ( m
m x) = f (m m ) = mf ( m ), we have that f ( m ) = m f (x), therefore
x
x
n
n
f ( m x) = f (n m ) = nf ( m ) = m f (x). Hence, f (rx) = rf (x), for all r ∈ Q and all
x ∈ R.
Letting c = f (1), we get f (r) = cr for all r ∈ Q.
We conclude that a function f : Q → R that satisfies equation (C1 ) should have the
form f (r) = cr, for all r ∈ Q, with c = f (1) a fixed constant. And a function of this
type f (x) = cx satisfies such a type of Cauchy equation, since c(x + y) = cx + cy,
for any x, y ∈ Q.
Additional hypothesis to the Cauchy equation f (x + y) = f (x) + f (y)
We would like to determine functions f : R → R that satisfy the first of the Cauchy
type equations, which are also known as additive functions. We will see that with
an additional hypothesis (we will analyze several of them), the conclusion is that
f should be linear, that is, of the form f (x) = ax, for all x ∈ R and with a = f (1).
(H1 )
The function is continuous in all R.
We know that f (r) = cr, for all r ∈ Q. Since f (x) and cx are continuous in R and
coincide in Q, we have, by Theorem 6.2.16, that the functions coincide in all R.
Hence f (x) = cx for all x ∈ R.
Example 6.3.1 (Jensen’s equation). Find all continuous functions f : R → R that
satisfy the equation,
f
x+y
2
=
f (x) + f (y)
, for all x, y ∈ R.
2
Note that by letting x = y = 0, we do not obtain information about what is the
value of f (0). Define g(x) = f (x) − f (0), which is also continuous. A straightforward calculation shows that
x+y
2
g
=
g(x) + g(y)
,
2
but now the function g satisfies g(0) = 0. Taking y = 0 in the new equation, we
have that
$ x % g(x)
,
=
g
2
2
and after substituting in this last equation x = u + v, we get
g
u+v
2
=
g(u + v)
.
2
104
Chapter 6. Functions and Functional Equations
Hence, we can affirm that g(u + v) = g(u) + g(v), for all u, v ∈ R, that is, g
is a continuous function that satisfies the first equation of Cauchy. Therefore,
g(x) = ax, with a = g(1), and then f (x) = ax + b, for all x ∈ R and b = f (0).
(H1′ )
The function is continuous only in x = 0.
To reduce this to the previous case, it will be enough to show the following result.
Lemma 6.3.2. If f : R → R is an additive function, that is, if it satisfies equation
(C1 ) and it is continuous at 0, then it is continuous at every real number a.
Proof. Let {an } be a sequence with limn→∞ an = a, then the sequence {an − a}
satisfies limn→∞ (an −a) = 0. Since f is continuous in 0, Theorem 6.2.14 guarantees
that limn→∞ f (an − a) = f (0) = 0. But equation (C1 ) implies that f (an ) =
f (an − a) + f (a), then limn→∞ f (an ) = limn→∞ f (an − a) + limn→∞ f (a) = f (a),
which implies f is continuous in a.
(H2 )
The function is monotone.
If the function f , besides being additive, is monotone (without loss of generality, we
can assume that it is non-decreasing), then f (x) should have the form f (x) = cx.
To support this claim consider a real number x. Let {rn } and {sn } be sequences
of rational numbers converging to x, with rn < x < sn for all n.
By the monotonicity of the function f , crn = f (rn ) ≤ f (x) ≤ f (sn ) = csn .
Taking the limit, we obtain cx = limn→∞ crn ≤ f (x) ≤ limn→∞ csn = cx. Thus,
f (x) = cx.
The non-increasing case is similar. Moreover, we can change ≤ to < and reach the
same conclusion, and similarly in the case ≥.
(H3 )
The function is positive (for positive numbers).
If f (x) > 0 for x > 0 and if in addition it is additive, then it is increasing. In fact,
if x < y then y − x > 0, and f (y − x) > 0. Hence, f (x) < f (x) + f (y − x) =
f (x + (y − x)) = f (y). Similarly, we can consider the decreasing, non-decreasing
and non-increasing cases, and in each one of them we can conclude that f is linear.
(H4 )
The function is bounded.
If the additive function f is bounded in an interval of the form [a, b], that is, if
there exists a constant M > 0 such that |f (x)| ≤ M for all x ∈ [a, b], then the
function must have the form f (x) = cx.
First note that x ∈ [0, b − a] if and only if x + a ∈ [a, b] and for x ∈ [0, b − a]
we have |f (x)| = |f (x + a) − f (a)| ≤ |f (x + a)| + |f (a)| ≤ 2M . This guarantees
that f is bounded by 2M in [0, b − a]. Let α = b − a, c = f (α)
α and g(x) = f (x)− cx.
We then have
(a) g(x + y) = f (x + y) − c(x + y) = f (x) − cx + f (y) − cy = g(x) + g(y), that
is, g is additive.
6.3 Equation Cauchy type
105
(b) g(α) = f (α) − cα = 0.
(c) g(x + α) = g(x) + g(α) = g(x), the function g is periodic with period
α.
f (α)
(d) For x ∈ [0, α], we have |g(x)| = |f (x) − cx| ≤ |f (x)|+|cx| ≤ 2M + α |α| ≤
3M , that is, g is bounded in the interval [0, α] and then, because it is periodic,
it is bounded in all R.
If g(x0 ) = 0 for some real number x0 , then since g(nx0 ) = ng(x0 ) for every integer
number n, we can make |g(nx0 )| as large as we wish, then g would not be bounded,
which would be a contradiction to the part (d). Therefore, g(x) = 0 for any real
number x and then f (x) = cx, for all x in R.
(H4′ )
The function is bounded on a neighborhood of 0.
By (H1′ ), it will be enough to show that f is continuous in 0.
Let {an } be a sequence that converges to 0. We will use Theorem 6.2.14 to show
that f (an ) converges to 0. Let ǫ > 0, we will see that |f (an )| < ǫ for large n. If
M > 0 is the bound for f in the interval (−a, a), let us choose an integer N such
that M
N < ǫ.
Since limn→∞ an = 0, there exists n0 such that |an | < Na , for all n ≥ n0 . Since
|N an | < a, it follows that |f (N an )| ≤ M for n ≥ n0 . But since f (N an ) = N f (an ),
we have that |f (an )| ≤ M
N < ǫ, for n ≥ n0 , as we wanted.
There are several conditions that can be added to the equation of Cauchy to make
sure that the function that satisfies the equation is a linear function. Many of these
conditions are the source of problems of the kind that appear in the mathematical
olympiad. Let us see an example.
Example 6.3.3. Find all the functions that satisfy the following equations:
(a) f (x + y) = f (x) + f (y), for all x, y ∈ R.
(b) f (xy) = f (x)f (y), for all x, y ∈ R.
First note that if x ≥ 0, then
√ √
√
√
√
f (x) = f ( x · x) = f ( x)f ( x) = (f ( x))2 ≥ 0.
But then, by (H3 ), we have that f is linear, that is, it has the form f (x) = cx
with c = f (1). Taking x = y = 1 in the equation (b), we get that c = c2 , hence
c = 0 or c = 1. Then, f (x) = 0 or f (x) = x, are the only solutions to the problem.
6.3.2 The other Cauchy functional equations ⋆
The Cauchy equation f (x · y) = f (x) + f (y)
We will find continuous solutions to this functional equation. If y = 0 belongs
to the domain of f , then f (x) = 0. Now, suppose that the function is defined
for x = 0. If we take x = y = 1 in the equation, we have that f (1) = 0. Also,
106
Chapter 6. Functions and Functional Equations
considering x = y = −1, we get that f (−1) = 0. Now, taking y = −1, we obtain
f (−x) = f (x), that is, the function must be even and it will be determined by its
behavior when x is positive. But if x, y are positive, there are u, v ∈ R such that
x = eu , y = ev , and with them the equation14 can be expressed as
f (eu · ev ) = f (eu ) + f (ev ).
If we let g(u) = f (eu ), then g(u + v) = g(u) + g(v), which is the first Cauchy
equation, and we know that its solution is g(u) = cu, with c = g(1) = f (e), then
f (x) = g(u) = f (e) log x for x > 0, and f (x) = f (e) log |x| for x = 0.
The Cauchy equation f (x + y) = f (x) · f (y)
First note that if for some y, f (y) = 0, then f is constant. This follows from
f (x) = f (x − y + y) = f (x − y)f (y) = 0. If f is never zero, then it is positive
since f (x) = f ( x2 + x2 ) = (f ( x2 ))2 > 0. Hence, since f is always positive we can
take logarithms on both sides of the equation in order to satisfy the functional
equation,
log f (x + y) = log f (x) + log f (y),
which is a functional equation of the first type, then log f (x) = cx, with c =
log f (1). Applying the exponential function, we have that f (x) = elog f (1)x =
f (1)x , for all x ∈ R. Note that here we have found only the continuous solutions.
The Cauchy equation f (x · y) = f (x) · f (y)
As in the previous equation, if for some y = 0, f (y) = 0, then f is constant. This
follows since f (x) = f ( xy ·y) = f ( xy )f (y) = 0. If f is never zero, then for x positive,
√
√ √
f (x) is positive, since f (x) = f ( x · x) = (f ( x))2 > 0. For x = 1, we have that
f (1) = (f (1))2 , therefore f (1) = 0 or f (1) = 1. The first option has been studied
before, and therefore f (1) = 1. Since f (x2 ) = (f (x))2 , f (−1) = ±1 and taking
y = −1 in the original equation, we have that f (−x) = ±f (x). Then it will be
enough to see what happens with x > 0. After that we will have two options to
extend to the negative real numbers, that is, whether making the function even
or odd. Since the function f is positive for x > 0, we can apply the logarithmic
function on both sides of the equality to get
log f (x · y) = log f (x) + log f (y).
Considering g(x) = log f (x), we have that g satisfies the second equation of
Cauchy, then g(x) = g(e) log x, hence f (x) = xg(e) = xlog f (e) . Thus the continuous solutions are f (x) = 0, f (x) = ±xlog f (e) . Observe that the point x = 0
remains outside the analysis we have made (when we take a logarithm), but at the
end we include 0 and necessarily it has to happen that f (0) = 0, in order to have
14 For
the definition of the exponential function, ex , and the function logarithm, log x, see [21].
107
6.3 Equation Cauchy type
continuity there. However, there exists an exception, if f (e) = 1 there are two
solutions. One of them is f (x) = 1, which is continuous, and the other solution is
f (x) = sign (x), which is not continuous at x = 0.
Exercise 6.16. Find all functions f : R → R that satisfy
f (x2 ) − f (y 2 ) = (x + y)(f (x) − f (y)), for all x, y ∈ R.
Exercise 6.17. Find all functions f : R+ → R+ that satisfy
f (x)f (y) − f (xy) =
x y
+ , for all x, y ∈ R+ .
y
x
Exercise 6.18.
(i) Find all functions f : R+ → R+ that satisfy the condition
f (xf (y)) + f (yf (x)) = 2xy, for all x, y ∈ R+ .
(ii) (Short list IMO, 2002) Find all functions f : R → R such that
f (f (x) + y) = 2x + f (f (y) − x), for all x, y ∈ R.
Exercise 6.19. Let f be a function such that for some number a ∈ R it satisfies
f (x + a) =
f (x) − 3
, for all x ∈ R.
f (x) − 2
Prove that f is periodic.
Exercise 6.20. Let f : R → R be a periodic function such that the set {f (n) | n ∈ N}
has an infinite number of elements. Prove that the period of f is an irrational
number.
Exercise 6.21 (Long list IMO, 1977). Determine all the real continuous functions
f (x) defined on the interval (−1, 1), that satisfy the functional equation
f (x + y) =
f (x) + f (y)
,
1 − f (x)f (y)
for x + y, x, y ∈ (−1, 1).
Exercise 6.22. Find all the continuous functions f : R → R such that
f (x) + f (y) = f
x+y
1 − xy
, with x, y = 1.
Exercise 6.23. Find all the continuous functions f : R → R that satisfy
f (x + y) = f (x) + f (y) + f (x)f (y), for all x, y ∈ R.
(6.7)
108
Chapter 6. Functions and Functional Equations
6.4 Recommendations to solve functional equations
Next we will present a series of recommendations of the things we need to do
in order to find solutions of functional equations. Moreover, we will show some
examples where we use these observations.
Substituting the variables for values. One of the first steps that we need
to follow is to see if it is possible to determine some values of the function we
are looking for, for instance f (0), f (1), etc. In some cases, the values can be
found through direct substitution. But sometimes we may need to make a variable
interchange. For instance, if we found something like f (x + y), it is natural to
make y = −x, to obtain f (0).
Mathematical induction. We should have in mind that the principle of mathematical induction can help us. In these cases it is important to remember the
induction basis. For instance, to know what is f (1) or f (j), and then later to be
able to conjecture something more specific that could be a relation
that allows
us to go from n to n + 1. Also, try to find expressions like f n1 , and afterwards
expressions of the form f (r), with r ∈ Q. These situations, in general, can arise
when dealing with equations with variables in Q or in Z.
Basic properties of functions. It is important to know if the function is injective, surjective, bijective, periodic, even, odd, or with some kind of symmetry.
This can help us to reduce the cases and to concentrate only on the set of numbers
where the equation is valid.
Substitutions. Beside substitutions by specific values, we can try other more
general substitutions, for instance, x1 , x + 1, x + y, x − y.
Symmetry in the variables. If the equation has two (or more) variables, for
instance x, y, we will always try to substitute the y by the x (and vice versa), and
look always for symmetries in the variables.
Compare with the Cauchy equations. If our equation can be reduced or simplified to an equation of Cauchy type, then we have made good progress, since we
already know the solutions to this type of equations.
Continuity, monotonicity. Investigate if the unknown function is monotone
or continuous. This is very useful, since the problem could then be reduced to be
solved on the rational numbers or on some dense subset of the real numbers.
Other numeral systems. In functional equations where the natural numbers
are present, it can help us to work in another numeral system different from the
10-base system, for instance, moving to the binary system or the 3-base system.
Check. It is important to always check that the function we proposed as
solving the equation, really does it. We should never forget this part.
Now, we will exhibit examples where these recommendations are put to use.
6.4 Recommendations to solve functional equations
109
Example 6.4.1. Find all functions f : Q → R that satisfy the following conditions,
f (xy) = xf (y) + yf (x)
and
f (x + y) = f (x2 ) + f (y 2 ), for x, y ∈ Q.
If in the first equation we set x = y = 0, we obtain f (0) = 0, and taking
x = y = 1, we have that f (1) = 2f (1), that is, f (1) = 0.
On the other hand, taking x = 0 in the second equation, we get f (y) = f (y 2 ),
then the second equation becomes f (x + y) = f (x) + f (y) and we know, by the
Cauchy equation, that the function should be such that f (x) = f (1)x, for all
x ∈ Q. Moreover, since f (1) = 0 the only solution is f (x) = 0.
Example 6.4.2 (Short list IMO, 1988). Let f : N → N be a function that satisfies
f (f (m) + f (n)) = m + n, for all m, n.
Find all possible values of f (1988).
In Example 6.2.1, we showed that the function is injective. Moreover, for
l < n, we have that
f (f (m + l) + f (n − l)) = m + l + n − l = m + n = f (f (m) + f (n)).
(6.8)
The injectivity property tells us that
f (m + l) + f (n − l) = f (m) + f (n), for l, m, n ∈ N and l < n.
(6.9)
Now, by induction we will see that f (n) = n. First, for n = 1 let us see that
f (1) = 1. If b = f (1), then the following two equalities are true:
f (2b) = f (f (1) + f (1)) = 2
and f (b + 2) = f (f (1) + f (2b)) = 1 + 2b.
Then, b = 2 is not possible, since, on the one hand we would have that f (2 · 2) = 2
and on the other hand f (2 + 2) = 1 + 2 · 2 = 5.
Neither is b > 2 possible, since using f (2b) = 2, f (1) = b and equation (6.9),
it follows that
b + 2 = f (1) + f (2b)
= f (1 + b − 2) + f (2b − (b − 2))
= f (b − 1) + f (b + 2)
= f (b − 1) + 1 + 2b,
then f (b − 1) = 1 − b < 0, which is not possible. Therefore, b = f (1) = 1. Suppose now that f (n) = n, then from the original equation and from the induction
hypothesis, it follows that
n + 1 = f (f (n) + f (1)) = f (n + 1).
Therefore, the only possible value of f (1988) is 1988.
110
Chapter 6. Functions and Functional Equations
Example 6.4.3. Find all increasing or decreasing functions f : R → R such that
f (x + f (y)) = f (x) + y, for x, y ∈ R.
Letting x = y = 0, we get f (f (0)) = f (0). Since f is increasing or decreasing,
it follows that f is injective, then f (0) = 0.
Taking x = 0, we get f (f (y)) = y for all y ∈ R. Letting a = f (y), it follows
that f (f (y)) = y = f (a), then
f (x + y) = f (x + f (a)) = f (x) + a = f (x) + f (y),
hence, f satisfies the additive Cauchy equation. Moreover, with the condition
f (f (y)) = y, we have that the only solutions are f (x) = x and f (x) = −x, which
verify the functional equation.
Example 6.4.4. Find all continuous functions f : R → R that satisfy the equation
f (x + y) + f (x − y) = 2f (x), for x, y ∈ R.
Letting x = y, it follows that f (2x) = 2f (x) − f (0). Then, from the original
equation we obtain f (x + y) + f (x − y) = f (2x) + f (0); subtracting 2f (0) on both
sides, we get f (x + y) − f (0) + f (x − y) − f (0) = f (2x) − f (0).
Hence, f (x) − f (0) is additive, since taking u = x + y, v = x − y in the last
equation, leads to f (u) − f (0) + f (v) − f (0) = f (u + v) − f (0).
Since f (x)−f (0) is continuous, it follows that f (x) = f (0)+ax, for all x ∈ R.
Example 6.4.5. Let f : R → R be a continuous function such that f (x) = x has no
real solutions. Then f n (x) = x has no real solutions, where f n is the composition
of f with itself, n times, for any n ∈ N.
Since f (x) = x has no real roots, that is, there is no x ∈ R such that
f (x) − x = 0, then it is true that either f (x) > x or f (x) < x, for all x ∈ R. In
fact, since f (x) − x is continuous, and if f were positive on a point d and negative
on another point e, then, by the intermediate value theorem15 , there would be a
point x0 between d and e such that f (x0 ) − x0 = 0, which is impossible. Therefore,
f (x) > x for all x ∈ R or f (x) < x for all x ∈ R.
If f (x) > x for all x ∈ R, then
x < f (x) < f (f (x)) < · · · < f (f (· · · f (x) · · · )) < · · · ,
and therefore f (f (· · · f (x) · · · )) = x has no real solutions.
Similarly, if f (x) < x, we get that f (f (· · · f (x) · · · )) = x has no real solutions.
Exercise 6.24. Prove that there are no functions f : N → N that satisfy
f (f (n)) = n + 1, for all n ∈ N.
15 See
[21].
111
6.5 Difference equations
Exercise 6.25 (IMO, 1986). Find functions f : R+ ∪ {0} → R+ ∪ {0} such that
satisfy:
(a) f (xf (y))f (y) = f (x + y), for x, y ≥ 0.
(b) f (2) = 0.
(c) f (x) =
0, for all x such that 0 ≤ x < 2.
Exercise 6.26. Find all functions f : R → R such that
f (x − y) = f (x + y)f (y), for all x, y ∈ R.
6.5 Difference equations. Iterations
In this section we will study two kinds of functional equations: those relating the
values of f (x) and f (x+h) or, more generally, with those of f (x+nh) for some n ∈
N, which are called difference equations, and the functional equations relating f (x)
with its iterations, that is, with f 2 (x) = f (f (x)), . . . , f n (x) = f (f (. . . f (x) . . . )).
n times
For the difference equations, we will use the difference operator, denoted by
∆, and which is defined, for a function f : R → R, as
∆f (x) = f (x + h) − f (x),
(6.10)
for x ∈ R, and where h is a fixed real number. Also, we will use the operator E
defined by Ef (x) = f (x+ h), and the identity operator I defined by If (x) = f (x),
so that ∆ = E − I, that is, ∆f (x) = Ef (x) − If (x) = f (x + h) − f (x).
The following properties of the operators are seen to hold trivially.
Properties 6.5.1.
(a) ∆(af (x) + g(x)) = a∆f (x) + ∆g(x), for a fixed real number a.
(b) ∆(f (x) · g(x)) = Ef (x)∆g(x) + g(x)∆f (x).
g(x)∆f (x) − f (x)∆g(x)
f
, if g(x) = 0.
(c) ∆ (x) =
g
g(x)Eg(x)
(d) ∆m f (x) = ∆m−1 (∆f (x)), and also ∆m ∆n = ∆n ∆m = ∆m+n .
Lemma 6.5.2. For each integer number n ≥ 1, it follows that ∆n xn = hn n! and
∆m xn = 0 for m > n.
Proof. The proof is by induction on n.
If n = 1, we have that ∆x = (x + h) − x = h and ∆2 x = ∆h = 0, then
m
∆ x = 0 for m > 1.
112
Chapter 6. Functions and Functional Equations
Suppose that the result is true for all j < n, then
∆n xn = ∆n−1 (∆xn ) = ∆n−1 ((x + h)n − xn )
n−1
n−1
# n
# n
= ∆n−1
hi ∆n−1 (xn−i )
xn−i hi =
i
i
i=1
i=1
=
n
h∆n−1 (xn−1 ) =
1
n
hhn−1 (n − 1)! = hn n!,
1
and since ∆ applied to a constant is zero, we have that ∆m xn = 0 for m > n, so
the induction step is true and the result holds.
Example 6.5.3. If P (x) = a0 + a1 x + · · · + an xn is a polynomial of degree n, it
follows that ∆n P (x) = an n!hn and ∆m P (x) = 0 for m > n.
In fact, by the previous lemma
∆n P (x) = ∆n (a0 + a1 x + · · · + an xn )
= ∆n (a0 ) + ∆n (a1 x) + · · · + ∆n (an xn )
= an ∆n (xn )
= an n!hn .
It is clear that if m > n,
∆m P (x) = ∆m−n (∆n P (x))
= ∆m−n (an n!hn ) = 0.
In general, we have the following theorem.
Theorem 6.5.4. For any function f , it follows that
∆n f (x) = (E − I)n f (x) =
n
#
j=0
(−1)j
n
f (x + (n − j)h).
j
Proof. The proof is by induction. The case n = 1 has been already validated.
Suppose now the result holds true for n and let us see what happens for n + 1.
∆n+1 f (x) = ∆n (f (x + h) − f (x))
n
n
#
#
n
n
(−1)j
(−1)j
=
f (x + (n − j)h)
f (x + (n − j + 1)h) −
j
j
j=0
j=0
=
n
#
(−1)j
j=0
n
#
+
j=0
n
f (x + (n − j + 1)h)
j
(−1)j+1
n
f (x + (n + 1 − (j + 1))h)
j
113
6.5 Difference equations
n
#
n
n
(−1)j
f (x + (n + 1 − j)h)
f (x + (n + 1 − j)h) +
j
−
1
j
j=1
j=0
+
,
n
#
n
n
n
j
+
(−1)
=
f (x + (n + 1)h) +
j
j−1
0
j=1
=
n
#
(−1)j
f (x + (n + 1 − j)h) + (−1)n+1
=
n+1
#
(−1)j
j=0
n
f (x)
n
n+1
f (x + (n + 1 − j)h).
j
When difference equations are applied to functions with variables among the
non-negative integers and with h = 1, we get expressions of the form
∆f (0) = f (1) − f (0), ∆f (1) = f (2) − f (1), ∆f (2) = f (3) − f (2), . . . ,
which are known as sequences in finite differences or recurrent sequences, notions
that will be studied more carefully in Chapter 7.
Let us see an example dealing with iterations in the functional equation.
Example 6.5.5. Find functions f : N → N that satisfy
f (f (n)) + f (n)2 = n2 + 3n + 3, for n ∈ N.
It is easy to verify that f (n) = n + 1 satisfies the equation. Let us see that
this function is the only one that satisfies the equation.
If f (n) > n + 1, then f (n)2 ≥ (n + 2)2 , hence f (f (n)) = n2 + 3n + 3 − f (n)2 ≤
2
n + 3n + 3 − (n + 2)2 = −n − 1 < 0 which is absurd. Therefore f (n) ≤ n + 1.
If f (n) < n + 1, we have that f (n)2 < (n + 1)2 , and then f (f (n)) = n2 + 3n +
3 − f (n)2 > n2 + 3n + 3 − (n + 1)2 = n + 2 > f (n) + 1. Hence, f (f (n)) > f (n) + 1,
which is impossible as follows from the previous case. Therefore f (n) < n + 1
cannot hold and f (n) = n + 1 is the only solution.
Example 6.5.6. Find continuous functions f : R → R that satisfy the following:
For each x ∈ R, there exists an integer n ≥ 1 such that f n (x) = x.
(6.11)
First, let us see that the function is bijective. Suppose that for x, y ∈ R, we
have that f (x) = f (y), by property (6.11), there exist n, m ∈ N with f n (x) = x
and f m (y) = y. It is clear that f nm (x) = x and f nm (y) = y. But if f (x) = f (y),
then f nm (x) = f nm (y), hence x = y. Therefore, f is injective.
Next, we will prove that the function is surjective. For each x, there exists
n ∈ N with x = f n (x) = f (f n−1 (x)), remember that f 0 (x) = x.
Now, we will show that f is increasing or decreasing, which is highlighted in
the following more general lemma, not just for the function in the example.
114
Chapter 6. Functions and Functional Equations
Lemma 6.5.7. If f : R → R is continuous and bijective, then f is increasing in all
R or f is decreasing in all R.
Proof. Let us see that in every open interval the function is increasing or decreasing.
Let a, b ∈ R with a < b. Since f is injective, then f (a) < f (b) or f (a) > f (b).
If f (a) < f (b), we will prove that f is increasing in (a, b), consider x, y with
a < x < y < b.
(a) It must happen that f (a) < f (x) < f (b), otherwise f (x) < f (a) or
f (b) < f (x). In the first case, since f (x) < f (a) < f (b), by the intermediate value
theorem there exists x1 ∈ (x, b) with f (x1 ) = f (a), which is a contradiction to the
fact that f is injective. Similarly, if f (a) < f (b) < f (x), by the intermediate value
theorem there exists x2 ∈ (a, x) with f (x2 ) = f (b) which is impossible, since f is
injective. Thus, f (a) < f (x) < f (b).
(b) Similarly, we have that for y with x < y < b, it follows that f (x) <
f (y) < f (b). Therefore, f is increasing in (a, b).
The case when f (a) > f (b) is similar, except that in this situation f will be
decreasing.
Let us come back to the example. We will show that if f is increasing, then
f (x) = x for all x. In fact, if for some x0 , we have that f (x0 ) = x0 , then f (x0 ) > x0
or f (x0 ) < x0 . But, since f is increasing we have that
or
x0 < f (x0 ) < f 2 (x0 ) < · · · < f n (x0 ) < . . .
x0 > f (x0 ) > f 2 (x0 ) > · · · > f n (x0 ) > . . . ,
in any case, f n (x0 ) = x0 for every n ≥ 1, which contradicts (6.11).
But if f is decreasing, then f 2 (x) = x for all x, in fact, if f is decreasing, we
have that f 2 is increasing and then f 2 (x) = x.
Finally, we point out some properties that the function has: f 2 (x) = x, for
all x ∈ R, and it can be proved that the number n in (6.11) is 1 or 2.
Exercise 6.27. Find the sum
n
#
k=0
(−1)k k n
n
.
k
Exercise 6.28. Find all functions f : N ∪ {0} → N ∪ {0} that satisfy
f (f (f (n))) + f (f (n)) + f (n) = 3n, for all n ∈ N ∪ {0}.
Exercise 6.29. Find all the continuous functions f : [0, 1] → [0, 1] that satisfy
f (0) = 0, f (1) = 1 and f n (x) = x, for all x ∈ (0, 1) with n ∈ N fixed.
Exercise 6.30. Find all the continuous functions f : R → R such that there is a
natural number n ≥ 1 with f n (x) = −x, for all x ∈ R.
Chapter 7
Sequences and Series
7.1 Definition of sequence
A sequence of numbers {an } can be thought of as a function f defined on the set
of positive integers and whose images are a set of numbers A. This set can be:
natural, integer, rational, real or complex numbers, that is,
f : N
n
→ A
→ f (n) = an .
Sometimes it is useful to start the sequence with a0 . We call every element an of
the sequence a term of the sequence. We can also think of a sequence as an infinite
collection of ordered numbers.
In the mathematical olympiad, the problems related to sequences are of different kinds. In some of them it is asked to find specific terms of the sequence,
in others to prove that the terms are related in some particular way or that they
satisfy certain identities. Also, there are some problems that require one to find
a closed formula of the nth term or to prove that the nth term satisfies some
property. In the following examples, we present a variety of these situations, and
in the next section we will give some properties and characteristics that will help
us to achieve the goal we have just described.
The set of points an , with n = 1, 2, . . . , is called the range of the sequence. The
range of the sequence can be finite or infinite.
Example 7.1.1.
)
(
(a) For the sequence an = n1 , the range is infinite;
(
)
(b) If an = n2 , then this sequence has infinite range;
n
, its range is infinite.
(c) If an = 1 + (−1)
n
(d) The sequence {an = (−1)n } has finite range.
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5_7
115
116
Chapter 7. Sequences and Series
In this first example, the sequences exhibit certain orders or patterns, but not
all sequences are like these. In Chapter 2 we studied the arithmetic and geometric
progressions, which are examples of sequences that have a pattern or a rule that
can be given in an explicit way, but a sequence {an } such that an is the nth digit
in the decimal expression of π has no explicit rule.
Let us analyze several examples to get more familiar with sequences.
Example 7.1.2. The sequence a0 , a1 , a2 , . . . , is defined as a0 = 0, a1 = 1 and, for
2n
. Find the value of a1000 .
m ≥ n ≥ 0, am+n + am−n = a2m +a
2
If n = 0, we have that 2am = a2m2+a0 , then a2m = 4am . If m = 1 and n = 0,
2
= 4a22+4 = 10,
a2 = 4a1 = 4 = 22 . If m = 2 and n = 1, then a3 + a1 = a4 +a
2
2
hence a3 = 9 = 3 .
This suggests that an = n2 . We will use induction to prove this claim. In
fact, it is only left to check the inductive step,
4an + a2
a2n + a2
=
2
2
= 2n2 + 2 − (n − 1)2 = (n + 1)2 .
an+1 + an−1 =
an+1
Thus, a1000 = 10002.
Example 7.1.3. Define the sequence {an } as a1 = a2 = 1 and, for n ≥ 1,
an+2 = an+1 an + 1. Which elements of the sequence are even and which ones
are multiples of 4?
By induction we can prove that an is a positive integer. If an−1 and an−2 are
positive integers, then an = an−1 an−2 + 1 is also a positive integer. Let us see
which terms are even. We have that a3 = a2 a1 + 1 = 2 is even, but a4 and a5 are
not even; since by definition each of them is the sum of an even number and 1,
then a4 and a5 are odd. However, a6 is even and the formula for a7 and a8 , tell us
both are odd. Then, the sequence modulo 2 is 1, 1, 0, 1, 1, 0, . . . . The recursive
relation an+2 = an+1 an + 1 generates an odd number if one of the factors is even,
and an even number if both factors are odd. Then, the terms of the form a3k are
even.
If now we consider the sequence modulo 4, we see that it is given by 1, 1,
2, 3, 3, 2, 3, 3, . . . ; after the third term in the sequence, the numbers 2, 3, 3 are
repeated, which shows that there are no multiples of 4.
Example 7.1.4. The sequence {an } is defined by a1 = a2 = a3 = 1 and, for n ≥ 3,
n an−1
. Then every element of the sequence is an integer.
by an+1 = 1+aan−2
Observe that for n ≥ 3, the elements of the sequence satisfy an+1 an−2 =
1 + an an−1 , therefore an+2 an−1 = 1 + an+1 an .
Subtracting the first equation from the second, we get
an+2 an−1 − an+1 an−2 = an+1 an − an an−1 .
117
7.1 Definition of sequence
After factoring and rearranging the terms, we obtain
(an+2 + an )an−1 = (an + an−2 )an+1
an+2 + an
an + an−2
=
.
an+1
an−1
n−2
If we define bn = ana+a
, it follows that bn+2 = bn . That is, it happens that the
n−1
even terms of {bn } are all equal and the odd terms are also equal to each other.
Then, since
b3 =
a3 + a1
=2
a2
and b4 =
we have that
an =
a4 + a2
=
a1
1+a3 a2
a1
+ a2
a1
=
1+1
1
+1
= 3,
1
3an−1 − an−2 , if n is even
2an−1 − an−2 , if n is odd.
By induction, we can conclude that an is an integer number.
Example 7.1.5. The sequence {an } defined by a1 = 1 and an+1 = a2n + an , for
1
1
+ · · · + 1+a
< 1.
n ≥ 1, satisfies that for any n, 1+a
1
n
1
Since an+1 = a2n + an , it follows that an+1
= an (a1n +1) =
1
1
1
equivalent to an +1 = an − an+1 . Adding, we obtain
n
#
j=1
1
=
1 + aj
1
1
−
a1
a2
+ ···+
1
1
−
an
an+1
1
an
=1−
− an1+1 , which is
1
< 1.
an+1
Exercise 7.1 (Croatia, 2009). The sequence {an } is defined by
a1 = 1, a2 = 3, an = an−1 + an−2 , for n ≥ 3.
n
Prove that an < 74 , for all n.
Exercise 7.2 (Croatia, 2009). The sequence {an } is defined by
a1 = 1, an = 3an−1 + 2n−1 , for n ≥ 2.
Find a formula for the general term an in terms of n.
Exercise 7.3. The sequence {an } is defined by
a1 = 1 and an+1 = 1 + a1 a2 . . . an , for n ≥ 1.
Prove that for all n ≥ 1,
1
a1
+ ··· +
1
an
< 2.
118
Chapter 7. Sequences and Series
Exercise 7.4. The sequence {an } is defined by
a1 = a2 = 1
and
an+1 =
a2n + 1
,
an−1
for
n ≥ 2.
Prove that every term of the sequence is a positive integer.
Exercise 7.5 (MEMO, 2008). Let {an } be a sequence of positive integers such that
an < an+1 for n ≥ 1. Suppose that for all 4-tuples (i, j, k, l) of indices, such that
1 ≤ i < j ≤ k < l and i + l = j + k, it follows that ai + al > aj + ak . Find the
smallest possible value of a2008 .
Exercise 7.6 (China, 2008). A sequence of real numbers {an } is defined by a0 =
0, 1, a1 = 1 − a0 and an+1 = 1 − an (1 − an ), for n = 1, 2, . . . . Prove that for each
positive integer n,
(a0 a1 . . . an )
1
1
1
+
+ ···+
a0
a1
an
= 1.
Exercise 7.7. Let {xn } and {yn } be sequences defined by the equations
xn+1 = x3n − 3xn
and
yn+1 = yn3 − 3yn .
If x20 = y0 + 2, prove that x2n = yn + 2, for all n.
Exercise 7.8. The sequence {an } is defined by
a1 = 1, a2 = 12, a3 = 20 and an+3 = 2an+2 + 2an+1 − an , for n ≥ 1.
Prove that 1 + 4an an+1 is a perfect square, for n ≥ 1.
7.2 Properties of sequences
In this section we study some properties of the sequences that are useful to find
specific relations among terms of the sequences, find closed formulas, etc.
7.2.1 Bounded sequences
We say that a sequence {an } is bounded if there exist K > 0 such that |an | ≤ K,
for all n ∈ N. That is, we say that a sequence is bounded if its range is bounded.
For instance, it is clear that the sequences {an = n1 } and {an = (−1)n } are
bounded by 1. However, there are sequences for which the bound has to be found.
The most important example of a sequence that is not bounded is the following.
119
7.2 Properties of sequences
Example 7.2.1. The sequence {an = n} is not bounded, since the set of natural
numbers is not bounded.
Suppose that N is bounded above. Then, there exists M > 0 such that n ≤ M ,
for all n ∈ N. Take ⌊M ⌋ the greatest integer less than or equal than M , then the
integer ⌊M ⌋ + 1 satisfies that it is a positive integer with M < ⌊M ⌋ + 1, hence M
is not an upper bound for N, which is a contradiction.
Example 7.2.2. The sequence {an } given by 0 < a0 < a0 + a1 < 1 and
an − 1
= 0,
an−1
an+1 +
for n ≥ 1,
is a bounded sequence.
Let us find a few terms of the sequence:
a0 + a1 − 1
1 − a0
1 − a1
a2 =
, a3 =
, a4 =
, a5 = a0 , a6 = a1 .
a0
a0 a1
a1
Therefore, we see that the terms of the sequence are repeated every five terms,
then it is bounded.
In the last example, we can observe that the term a5 is equal to the term
a0 , and in general we have that an+5 = an , for all n ∈ N. The sequences with this
property have a special name, which will be studied next.
7.2.2 Periodic sequences
A sequence {an } is periodic, with period p ≥ 1, if it satisfies that an+p = an , for
all n ∈ N.
If a sequence is periodic with period p, then we can find all the values of the
sequence if we know the values of the first p terms of the sequence. Actually, if
{an } is a sequence with period p and n is a positive integer, by Euclid’s algorithm,
we can express n as n = pq + r, with 0 ≤ r < p. Then an = ar if r = 0, and
an = ap if r = 0. Also, observe that every periodic sequence is bounded, moreover,
the sequences with finite rank are clearly bounded.
Example 7.2.3. The sequence {an } is defined by an+2 =
a2013 .
1+an+1
.
an
Find the value of
We analyze the first terms of the sequence
a3 =
1 + a2
,
a1
a4 =
2
1 + 1+a
1 + a3
1 + a1 + a2
a1
=
=
,
a2
a2
a1 a2
1 +a2
1 + 1+a
(1 + a1 )(1 + a2 )
1 + a4
1 + a1
a1 a2
=
=
,
a5 =
=
1+a2
a1 a2 (1+a2 )
a3
a2
a
a1
1
1
1 + 1+a
1 + a5
a2
a6 =
= a1 ,
= 1+a1 +a
2
a4
a a
1 2
a7 =
1 + a6
1 + a1
= 1+a1 = a2 .
a5
a
2
120
Chapter 7. Sequences and Series
Then, the sequence is periodic with period 5, that is, an+5 = an for all n. Therefore,
2
a2013 = a3 = 1+a
a1 .
In some of the examples we have seen so far, we can notice that in order to
define the general term of a sequence it is necessary to know some of the previous
terms. We will come to this in the following section.
7.2.3 Recursive or recurrent sequences
Some of the sequences we have thus far studied satisfy the condition that the term
an+1 is a function of some of the previous terms, that is, an+1 = f (a1 , . . . , an ).
Sequences of this sort are known as recurrent sequences or recursive sequences.
More precisely, we will say that {an } satisfies the recursive equation
an+1 = f (a1 , . . . , an ),
(7.1)
if for every n, the terms of {an } satisfy the last identity. Note that the function f
is not the same for each n, for instance, if f (a1 , . . . , an ) = a1 + a2 + · · · + an we
have that a2 = f (a1 ) = a1 , a3 = f (a1 , a2 ) = a1 + a2 , . . . .
The simplest examples of recursive sequences are the arithmetic progressions
an = a1 + (n − 1)d that satisfy the recurrent equation an+1 = an + d and the
geometric progressions an = rn−1 a1 , that solve the recurrent equation an+1 = ran .
Example 7.2.4. A sequence that generalizes the arithmetic and geometric progressions is the sequence that solves the recursive equation, an+1 = rn an + dn , where
{rn } and {dn } are sequences independent of the terms an . Let us find a closed
formula for an .
It is evident that if for every integer n the equality an = rn−1 an−1 + dn−1
holds, then
an = rn−1 an−1 + dn−1
rn−1 an−1 = rn−1 rn−2 an−2 + rn−1 dn−2
rn−1 rn−2 an−2 = rn−1 rn−2 rn−3 an−3 + rn−1 rn−2 dn−3
..
..
.
.
rn−1 · · · r2 a2 = rn−1 · · · r2 r1 a1 + rn−1 · · · r2 d1 .
After adding and canceling terms, we get that
an = (rn−1 · · · r1 ) a1 +
n−2
#
j=1
rn−1 · · · rj+1 dj + dn−1 .
In particular, for an+1 = ran +d, it follows that an = rn−1 a1 +(1+r+· · ·+rn−2 )d.
In Example (3.1.4) of the Hanoi’s towers, we notice that the number of necessary movements hn to move n disks from one stick to another, satisfies the
121
7.2 Properties of sequences
recursive formula hn+1 = 2hn + 1. This formula confirms the value that we found
for hn , because it implies that,
hn = 2n−1 · h1 + (1 + 2 + · · · + 2n−2 ) · 1 = 2n−1 + 2n−1 − 1 = 2n − 1.
We say that {an } is a recurrent linear sequence of order k ≥ 1, if it satisfies
the recursive equation
an+k = c1 an+k−1 + c2 an+k−2 + · · · + ck an ,
where c1 , . . . , ck are constant numbers.
For instance, the Fibonacci sequence {fn }, defined as the sequence that satisfies the Fibonacci recursion formula fn+1 = fn−1 + fn , with f1 = f2 = 1, is
a recurrent linear sequence of order 2. A geometric progression is a recurrent
linear sequence of order 1, since an+1 = ran , and these sequences are the only
ones of order 1, where r can be any number. An arithmetic progression satisfies
an+1 = an + d, which is not a linear recursion because of the constant term d.
However, since an+2 = an+1 + d, it follows that an+2 − an+1 = an+1 − an , so
that an+2 = 2an+1 − an . Thus the arithmetic progressions are recurrent linear
sequences of order 2.
Our next objective is to solve the linear recursions of second order, that is,
we want to recognize the sequences that satisfy the recursive equation
an+2 = ban+1 + can ,
(7.2)
where b and c are fixed constants.
For instance, the recursive equation an+1 = 5an − 6an−1 is solved by the
sequences {2n } and {3n }, which shows that there is not always only one sequence
that solves the equation. Moreover, if {an } and {bn } are sequences that solve
a recursive linear equation, then also {Aan + Bbn } solves the equation, for any
numbers A and B. Hence there could be several sequences solving equation (7.2).
However, if we have that the first two terms of each of two sequences that
solve a linear recursion of order 2 are equal, then the two solutions coincide. This
follows because, if the first two terms of each solution coincide, then the third
terms coincide too and, by induction, all the terms of the two sequences coincide.
An application of this remark can be found in the following example.
Example 7.2.5. If {fn } is the Fibonacci sequence, then fm+n = fm fn−1 + fm+1 fn ,
for m ≥ 0 and n ≥ 1, where f0 = 0.
Define the sequences am = fm+n and bm = fm fn−1 + fm+1 fn , for m ≥ 0 and
with n ≥ 1 fixed. It is easy to show that am and bm satisfy the Fibonacci recursion
formula. For instance, am+2 = fm+2+n = fm+1+n + fm+n = am+1 + am and
bm+2 = fm+2 fn−1 + fm+3 fn = (fm+1 + fm )fn−1 + (fm+2 + fm+1 )fn
= (fm+1 fn−1 + fm+2 fn ) + (fm fn−1 + fm+1 fn ) = bm+1 + bm .
122
Chapter 7. Sequences and Series
On the other hand, a0 = fn , a1 = fn+1 , b0 = f0 fn−1 + f1 fn = fn and b1 =
f1 fn−1 + f2 fn = fn−1 + fn = fn+1 . Since both sequences satisfy the same linear
recurrence of order 2 and coincide in the first two terms, we have that the sequences
are equal. Therefore the identity holds.
Let us go back to see how to solve the linear recursions of order 2. Following
the idea that the linear recursions of order 1 are solved by sequences of the form
an = Aλn , let us see what a sequence of the form {an = Aλn } should satisfy in
order to be a solution of (7.2). Substituting in equation (7.2), we get
Aλn+2 = bAλn+1 + cAλn .
If A = 0, it is clear that the constant sequence
(7.2). If A = 0, we
an = 0 satisfies
can cancel A and after factoring we get λn λ2 − bλ − c = 0.
Now, if for some integer n we have that λn = 0, that is, λ = 0 and, therefore
an = 0, which we know solves the equation. Now, suppose λ = 0, hence
λ2 − bλ − c = 0,
so that λ =
an = Aλn .
√
b± b2 +4bc
2
(7.3)
are the only possible values if the solution is of the form
Equation (7.3) is known as the characteristic equation of the recursion formula (7.2) and the polynomial on the left is known as the characteristic polynomial. To conclude we analyze two cases. The first corresponding to the roots of
the characteristic equation being different and then the case when they are equal.
Case A. λ1 and λ2 are the solutions of equation (7.3), with λ1 = λ2 .
In this case, we notice that an = Aλn1 + Bλn2 solves equation (7.2). Now, let
us see that, if {bn } is a sequence that satisfies the equation, then bn = Aλn1 + Bλn2
for some numbers A and B.
For this, we know that it is enough to see that a0 = b0 and a1 = b1 . Then
we have to solve
a0 = A + B
a1 = Aλ1 + Bλ2 .
But this system of two equations with two unknowns can be solved in a unique
way for A and B. In fact,
A=
a0 λ2 − a1
λ2 − λ1
and
B=
a1 − λ1 a0
,
λ2 − λ1
and this is the only solution of the system when λ2 − λ1 = 0.
123
7.2 Properties of sequences
Case B. The roots of the characteristic polynomial λ1 and λ2 coincide.
In this case an = Aλn1 + Bλn2 is not a general solution anymore, since an =
(A + B)λn1 , and it is not always possible to choose A and B such that A + B = a0
and (A + B)λ1 = a1 .
However, there is another solution of the recursion formula different from λn1 ;
this happens to be sequence bn = (n + 1)λn1 . In order to see that this sequence
satisfies the recursion, first note that if λ1 = λ2 are the roots of λ2 − bλ − c = 0,
by Vieta, b = 2λ1 and c = −λ21 . Then, the recursion is
an+2 = 2λ1 an+1 − λ21 an
and we can verify that
(n + 3)λn+2
= 2λ1 (n + 2)λn+1
− λ21 (n + 1)λn1 ,
1
1
which proves that bn = (n + 1)λn1 solves the recursion formula.
Now, the two known solutions bn = λn1 and cn = (n + 1)λn1 generate the general solution an = Aλn1 + (n + 1)Bλn1 . In this case the initial conditions determine
A and B, that is, there is only one pair of numbers A and B with
a0 = A + B
a1 = (A + 2B)λ1 .
0 λ1
Actually, A = 2a0 λλ11−a1 and B = a1 −a
. Then, in this case, the solution with
λ1
initial conditions is unique.
We can summarize both cases in the following result.
Theorem 7.2.6.
(a) If the roots of the equation λ2 − bλ − c = 0 are different (b2 + 4c = 0), then all
the solutions an+2 = ban+1 + can , of the recursion formula are of the form
an = Aλn1 + Bλn2 ,
where A and B are any real numbers.
(b) If the equation λ2 − bλ − c = 0 has only one double real root equal to λ = 2b ,
then all solutions of the recursion are of the form
an = (A + (n + 1)B)λn ,
where A and B are any real numbers.
(c) If a0 and a1 are given numbers, then A and B are determined by a0 = A + B
and a1 = Aλ1 + Bλ2 in case (a), and by a0 = A + B and a1 = (A + 2B)λ in
case (b).
For instance, the recursion xn+2 = 2xn+1 − xn , has characteristic polynomial
λ2 − 2λ + 1, with λ = 1 as the only root. Then, the solutions are of the form
xn = (c + dn)1n = c + dn, something we already know as arithmetic progressions.
124
Chapter 7. Sequences and Series
Example 7.2.7. Find the solutions of the Fibonacci recursion, f0 = 0, f1 = 1 and
fn+2 = fn+1 + fn , for n ≥ 0.
The characteristic
equation is given by λ2 − λ − 1 = 0 and its roots are
√
λ1 , λ2 = 1±2 5 , which are different. Then, the solutions of the recursion are of the
$ √ %n
$ √ %n
form fn = Aλn1 + Bλn2 = A 1+2 5
+ B 1−2 5 . Because the first terms are
f0 = 0 and f1 = 1, then A =
1
fn = √
5
√1
5
= −B. Hence the Fibonacci numbers fn are
√ n
√ n 5
1+ 5
1− 5
.
−
2
2
Example 7.2.8. Find the solutions of the recursion defined by a0 = 0, a1 = sin α
and an+2 = 2 cos α · an+1 − an , for n ≥ 0 and α = nπ.
The characteristic polynomial of the given recursion is λ2 − 2 cos α λ + 1 = 0,
which has solutions
√
2 cos α ± 4 cos2 α − 4
= cos α ± i sin α.
λ1 , λ2 =
2
Hence, the solutions of the recursion are of the form an = Aλn1 + Bλn2 . From the
initial conditions we obtain that a0 = A+B = 0 and a1 = Aλ1 +Bλ2 = sin α. From
the first equation we have that B = −A, so that A(λ1 − λ2 ) = A(2i sin α) = sin α,
1
. Therefore,
and then, using the fact that sin α = 0, A = 2i
1
{(cos α + i sin α)n − (cos α − i sin α)n }
2i
1
= (cos nα + i sin nα − cos nα + i sin nα)
2i
= sin nα.
an =
Example 7.2.9. Analyze the non-linear recurrent equation an+1 = a2n − 2.
n
n
It is clear that if a0 = a + a1 , then an = a2 + a−2 solves the recurrence,
n
n 2
n+1
n+1
since a2n − 2 = a2 + a−2
− 2 = a2
+ a−2
= an+1 . If |a0 | > 2, then
√ 2
a0 + a0 −4
1
a=
satisfies a0 = a + a , if |a0 | ≤ 2, so that we can take a0 = 2 cos θ,
2
and therefore a = eiθ = cos θ + i sin θ.
7.2.4 Monotone sequences
A sequence {an } of real numbers is monotone increasing if
an ≤ an+1 ,
for all n ∈ N.
The sequence is increasing if an < an+1 , for all n ∈ N.
125
7.2 Properties of sequences
Similarly, we say that a sequence {an } of real numbers is monotone decreasing if
an ≥ an+1 ,
for all n ∈ N.
The sequence is decreasing if an > an+1 , for all n ∈ N.
Example 7.2.10.
(a) Any arithmetic progression with difference d > 0 is an increasing sequence.
(b) Any geometric sequence with a1 > 0 is monotone increasing if r ≥ 1.
In (a), if {an } is the arithmetic progression with difference d, then an+1 −an =
d > 0, where clearly an < an+1 .
In (b), if {an } is a geometric progression with ratio r ≥ 1, then is clear that
an+1 = ran ≥ an .
Example 7.2.11. The sequence defined by 0 < a1 <
increasing.
1
2
and an+1 = 2an (1 − an ) is
In order to see that it is increasing, we will need to show that 0 < an and
= 2(1 − an ), that is, 0 < an < 12 for all n.
Let us proceed by induction. First note that if 0 < an < 21 , then an+1 =
2an (1 − an ) > 0. Now using the geometric mean and the arithmetic mean inequality, we have that
1<
an+1
an
an+1 = 2an (1 − an ) ≤ 2
an + 1 − an
2
2
=
1
.
2
The inequality is strict since the equality holds when an = 1 − an = 12 , which is
not the case in the previous expression.
7.2.5 Totally complete sequences
A sequence {an } of positive integers is called totally complete if every positive
integer can be expressed as a sum of one or more different terms of the sequence.
Clearly the sequence of positive integers {1, 2, 3, . . . , n, . . . } is totally complete.
Example 7.2.12. The sequence of powers of 2, {20 , 21 , 22 , . . . , 2n , . . . } is totally
complete.
We will give a proof using strong induction. The basis cases are evident, since
1 = 20 and 2 = 21 . Suppose that any integer less than n can be written as a sum
of different powers of 2. Let 2m be the greatest power of 2 that is less than or equal
to n, then 2m ≤ n < 2m+1 . Set d = n − 2m , which is clearly less than or equal to
n, and also less than 2m (n < 2m+1 implies that n − 2m < 2m ). By the induction
hypothesis, d can be expressed as the sum of different powers of 2 and, since d is
less than 2m , this power is not included in the representation of d. Adding 2m to
the representation of d we get the representation of n.
126
Chapter 7. Sequences and Series
Proposition 7.2.13. A sequence of positive integers {an } that satisfies a1 = 1 and
an+1 ≤ 1 + a1 + a2 + · · · + an , for all n = 1, 2, . . . ,
is totally complete.
Proof. We will show, using induction on n, that every integer k less than or equal
to a1 + a2 + · · · + an , is the sum of different terms of {a1 , a2 , .., an }.
If n = 1, the only k ≤ a1 = 1 is k = 1 and clearly k = a1 = 1.
If n = 2, the numbers k to consider are the ones that satisfy k ≤ a2 ≤ 1 + a1 . Since
a2 ≤ 1 + a1 = 2, it follows that a2 = 1 or 2. If a2 = 1, k = 1 or 2, then k = 1 = a1
and k = 2 = a1 + a2 . If a2 = 2, k = 1, 2 or 3, then k = 1 = a1 , k = 2 = a2 and
k = 3 = a1 + a2 .
For the inductive step, suppose the statement true for n and let us prove it for
n + 1. Consider a positive integer k that satisfies k ≤ a1 + · · · + an + an+1 . If k ≤
a1 +· · ·+an , by the induction hypothesis, such k is the sum of different elements of
{a1 , a2 , . . . , an }. Suppose then that 1+a1 +· · ·+an ≤ k ≤ a1 +· · ·+an +an+1 . Then,
since an+1 ≤ 1 + a1 + a2 + · · · + an , it follows that an+1 ≤ k ≤ a1 + · · · + an + an+1 ,
hence 0 ≤ k − an+1 ≤ a1 + · · · + an . Now, if k − an+1 = 0 the proof is finished, and
if 0 < k − an+1 ≤ a1 + · · · + an , then, by the induction hypothesis, k − an+1 is the
sum of different elements of the set {a1 , a2 , . . . , an }. Adding an+1 , we have that k
is the sum of different elements of {a1 , . . . , an , an+1 }, as we wanted to prove.
Example 7.2.14. The Fibonacci sequence 1, 1, 2, 3, 5, . . . , fn , . . . is totally complete.
By the last proposition, it is enough to see that fn+1 ≤ 1 + f1 + f2 + · · · + fn , for
every n ≥ 1. For n = 1, it is clear since 1 = f2 ≤ 1 + f1 = 2. For n = 2, the result
follows because 2 = f3 ≤ 1 + f1 + f2 = 3. For n ≥ 3, it is enough to show that
1 + f1 + f2 + · · · + fn − fn+1 ≥ 0. But since fn+1 = fn−1 + fn , it follows for n ≥ 3
that
1 + f1 + f2 + · · · + fn − fn+1 = 1 + f1 + f2 + · · · + fn − (fn−1 + fn )
= 1 + f1 + f2 + · · · + fn−2 ≥ 0.
7.2.6 Convergent sequences
A sequence {an } converges to or has limit a if for all ǫ > 0 there exists a natural
number N such that, for all n ≥ N , it follows that |an − a| < ǫ. This can be
written as,
∀ ǫ > 0, ∃ N ∈ N such that, ∀ n ≥ N, |an − a| < ǫ.
We say that a is the limit of the sequence and write limn→∞ an = a.
A sequence diverges if it does not converge to any point a.
127
7.2 Properties of sequences
Example 7.2.15. The sequence {an =
1
n}
converges to 0.
Given ǫ > 0 we would like to show the existence of N ∈ N such that, for
all n ≥ N , it implies that 0 < n1 < ǫ. But if n ≥ N , then n1 ≤ N1 , therefore it is
enough to show the existence of N with N1 < ǫ. If such N does not exist, that is
if N1 ≥ ǫ, then N ≤ 1ǫ for all N ∈ N. This is a contradiction, because the natural
numbers are not bounded.
Example 7.2.16. The sequence {an = an } converges to 0 if |a| < 1.
If |a| < 1, then
1
|a|
> 1, hence
1
|a|
n
= 1 + p with p > 0. Thus, (1 + p)n =
1
1
1 + np + n(n−1)
p2 + · · · ≥ np, and |a| = (1+p)
n ≤ np . Choose N ∈ N such that if
2
1
1
n
n ≥ N , then n1 < pǫ, so that |a|n = (1+p)
n ≤ np < ǫ. Thus {a } converges to 0.
7.2.7 Subsequences
Given a sequence {an }, consider a sequence {nk } of positive integers, such that
n1 < n2 < n3 < · · · . The sequence {ank } is called a subsequence of {an }. Note
that given a sequence, we can obtain an infinite number of subsequences from it.
Example 7.2.17. The sequence {an } defined by
1
if n is even,
n,
an =
2
n , if n is odd,
has subsequences that are convergent, not convergent, bounded, not bounded, increasing and decreasing.
First note that the sequence is not convergent, and neither bounded nor monotone,
but we can find the following subsequences:
(a) If we take {a2n }, the sequence is convergent, decreasing and bounded, since
1
for all n ∈ N.
the terms are given by 2n
(b) If we take {a2n−1 }, the sequence is not convergent, increasing and not bounded, since the terms are given by (2n − 1)2 , for all n ∈ N.
Exercise√7.9. Prove that the sequence {an } defined by a0 = 0 and for n ≥ 0,
an+1 = 4 + 3an , is a bounded sequence.
Exercise 7.10. A sequence {an } is defined by
a1 = 1, an+1 = an +
1
, for n ≥ 1.
a2n
Determine if the sequence is bounded or not and prove that a9000 > 30.
Exercise 7.11. The terms of the sequence {an } are positive and a2n+1 = an + 1,
for all n. Prove that the sequence contains at least one irrational number.
128
Chapter 7. Sequences and Series
Exercise 7.12. Find, in each case, the solutions of the recursive equation of degree
3, an+3 = 3an+2 − 3an+1 + an , if:
(i) a1 = a2 = a3 = 1.
(ii) a1 = 1, a2 = 2, a3 = 3.
(iii) a1 = 1, a2 = 4, a3 = 9.
Exercise 7.13. The positive integers a1 , a2 , . . . , are bounded and form a sequence
that satisfies the following condition: if m and n are positive integers, then am +an
is divisible by am+n . Prove that the sequence is periodic after some terms.
Exercise 7.14. Prove that an = n!, the number of permutations of n elements,
and dn , the number of permutations of n elements without fixed points, satisfy the
recursive equation xn+1 = n(xn + xn−1 ). Why is an = dn for all n?
Note: a permutation without fixed points is called a derangement.
Exercise 7.15.
(i) Use the recursive formula for the number of derangements of a set of n elements, given by dn = (n−1)(dn−1 +dn−2 ), to prove that dn = ndn−1 +(−1)n .
(ii) Use Example 7.2.4 to justify the formula
dn = n! 1 +
(−1)2
(−1)n
(−1)1
+
+ ···+
1!
2!
n!
.
Exercise 7.16. Lucas’ numbers are defined by the recurrence L1 = 1, L2 = 3 and
Ln+1 = Ln + Ln−1 , for n ≥ 2. Prove that
√ n
√ n
1+ 5
1− 5
Ln =
+
.
2
2
Exercise 7.17. Solve the recursive equation bn+1 =
b0 is a fixed positive number.
bn
1+bn ,
for n = 0, 1, 2, . . . , where
Exercise 7.18. Prove that the sequence defined by a0 = a = 1 and an+1 =
is periodic.
1
1−an ,
Exercise 7.19. Solve the recursive equation an+1 = 4− a4n . Prove that an converges
to 2.
Exercise 7.20. Prove that a sequence {an } that satisfies a1 = 1 and an+1 ≤ 2an ,
for n ≥ 1, is totally complete.
Exercise 7.21. Prove that the sequence {1, 2, 3, 5, 7, 11, 13, . . .} of all prime numbers and the number 1, is totally complete.
129
7.3 Series
Exercise 7.22. Two brothers inherit n golden pieces with total weight 2n. Each
piece has an integer weight and the heaviest of all the pieces does not weigh more
than all the others together. Prove that if n is even, then the brothers can divide
the golden pieces into two parts with equal weight.
Exercise 7.23 (Romania, 2009). A sequence {an } is defined by
a1 = 2, an+1 =
an +
1
, for n ≥ 1.
n
Prove that the limit of the sequence exists and is 1.
7.3 Series
Given a sequence {an }, to denote the sum ap + ap+1 + · · · + aq with p ≤ q, we use
the notation
q
#
an .
n=p
To a sequence {an } we associate the sequence {sn }, called the sequence of partial
sums, given by
n
#
ak .
sn =
k=1
The infinite sum a1 + a2 + a3 + · · · can be written in short form as
∞
#
an = lim
n=1
N →∞
N
#
an = lim sN .
n=1
N →∞
This last expression is called infinite series or, more simply, series. If {sn } converges
to s, we say that the series converges to s and we write
∞
#
an = s.
n=1
The number s is called the sum of the series.
We say that the series diverges if {sn } diverges.
Example 7.3.1. Find the sum
∞
#
1
.
n(n + 1)
n=1
130
Chapter 7. Sequences and Series
The partial sum is
1
1
1
+
+ ··· +
1·2 2·3
n · (n + 1)
1 1
1
1
1
−
−
+
+ ···+
= 1−
2
2 3
n n+1
sn =
=1−
1
.
n+1
1
Since { n+1
} → 0, when n → ∞, then {sn } → 1. Thus
∞
*
n=1
1
= 1.
n(n + 1)
Exercise 7.24. If {an } is an arithmetic progression with difference d = 0, prove
that:
n
*
1
1 1
1
(i)
=
−
.
a
·
a
d
a
a
i
i+1
0
n+1
i=0
n
*
1
1
1
1
=
−
.
(ii)
2d a0 · a1
an+1 · an+2
i=0 ai · ai+1 · ai+2
∞
∞
*
*
1
1
1
1
=
.
(iv)
=
.
(iii)
da0
2d · a0 · a1
i=0 ai · ai+1
i=0 ai · ai+1 · ai+2
Exercise 7.25. Let fn be the Fibonacci sequence (f1 = 1, f2 = 1, fn+1 = fn +fn−1 ).
Find the sum of the following series:
∞
∞
*
*
1
fn
(ii)
.
(i)
n=2 fn−1 fn+1
n=2 fn−1 fn+1
Exercise 7.26. The Koch snowflake is obtained by means of the following process:
(i) In step 0, the curve is an equilateral triangle of side 1.
(ii) In step n + 1, the curve is obtained from the curve in step n, dividing each
one of the sides in 3 equal parts, and constructing externally in the middle
part of the divided side, an equilateral triangle erasing the side in which it
was constructed. In the following diagram steps 0 and 1 are shown.
If Pn and An are the perimeter and the area, respectively, of the curve of step n,
find:
(i) Pn
(ii) An
(iii) lim Pn
n→∞
(iv) lim An .
n→∞
131
7.3 Series
Exercise 7.27. Consider the harmonic progression
∞
(1)
*
n . The series
n=1
as the harmonic series.
1
n
is known
1
1
1
1
+
+ · · · + n+1 > .
2n + 1 2n + 2
2
2
1
1
1
(ii) Prove that for n ≥ 2, +
+ · · · + 2 > 1.
n n+1
n
1
1
3
1
+ +
> .
(iii) Prove that for n ≥ 2,
n−1 n n+1
n
(iv) Use any of the previous inequalities to conclude that the harmonic series is
divergent.
(i) Prove that for n ≥ 1,
7.3.1 Power series
The following expression is known as formal power series in the variable x with
center at zero,
f (x) = a0 + a1 x + a2 x2 + · · · + an xn + · · · ,
(7.4)
where a0 , a1 , a2 , . . . is an arbitrary sequence of numbers. The power series can be
written as
∞
#
f (x) =
an xn .
n=0
Let us see some examples of how to calculate these series.
Example 7.3.2. The sum of the geometric series is given by
∞
#
axn =
n=0
a
, if
1−x
|x| < 1.
Consider the partial sum
sn = a + ax + ax2 + · · · + axn
= a(1 + x + · · · + xn )
=a
1 − xn+1
1−x
Then, limn→∞ sn = limn→∞
a
n
n=0 ax = 1−x .
*∞
16 See
Example 7.2.16.
a(1−xn+1 )
1−x
.
=
a
1−x ,
since16 |x| < 1. Therefore
132
Chapter 7. Sequences and Series
Example 7.3.3. The following series, known as the derivative series of the geometric series, converges to the given value
∞
#
nxn−1 =
n=1
1
, for |x| < 1.
(1 − x)2
*n
Consider the partial sum k=0 kxk = x + 2x2 + 3x3 + · · · + nxn . This partial
sum is the sum of the following partial sums:
x + x2 + x3 + · · · + xn = x
x2 + x3 + · · · + xn = x2
x3 + · · · + xn = x3
..
.
..
.
xn = xn
1 − xn
1−x
1 − xn−1
1−x
1 − xn−2
1−x
1−x
1−x
.
Adding these sums, we get
x(1 − xn ) + x2 (1 − xn−1 ) + · · · + xn (1 − x)
1−x
x + x2 + · · · + xn − nxn+1
=
1−x
$
%
1−xn
x 1−x − nxn+1
=
1−x
x(1 − xn ) nxn+1
.
(7.5)
−
=
(1 − x)2
1−x
x + 2x2 + 3x3 + · · · + nxn =
The value of the infinite sum is
∞
#
nxn = lim
n→∞
n=1
x(1 − xn ) nxn+1
−
(1 − x)2
1−x
=
x
,
(1 − x)2
(7.6)
since xn+1 and nxn+1 go to zero as n → ∞, because |x| < 1. Therefore, canceling
one x on both sides, we have the desired equality.
Example 7.3.4. The sum of the following series is given by
∞
#
n=0
n(n − 1)xn =
2x2
, for |x| < 1.
(1 − x)3
133
7.3 Series
Consider the partial sum
sn =
n
#
k=0
k(k − 1)xk = 2x2 + 3 · 2x3 + 4 · 3x4 + · · · + n(n − 1)xn
= 2x x + 3x2 + 2 · 3x3 + · · · +
n(n − 1) n−1
x
.
2
We can write the factor in parentheses as
x + 3x2 + 6x3 + · · · +
n(n − 1) n−1
= x + x2 + x3 · · · + xn−1
x
2
+ 2x2 + 2x3 + · · · + 2xn−1
+ 3x3 + · · · + 3xn−1
..
.
=x
=
1 − xn−1
1−x
+ 2x2
1 − xn−2
1−x
+ (n − 1)xn−1
+ · · · + (n − 1)xn−1
1−x
1−x
[x + 2x2 + · · · + (n − 1)xn−1 ] − [xn + 2xn + · · · + (n − 1)xn ]
.
1−x
By equation (7.5) and the Gauss sum17 , we have that this sum is equal to
&
' & $
%'
x(1−xn−1 )
(n−1)xn
n n(n−1)
−
x
−
2
(1−x)
(1−x)
2
.
(7.7)
1−x
Therefore the value of the series is
⎛& x(1−xn−1 )
∞
−
#
(1−x)2
n(n − 1)xn = lim 2x ⎝
n→∞
n=0
(n−1)xn
(1−x)
'
1−x
& $
%'⎞
− xn n(n−1)
2
⎠=
2x2
.
(1 − x)3
7.3.2 Abel’s summation formula
The sums calculated in some of the previous examples in this section can be
simplified using what is known as the Abel’s summation formula.
Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be two finite sequences of numbers.
Then
n
#
ai b i =
i=1
n−1
#
i=1
(ai − ai+1 )(b1 + · · · + bi ) + an (b1 + b2 + · · · + bn ),
which can be proved simplifying the right-hand side of the identity.
17 See
Section 2.1.
134
Chapter 7. Sequences and Series
Example
7.3.5. Using the Abel’s summation formula, find the value of the sum
*n
k−1
, for q = 1.
k=1 kq
Using the Abel’s summation formula, we obtain that
n
#
kq k−1 =
k=1
n−1
#
(k − (k + 1))(1 + q + · · · + q k−1 ) + n(1 + q + · · · + q n−1 )
k=1
n−1
#
=−
k=1
qk − 1
+n
q−1
qn − 1
nq n
−
.
=
q − 1 (q − 1)2
qn − 1
q−1
=−
n−1
1 # k n−1
+n
q +
q−1
q−1
qn − 1
q−1
k=1
Example 7.3.6 (Rearrangement inequality). If a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤
· · · ≤ bn are two collections of real numbers in increasing order and (b′1 , b′2 , . . . , b′n )
is a permutation of (b1 , b2 , . . . , bn ), then it follows that
a1 b1 + a2 b2 + · · · + an bn ≥ a1 b′1 + a2 b′2 + · · · + an b′n .
Apply the Abel’s summation formula to the difference of the sums
n
#
i=1
ai b i −
n
#
ai b′i =
n
#
i=1
i=1
=
n−1
#
i=1
=
n−1
#
i=1
ai (bi − b′i )
(ai − ai+1 )
i
#
j=1
⎛
(ai − ai+1 ) ⎝
bj −
i
#
j=1
i
#
+ an
i
#
j=1
*i
n
#
j=1
j=1
bj −
since for every i = 1, . . . , n − 1, ai ≤ ai+1 and
b′j
⎞
bj −
n
#
b′j
j=1
b′j ⎠ ≥ 0,
j=1 bj
≤
*i
′
j=1 bj .
Exercise 7.28. Find, using Abel’s summation formula, the value of the sum
n
#
k=1
k 2 q k−1 , with q = 1.
Exercise 7.29. Prove that the following series converges to the given value:
∞
#
n=0
n2 xn =
2x2
x
+
, for |x| < 1.
3
(1 − x)
(1 − x)2
Exercise 7.30. Find the sum of the following series:
n
n
∞
∞ 2
∞ n
∞
*
*
*
*
−1
3
. (ii)
. (iii)
. (iv)
.
(i)
n
n
3
n=1
n=1 3
n=1 2
n=0 4
(v)
∞ n2
*
.
n
n=1 2
7.4 Convergence of sequences and series ⋆
135
7.4 Convergence of sequences and series ⋆
In this section, we shall present the proofs of the convergence theorems and their
properties. However, you can continue reading the book without studying this
section.
Remember that a sequence {an } converges to or has a limit a if
∀ ǫ > 0, ∃ N ∈ N such that ∀ n ≥ N ⇒ |an − a| < ǫ.
If a is the limit of the sequence, we write limn→∞ an = a or briefly an → a.
If for any a ∈ R the sequence {an } does not converge to a, we will say that the
sequence diverges.
Theorem 7.4.1. If a sequence {an } converges to a1 and to a2 , then a1 = a2 . That
is, the limit is unique.
Proof. If a1 = a2 , take ǫ = 21 |a1 − a2 | > 0. Since {an } converges to a1 and to a2 ,
there exist N1 and N2 ∈ N such that
|an − a1 | < ǫ, if n ≥ N1
and |an − a2 | < ǫ, if n ≥ N2 .
For n ≥ N = max(N1 , N2 ), it follows that
|a1 − a2 | ≤ |a1 − an | + |a2 − an | < 2ǫ = |a1 − a2 |,
which is a contradiction. Hence, a1 = a2 .
Next we present some properties of limits of sequences.
Theorem 7.4.2. If limn→∞ an = a, limn→∞ bn = b and α is any real number, then:
(a) lim (an + bn ) = a + b.
n→∞
(b) lim αan = α a.
n→∞
(c) lim an bn = ab.
n→∞
(d) If b = 0, then for bn = 0 and n large enough it happens that
lim
n→∞
1
1
= ,
bn
b
lim
n→∞
an
a
= .
bn
b
Proof. We will prove only parts (a) and (d); the rest is left as an exercise for the
reader.
(a) Since limn→∞ an = a and limn→∞ bn = b, then there exist N1 , N2 ∈ N such
that
ǫ
ǫ
|an − a| < , for n ≥ N1 and |bn − b| < , for n ≥ N2 .
2
2
If N = max(N1 , N2 ), then for n ≥ N it follows that
|an + bn − (a + b)| ≤ |an − a| + |bn − b| ≤
ǫ
ǫ
+ = ǫ.
2 2
136
Chapter 7. Sequences and Series
(d) Since b = 0, it follows that |b|
2 > 0. Since bn → b, there exists N1 ∈ N such
|b|
|b|
that, for all n ≥ N1 , |bn − b| < |b|
2 , then |b| − |bn | ≤ |bn − b| < 2 and |bn | > 2 .
2
1
Hence |bn | < |b| , for all n ≥ N1 .
Let ε > 0, since bn → b, considering
all n ≥ N , |bn − b| <
Observations 7.4.3.
|b|2
2 ε.
|b|2 ε
2
there exists N ≥ N1 such that, for
Hence, for all n ≥ N ,
1
− 1 = |bn − b| ≤ 2 |bn − b| < ε.
2
bn
b
|b| |bn |
|b|
(a) If {an } converges to a, then any open interval that contains the number a has
an infinite number of terms of the sequence. Moreover, outside this interval
there are only a finite number of terms of the sequence.
(b) If an converges to a, then the sequence is bounded.
Part (a) follows since there exists ǫ > 0, with I = (a − ǫ, a + ǫ), contained in
the open interval, but in I there are an infinite number of terms of {an }.
In order to prove (b), observe that the interval (a − ǫ, a + ǫ) contains all the
terms an with n ≥ N for some N , then |an | < |a| + ǫ. Therefore, if we define
K = max{|a| + ǫ, |a1 |, . . . , |an |}, it is clear that K is a bound for the sequence,
that is, |an | ≤ K for all n ∈ N.
Theorem 7.4.4. Let {an }, {bn } and {cn } be three sequences of real numbers. Suppose that there is N ∈ N such that, for all n ∈ N with n ≥ N , an ≤ bn ≤ cn holds.
If {an } and {cn } converge to the same limit a, then {bn } converges to a.
Proof. Let ǫ > 0, by hypothesis there exist N1 , N2 ∈ N such that:
|an − a| < ǫ, for n ≥ N1
and |cn − a| < ǫ, for all n ≥ N2 .
If N0 = max(N, N1 , N2 ), then for all n ≥ N0 , it follows that
−ǫ < an − a ≤ bn − a ≤ cn − a < ǫ,
which implies that |bn − a| < ǫ, that is, {bn } converges to a.
Theorem 7.4.5. A sequence {an } converges to a if and only if any subsequence of
{an } converges to a.
Proof. If the sequence converges to a, then limk→∞ ak = a, that is, for all ǫ > 0
there exists N ∈ N such that if k > N , then |ak − a| < ǫ. Let nk be an increasing
sequence of positive integers and consider the subsequence {ank }. Since nk ≥ k
and if k > N , then |ank − a| < ǫ, hence the subsequence converges.
Suppose now that the sequence does not converge to a, then there exists
ǫ > 0 such that, for all N ∈ N, there exists n > N such that |an − a| > ǫ. For
7.4 Convergence of sequences and series ⋆
137
N = 1, there is n1 > 1 such that |an1 − a| > ǫ. For n1 , there is n2 > n1 such that
|an2 − a| > ǫ. For n2 , there is n3 > n2 such that |an3 − a| > ǫ. Proceeding in the
same way, we construct a subsequence {ank } that does not converge to a, which
is a contradiction.
Theorem 7.4.6. If {an } is bounded, then there exists a subsequence of {an } that
converges.
Proof. In order to prove this theorem, we need to construct a convergent subsequence. Since the sequence is bounded, we know that there is M > 0 such that
|an | ≤ M , for all n, that is, −M ≤ an ≤ M . Divide the closed interval18 [−M, M ]
into two intervals [−M, 0] and [0, M ]. Consider the interval where there are a infinite number of terms of the sequence; suppose that this interval is [0, M ]. To
construct the subsequence choose one of the terms in the interval [0, M ], say an1 .
M
Again, divide the interval [0, M ] into two intervals [0, M
2 ], [ 2 , M ] and choose the
interval that contains an infinite number of terms of the sequence. Without loss of
generality, we can assume that the interval is [ M
2 , M ]. Choose as a second element
of the subsequence one term an2 such that n2 > n1 in the interval [ M
2 , M ]. Continuing this process, we will get a subsequence {ank } and a collection of nested
closed intervals of lengths 2Mn . The infinite intersection of these closed intervals is
not empty, in fact, it is a unique point19 , say a. This point a is the limit of the
M
subsequence, since |a − ank | < M
2k , moreover |a − anl | < 2k , for l ≥ k.
Theorem 7.4.7. Every upper bounded, increasing sequence of real numbers (monotonically increasing) is convergent. Similarly, every lower bounded decreasing sequence of real numbers (monotonically decreasing) is convergent.
Proof. Let {an } be an upper bounded increasing sequence. Since the sequence
is bounded, Theorem 7.4.6 implies that there exists a subsequence {ank } that
converges to a point a, that is, given ǫ > 0 there exists K ∈ N such that for all
k ≥ K, it follows that |ank − a| < ǫ. Let N = nK , we would like to show that
for all n ≥ N it follows that |an − a| < ǫ. Since nk ≤ n there is j ≥ k such that
nK < · · · < nj ≤ n < nj+1 . Using that the sequence {an } is increasing, we have
that anj − a ≤ an − a < anj+1 − a, so that |an − a| < ǫ. The other cases are
similar.
Theorem 7.4.8. Let f be a function, then limx→a f (x) = b if and only if for every
sequence {an } such that limn→∞ an = a, it follows that limn→∞ f (an ) = b.
Proof. Suppose that limn→∞ an = a. Since limx→a f (x) = b, it follows that given
ǫ > 0 there exists δ > 0 such that if |x − a| < δ, then |f (x) − b| < ǫ. By
the convergence of {an } to a, for δ > 0, there is N ∈ N, such that if n ≥ N ,
18 See
19 See
Section 1.2, for the definition of interval.
[17].
138
Chapter 7. Sequences and Series
|an − a| < δ. Then, since |an − a| < δ if n ≥ N , hence |f (an ) − b| < ǫ, that is,
f (an ) → b.
Conversely, suppose that limx→a f (x) is not b, that is, there exists ǫ > 0,
such that for every δ > 0, there exists x with |x − a| < δ and |f (x) − b| ≥ ǫ.
In this way, for all δ = n1 with n ∈ N, there exists an with |an − a| < n1 and
|f (an ) − b| ≥ ǫ. Hence, {an } is a sequence that converges to a and it suffices that
{f (an )} does not converge to b, which is a contradiction.
Theorem 7.4.9. A function f is continuous at a if and only if for every sequence
{an } such that limn→∞ an = a, then limn→∞ f (an ) = f (a).
The proof follows directly from the previous theorem, setting b = f (a).
Theorem 7.4.10. The set of rational numbers is dense in the set of real numbers.
Proof. Let (a, b) be an open interval and let ǫ = b−a > 0. As we proved in Example
7.2.15, there exists a positive integer n with
0<
1
< ǫ.
n
(7.8)
For some m ∈ Z, it follows that a < m
n < b, otherwise a and b are between two
m+1
m
m+1
consecutive numbers of the form m
n and
n , that is, n ≤ a < b ≤ n . Hence,
m
1
m+1
ǫ = b − a ≤ n − n = n , which contradicts inequality (7.8).
Lemma 7.4.11. If D ⊂ R is dense, then for every x ∈ R there exists a sequence
{an } in D with limn→∞ an = x.
Proof. Let x ∈ R, for every n ∈ N it follows, since D is dense, that there exists
an ∈ x − n1 , x + n1 ∩D. It is clear that if |an −x| < n1 for all n, then limn→∞ an =
x.
Theorem 7.4.12. If f : R → R is continuous and f (x) = 0 for all x ∈ D, where D
is dense in R, then f (x) = 0 for all x ∈ R.
Proof. By the previous lemma, for x ∈ R there exists a sequence {an } in D with
limn→∞ an = x. Since f is continuous in x, by Theorem 7.4.8, it follows that
f (x) = limn→∞ f (an ) = 0.
Corollary 7.4.13. If the functions f and g are continuous and coincide in a dense
set, then they coincide in all points.
The proof of the corollary follows from the last theorem, using f − g.
Chapter 8
Polynomials
8.1 Polynomials in one variable
A polynomial P (x) in one variable x is an expression of the form
P (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 ,
where a0 , a1 , . . . , an are constants and n ∈ N. Every term of the polynomial is
called a monomial or simply a term. The constants ai are known as the coefficients
of the polynomial. We will denote by A[x] the set of polynomials with coefficients
in A and variable x. Usually, the set A is Z, Q, R or C. In this chapter we will
study polynomials with complex coefficients, unless otherwise stated.
If an = 0, we say that the polynomial has degree n. In this sense, an xn is
the most important term of the polynomial, because it defines the degree and it
is called the main term . The number a0 is the constant term. We write deg(P )
to denote the degree of P (x). A polynomial is constant if it has a unique term a0 .
If the constant a0 is different from 0 we say that the polynomial has degree zero.
If an = 1, we say that the polynomial is monic.
There are some special names for polynomials whose degree is small. A polynomial
is linear if it has degree 1. We have already studied the quadratic and cubic
polynomials, which have degrees 2 and 3, respectively. If the polynomial has degree
4, it is called quartic.
In the same way we did for quadratic and cubic polynomials, we say that two
polynomials are equal if its coefficients are equal term by term, that is, if the
coefficients of the monomials of the same degree are equal.
A zero of the polynomial P (x) is a number r such that P (r) = 0. When P (r) = 0,
we also say that r is a root or a solution of the equation P (x) = 0.
As we did with the quadratic and cubic polynomials, we can add, subtract, multiply and divide polynomials.
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5_8
139
140
Chapter 8. Polynomials
Let
P (x) = a0 + a1 x + a2 x2 + · · · + an xn ,
Q(x) = b0 + b1 x + b2 x2 + · · · + bm xm ,
be any two polynomials with n ≥ m.
We define the sum as (P + Q)(x) = (a0 + b0 ) + (a1 + b1 )x + (a2 + b2 )x2 + · · ·+ (am +
bm )xm + am+1 xm+1 + · · · + an xn . The difference as (P − Q)(x) = (a0 − b0 ) + (a1 −
b1 )x + (a2 − b2 )x2 + · · · + (am − bm )xm + am+1 xm+1 + · · · + an xn . The product of
a polynomial P (x) and a constant c is cP (x) = ca0 + ca1 x + ca2 x2 + · · · + can xn .
The product of the two polynomials is
(P Q)(x) = a0 b0 + (a0 b1 + a1 b0 )x + (a0 b2 + a1 b1 + a2 b0 )x2 + · · ·
+(a0 br + a1 br−1 + · · · + ai br−i + · · · + ar b0 )xr + · · · + (an bm )xn+m .
Example 8.1.1. In order to multiply two polynomials20 we can use the previous
definition or it is enough to multiply the coefficients. For instance, if we have the
polynomials x4 + 3x3 + x2 − 2x + 5 and 3x3 + 2x2 + 6, its product can be obtained
as follows:
1
3 1
−2 5
3 2
6
3
2 6
9 3
3 11 9
7
6 18 6 −12 30
2 −4 10
−6 15
2
6
29 16 −12 30
The product polynomial is 3x + 11x + 9x5 + 2x4 + 29x3 + 16x2 − 12x + 30.
Exercise 8.1. Multiply the polynomials P (x) = 4x3 + 2x2 + 7x + 1 and Q(x) =
2x2 + x + 8. Evaluate the two polynomials and their product at x = 2.
Exercise 8.2. Let P (x) = (1−x+x2 −· · ·+x100 )(1+x+x2 +· · ·+x100 ). Prove that
after simplifying the product, the only terms left are those that have even powers
of x.
8.2 The division algorithm
Let
P (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 , with an = 0,
Q(x) = bm xm + bm−1 xm−1 + · · · + b1 x + b0 , with bm = 0
20 See
[4], p. 4
141
8.2 The division algorithm
be polynomials of degree n and m, respectively, with m ≤ n and complex or real
coefficients.
The division algorithm says that given polynomials P (x) and Q(x) there exist
unique polynomials H(x) and R(x) with real or complex coefficients, according to
the case, such that
P (x) = Q(x)H(x) + R(x),
deg(R) < deg(Q) or R(x) = 0.
In order to show how to find H(x) and R(x), let us see an example. These polynomials are known as the quotient and the remainder, respectively.
Example 8.2.1. Let P (x) = x5 + x3 + 2x and Q(x) = x2 − x + 1, divide21 P (x) by
Q(x) and find H(x), R(x).
Dividing P (x) by Q(x) we get
x3 + x2 + x
x2 − x + 1
+ x3
x5
5
−x
4
+2x
3
−x
+x
x4
+2x
−x4
+x3 −x2
x3
−x3
−x2 +2x
+x2 −x
x
3
2
In this case, H(x) = x + x + x and R(x) = x.
If R(x) = 0, we say that Q(x) divides P (x) and we write Q(x)|P (x). Note
that the variable x must be the same in all polynomials; thus we can omit it
sometimes.
The division of a polynomial P (x) of degree n by a polynomial of the form x − a,
gives
P (x) = (x − a)Q(x) + r,
with
r∈R
and deg(Q) = n − 1.
Letting x = a, we get r = P (a), and therefore
P (x) = (x − a)Q(x) + P (a) or P (x) − P (a) = (x − a)Q(x).
(8.1)
It follows from equation (8.1) that
P (a) = 0 is equivalent to P (x) = (x − a)Q(x),
(8.2)
for some polynomial Q(x). Thus, we have proved the following theorem.
Theorem 8.2.2 (Factor theorem). The number a is a root of a polynomial P (x) if
and only if the polynomial P (x) is divisible by x − a.
21 See
[4] p. 58
142
Chapter 8. Polynomials
A polynomial H(x) is the greatest common divisor of P (x) and Q(x) if it satisfies:
(a) H(x) divides P (x) and Q(x).
(b) If K(x) is any other polynomial that divides P (x) and Q(x), then K(x)
divides H(x).
It can be proved that H(x) is unique, up to multiplication by a constant.
There is a method called Euclid’s algorithm, that is used to find the greatest
common divisor of two polynomials and which follows the same ideas of Euclid’s
algorithm to find the greatest common divisor of two integers. Let us see an example.
Example 8.2.3. Find the greatest common divisor of the polynomials x4 − 2x3 −
x2 + x − 1 and x3 + 1.
We perform the following divisions of polynomials
x−2
x3 + 1
x4 − 2x3 − x2 + x − 1
−x
−x4
−2x3 − x2
−1
3
+2
+2x
− x2
+1
−x
−x2 + 1
−x+ 1
x3
3
−x
+1
+x
x +1
x+1
− x2
+1
2
x +x
x+ 1
−x− 1
0
Then, as when dealing with integers, the greatest common divisor is x + 1,
which is the remainder before reaching 0 as a remainder.
We can express the greatest common divisor above as a combination of the
polynomials x4 − 2x3 − x2 + x − 1 and x3 + 1, following the inverse steps of the
divisions as shown:
x + 1 = (x3 + 1) − (−x2 + 1)(−x) = (x3 + 1) + x(−x2 + 1)
!
"
= (x3 + 1) + x (x4 − 2x3 − x2 + x + 1) − (x3 + 1)(x − 2)
= x(x4 − 2x3 − x2 + x + 1) + (1 − x(x − 2))(x3 + 1)
= x(x4 − 2x3 − x2 + x + 1) + (−x2 + 2x + 1)(x3 + 1).
143
8.2 The division algorithm
If a1 and a2 are two different zeros of P (x), then by the factor theorem, P (x) =
(x − a1 )Q1 (x), with Q1 (x) a polynomial. Since 0 = P (a2 ) = (a2 − a1 )Q1 (a2 ) and
a1 = a2 , then Q1 (a2 ) = 0. Again, by the factor theorem Q1 (x) = (x − a2 )Q2 (x),
with Q2 (x) a polynomial. Then,
P (x) = (x − a1 )(x − a2 )Q2 (x)
with deg(Q2 ) = n − 2.
In general, if a1 , a2 , . . . , am are different zeros of P (x) we can write
P (x) = (x − a1 )(x − a2 ) . . . (x − am )Q(x),
for some polynomial Q(x), with deg(Q) = deg(P ) − m.
If a is a zero of P (x), then the factor theorem guarantees that there exists
a polynomial Q1 (x) with P (x) = (x − a)Q1 (x). If Q1 (a) = 0, we say that a is a
zero of order 1, but if Q1 (a) = 0 we say that a is a zero of order greater than 1.
If there is m ∈ N and a polynomial Q(x) such that,
P (x) = (x − a)m Q(x), with Q(a) = 0,
(8.3)
then a is a root or a zero of P (x) of multiplicity m.
One of the important consequences of the factor theorem is the following result:
Theorem 8.2.4. If the polynomial P (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 has
n + 1 distinct roots, then the polynomial is identically zero.
Proof. We proceed by induction on n. For n = 1, the result is clear, since a polynomial of degree 1, has only one root. Suppose that the result is true for n − 1;
let us show that it is true for n. Suppose that r0 , r1 , . . . , rn are roots of the polynomial P (x). By the factor theorem, P (x) = (x − rn )Q(x), where the polynomial
Q(x) = an xn−1 + bn−2 xn−2 + · · · + b0 has n distinct roots r0 , r1 , . . . , rn−1 . By
induction, Q(x) is identically zero, hence P (x) also is identically zero.
Observation 8.2.5. The previous theorem guarantees that a polynomial of degree n
must have at most n distinct roots.
A polynomial P (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 , with an = 0 is called
reciprocal if ai = an−i , for all i = 0, 1, . . ., n.
Example 8.2.6. The polynomials xn + 1, x5 + 3x3 + 3x2 + 1 and 6x7 − 2x6 + 4x4 +
4x3 − 2x + 6 are reciprocal polynomials.
Theorem 8.2.7. A reciprocal polynomial P (x) of degree 2n, can be written as
P (x) = xn Q(z), where z = x + x1 and Q(z) is a polynomial in z of degree n.
Proof. Let P (x) = a0 x2n + a1 x2n−1 + · · · + a1 x + a0 , then
$
a0 %
a1
P (x) = xn a0 xn + a1 xn−1 + · · · + n−1 + n
x
x
1
1
P (x) = xn a0 xn + n + a1 xn−1 + n−1 + · · · + an .
x
x
144
Chapter 8. Polynomials
Using the recursive formula (3.3),
xk+1 +
1
=
xk+1
xk +
1
xk
x+
1
x
− xk−1 +
1
xk−1
,
it is clear that we can express each term xk + x1k as a polynomial in z = x+ x1 .
Exercise 8.3. Divide P (x) = x8 − 5x3 + 1 by Q(x) = x3 + x2 + 1. Using the division
algorithm, find the polynomials H(x) and R(x).
Exercise 8.4. Prove that, for n ≥ 1, (x − 1)2 divides nxn+1 − (n + 1)xn + 1.
Exercise 8.5. Let n be a positive integer. Find the roots of the polynomial
Pn (x) = 1 +
x(x + 1)
x(x + 1) . . . (x + n − 1)
x
+
+ ···+
.
1!
2!
n!
Exercise 8.6. Determine the polynomials with real coefficients P (x) that satisfy
P (0) = 0 and P (x2 + x + 1) = P 2 (x) + P (x) + 1 for all x ∈ R.
Exercise 8.7. Prove that any polynomial P (x) of degree n, with a0 = 0, is reciprocal
if and only if
1
xn P
= P (x), for every x = 0.
x
Exercise 8.8. Prove that every reciprocal polynomial P (x) of odd degree is divisible
by x + 1 and its quotient is a reciprocal polynomial of even degree.
Exercise 8.9. If a is a root of a reciprocal polynomial P (x), prove that
root of the polynomial.
1
a
is also a
Exercise 8.10. Determine for which integers n, the polynomial 1 + x2 + x4 + · · · +
x2n−2 is divisible by the polynomial 1 + x + x2 + · · · + xn−1 .
Exercise 8.11. Prove that the greatest common divisor of xn − 1 and xm − 1 is
x(n,m) − 1, where (n, m) denotes the greatest common divisor of m and n.
Exercise 8.12 (USA, 1977). Find all the pairs of positive integers (m, n) such that
1 + x + x2 + · · · + xm divides 1 + xn + x2n + · · · + xmn .
Exercise 8.13 (Canada, 1971). Let P (x) be a polynomial with integer coefficients.
Prove that if P (0) and P (1) are odd, then P (x) = 0 has no integer solutions.
145
8.3 Roots of a polynomial
8.3 Roots of a polynomial
8.3.1 Vieta’s formulas
Vieta’s formulas (4.1) and (4.2) can be generalized for polynomials of higher degree.
If a monic polynomial xn + an−1 xn−1 + · · · + a1 x + a0 has n roots x1 , x2 ,
. . . , xn , then
xn + an−1 xn−1 + · · · + a1 x + a0 = (x − x1 )(x − x2 ) · · · (x − xn )
= xn − (x1 + · · · + xn )xn−1 + (x1 x2 + · · · + x1 xn + x2 x3 + · · · + xn−1 xn )xn−2
+ · · · + (−1)n x1 · · · xn ,
hence,
an−1 = −(x1 + · · · + xn )
an−2 = (x1 x2 + · · · + x1 xn + x2 x3 + · · · + xn−1 xn )
..
.
#
an−j = (−1)j
xi1 xi2 · · · xij
..
.
(8.4)
1≤i1 <···<ij ≤n
a0 = (−1)n x1 x2 . . . xn .
The formulas (8.4) are known as Vieta’s formulas.
Example 8.3.1. Consider the polynomial P (x) = xn − (x − 1)n , where n is an odd
positive integer. Calculate the sum and the product of its roots.
The polynomial P (x) can be written as
n(n − 1) n−2
x
− · · · + (−1)n )
2
n(n − 1) n−2
x
= xn − (xn − nxn−1 +
− · · · − 1)
2
n(n − 1) n−2
= nxn−1 −
x
+ · · · + 1.
2
P (x) = xn − (xn − nxn−1 +
Then, the sum of its roots is
n(n−1)
2n
=
n−1
2
and the product of its roots is
1
n.
8.3.2 Polynomials with integer coefficients
Consider the polynomial P (x) = an xn + · · · + a1 x + a0 with integer coefficients.
The difference P (x) − P (y), can be written as
an (xn − y n ) + · · · + a2 (x2 − y 2 ) + a1 (x − y),
146
Chapter 8. Polynomials
where every term of the sum is a multiple of x − y. This leads us to the following
arithmetic property of the polynomials in Z[x].
Theorem 8.3.2. If P (x) is a polynomial with integer coefficients, then P (a) − P (b)
is divisible by a − b, for any pair of different integers a and b.
In particular, all integer roots of P (x) divide P (0).
There is a similar statement for the rational roots of polynomials P (x) with
integer coefficients.
Theorem 8.3.3 (Theorem of the rational root). Any rational root pq , with (p, q) = 1,
of a polynomial P (x) = an xn +an−1 xn−1 +· · ·+a0 with integer coefficients, satisfies
that p divides a0 and q divides an .
Proof. Let
p
q
be a root of P (x), then
qn P
p
q
= an pn + an−1 pn−1 q + · · · + a0 q n = 0.
All the terms of the sum, except possibly the first, are multiples of q, and all
the terms of the sum, except possibly the last, are multiples of p. Since p and q
divide 0, it follows that q|an pn and p|a0 q n , and then the assertion follows, since
(p, q) = 1.
Corollary 8.3.4. If P (x) is a polynomial with integer coefficients that takes values
{±1} in three different integers, then P (x) has no integer roots.
Proof. Suppose that there are integers a, b, c and d such that P (a), P (b), P (c) ∈
{−1, 1} and P (d) = 0. Then, since the integers a, b and c are different, a − d,
b − d and c − d are also different and, by Theorem 8.3.2, all divide 1, which is
impossible.
8.3.3 Irreducible polynomials
A polynomial P (x) with integer coefficients is called irreducible in Z[x], if it cannot
be written as a product of two non-constant polynomials with coefficients in Z.
Example 8.3.5. Any quadratic polynomial with at least one non-rational root is
irreducible in
Z[x]. For instance, x2 − x − 1 is irreducible in Z[x], since it has roots
√
given by 1±2 5 .
Similarly, we define irreducibility over the set of polynomials with coefficients
in Q, R. The next theorem claims that for polynomials with integer coefficients,
the fact that the polynomial could be factored in Q[x] is equivalent to the fact
that the polynomial could be factored in Z[x]. Moreover, a polynomial with real
coefficients always can be expressed as a product of linear polynomials and irreducible quadratic polynomials in R[x]. In the case of a polynomial with complex
coefficients, it can always be factored into linear factors over C[x].
8.3 Roots of a polynomial
147
Lemma 8.3.6 (Gauss’ lemma). If P (x) has integer coefficients and P (x) can be
factored over the rational numbers, then P (x) can be factored over the integers as
well.
Proof. Suppose that P (x) = an xn +· · ·+a0 has integer coefficients and that P (x) =
Q(x)R(x), where Q(x) and R(x) are non-constant polynomials with rational coefficients. Let q and r be the smallest natural numbers such that qQ(x) and rR(x) have
integer coefficients. Then, if d = qr it follows that P1 (x) = dP (x) = qQ(x)·rR(x) =
Q1 (x)R1 (x) is a factorization of the polynomial P1 (x) into two polynomials with
integer coefficients Q1 (x) = qk xk + · · · + q0 and R1 (x) = rm xm + · · · + r0 . Let a′j ,
with 0 ≤ j ≤ n, be the coefficients of P1 (x). Based on this, we will construct the
required factorization of P (x).
Let p be a prime divisor of d. Then all coefficients of P1 (x) are divisible by p.
Now let us show that p divides all coefficients of Q1 (x) or divides all coefficients
of R1 (x).
If p divides all coefficients of Q1 (x), we are done. Otherwise, let i be such
that p|q0 , q1 ,. . ., qi−1 , but p ∤ qi . We have that p|a′i and a′i = q0 ri + · · · + qi r0 ≡
qi r0 mod p, which implies that p|r0 . Moreover, p|a′i+1 and a′i+1 = q0 ri+1 + · · · +
qi r1 + qi+1 r0 ≡ qi r1 mod p, and then p|r1 . Proceeding in the same way, we can
deduce that p|rj , for all j. Then R1p(x) has integer coefficients. Then we have a
factorization of dp P (x) into two polynomials with integer coefficients. Taking all the
prime divisors of d, we will eventually finish with a factorization of the polynomial
P (x) into two polynomials with integer coefficients.
Example 8.3.7. If a1 , a2 , . . . , an are different integers, then the polynomial P (x) =
(x − a1 )(x − a2 ) · · · (x − an ) − 1 is irreducible over Z[x].
Suppose that P (x) = Q(x)R(x), for some non-constant polynomials R(x)
and Q(x) with integer coefficients. Since Q(ai )R(ai ) = −1 for i = 1, . . . , n, then
Q(ai ) = 1 and R(ai ) = −1 or Q(ai ) = −1 and R(ai ) = 1; in both cases, we have
that Q(ai ) + R(ai ) = 0. Also, the polynomial Q(x) + R(x) is not zero (because
otherwise P (x) = −Q(x)2 ≤ 0 for every real number x, but if x is very large, P (x)
is positive, a contradiction). Moreover, Q(x) + R(x) has n zeros a1 , . . . , an , which
is impossible given that its degree is less than n.
A polynomial with integer coefficients is primitive, if its principal coefficient
is positive and there is no integer number that divides all coefficients of the polynomial.
Theorem 8.3.8 (Eisenstein’s irreducibility criterion). Consider a polynomial
P (x) = an xn + an−1 xn−1 + · · · + a0 ,
with integer coefficients. Let p be a prime number such that
148
Chapter 8. Polynomials
(a) p does not divide an ,
(b) p divides every coefficient a0 , a1 , . . . , an−1 ,
(c) p2 does not divide a0 .
Then P (x) is irreducible over Q[x]. Moreover, if P (x) is primitive, then it is
irreducible over Z[x].
Proof. Suppose that P (x) is reducible over Q[x]. By Gauss’ lemma, P (x) =
Q(x)R(x), where Q(x) = qk xk + · · · + q0 and R(x) = rm xm + · · · + r0 are polynomials with integer coefficients. Since a0 = q0 r0 is divisible by p but not by p2 ,
exactly one of q0 or r0 is a multiple of p. Suppose that p|q0 and p ∤ r0 . Moreover,
since p|a1 = q0 r1 + q1 r0 it follows that p|q1 r0 ; then, p|q1 and so on. We conclude
that all coefficients q0 , q1 , . . . , qk are divisible by p, but then p|an since an = qk rm ,
which is a contradiction.
One of the most important applications of the Eisenstein criterion, is to show the
irreducibility of the cyclotomic polynomials, xp−1 + xp−2 + · · · + x + 1, with p a
prime number. Note that the roots of this polynomial are the pth roots of unity,
2πi
that is, the powers of e p .
Example 8.3.9. The polynomial P (x) = xp−1 + xp−2 + · · · + x + 1, with p a prime
number, is irreducible over Q[x].
Note that (x−1)P (x) = xp −1. With the substitution x = y +1 in the last product
we get
yP (y + 1) = (y + 1)p − 1 = y p +
p p−1
p p−2
p
y
+
y
+ ··· +
y.
1
2
p−1
, if i < p then, since the prime p is not
Since pi = p(p−1)···(p−i+1)
i!
a factor of i!,
i! divides the product (p − 1) · · · (p − i + 1). This implies that pi is divisible by
p. Dividing yP (y + 1) by y, it follows that P (y + 1) satisfies the conditions of the
Eisenstein criterion and therefore it is an irreducible polynomial, hence P (x) is
also irreducible.
Let us see some applications in number theory of the previous example. Let
p be an odd prime number and consider the polynomial P (x) = xp−1 − 1 with
coefficients22 in Zp . By Fermat’s little theorem23 , each of the numbers 1, 2, . . . ,
p − 1 is a root of the polynomial, then
xp−1 − 1 = (x − 1)(x − 2) · · · (x − p + 1).
(8.5)
(a) Comparing the constant coefficients in the last identity we get Wilson’s theorem:
(p − 1)! ≡ −1(mod p).
22 Z = {0, 1, . . . , p
p
23 See [8].
− 1} with the sum and product operations module p.
149
8.3 Roots of a polynomial
(b) If we expand the right-hand side of (8.5), the coefficient σj of xp−1−j is the
sum of all products of j elements of the set {1, 2, . . . , p − 1}. Comparing
coefficients, we get that p divides σj for j = 1, 2, . . . , p − 2. Now assume that
p ≥ 3.
(c) (Wolstenholme) The numerator of the (reduced) fraction
1
1
m
= 1 + + ···+
n
2
p−1
σ
p−2
is divisible by p. In fact, m
n = (p−1)! and, since p|σp−2 and p is relatively
prime to (p − 1)!, it follows that p|m.
(d) Let m be as in the previous part. If p ≥ 5, it follows that p2 |m.
Since (x − 1) . . . (x − p + 1) = xp−1 − σ1 xp−2 + σ2 xp−3 + · · · + σp−1 , it follows,
after evaluating in x = p, that
(p − 1)! = pp−1 − σ1 pp−2 + σ2 pp−3 + · · · − σp−2 p + σp−1 .
Since σp−1 = (p − 1)!, we can reduce the last identity to
σp−2 = pp−2 − σ1 pp−3 + · · · + σp−3 p.
This shows that σp−2 is divisible by p2 , since every σj is divisible by p and
since p2 and (p − 1)! are relatively prime, it follows that p2 |m.
Exercise 8.14. Find the solutions of the system
x+y+z =w
1 1
1
1
+ + = .
x y z
w
Exercise 8.15. Prove that the polynomial x4 − x3 − 3x2 + 5x + 1 is irreducible
over Q[x].
Exercise 8.16. Let P (x) be a polynomial with integer coefficients. Prove that if
P k (n) = n, for some integer number k ≥ 1, and for some integer number n, then
for such integer n it follows that P (P (n)) = n.
Exercise 8.17. Find all polynomials of the form an xn + an−1 xn−1 + · · · + a1 x + a0
with aj ∈ {−1, 1}, such that all its roots are real numbers.
Exercise 8.18 (USA, 1974). Let a, b, c be different integers and let P (x) be a
polynomial with integer coefficients. Prove that it is impossible that P (a) = b,
P (b) = c and P (c) = a.
Exercise 8.19. Prove that the polynomial with real coefficients P (x) = xn +
2nxn−1 + 2n2 xn−2 + an−3 xn−3 + · · · + a1 x + a0 , cannot have all its roots real.
150
Chapter 8. Polynomials
8.4 The derivative and multiple roots ⋆
For a polynomial of degree n,
P (x) = a0 + a1 x + a2 x2 + · · · + an−1 xn−1 + an xn =
n
#
ak xk ,
k=0
we define the derivative of P (x) as the polynomial*
of degree n−1 given by P ′ (x) =
n
n−2
n−1
a1 + 2a2 x + · · · + (n − 1)an−1 x
+ nan x
= k=1 kak xk−1 .
It can be shown that if P (x) = (x − a)Q(x), for some polynomial Q(x), then
P ′ (x) = Q(x) + (x − a)Q′ (x).
(8.6)
There is an important relationship between the roots of a polynomial P (x) and
the roots of its derivative P ′ (x), given by the following theorem.
Theorem 8.4.1. If for some positive integer m the polynomial P (x) is divisible by
(x − a)m+1 , then the polynomial P ′ (x) is divisible by (x − a)m . That is, if a is a
zero of multiplicity m + 1 of P (x), then a is a zero of multiplicity m for P ′ (x).
Proof. For the proof we will use induction over m. For m = 1, if P (x) is divisible
by (x − a)2 , then P (x) = (x − a)Q(x), where Q(x) is a polynomial divisible by
x − a, and P ′ (x) = Q(x) + (x − a)Q′ (x). Therefore, P ′ (x) is divisible by x − a.
Suppose the result is true for m − 1. Let P (x) be divisible by (x − a)m+1 . Then
P (x) = (x − a)Q(x), where Q(x) is divisible by (x − a)m , and from P ′ (x) =
Q(x)+(x−a)Q′ (x), it follows that P ′ (x) is divisible by (x−a)m , where it has been
used that Q′ (x) is divisible by (x−a)m−1 , this being the induction hypothesis.
Observe that if a is a zero of P (x) with multiplicity one, then P ′ (a) = 0.
Example 8.4.2. If m and n are integers, such that 0 < m < n, then
n
#
(−1)k k m
k=1
n
k
= 0.
The problem is equivalent to showing that the polynomial
Pm (x) =
n
#
k=0
km
n k
x ,
k
has x = −1 as a root. Let us prove the following stronger result: Pm (x) has a zero
of multiplicity at least n − m in x = −1. For this we will see that (x + 1)n−m
divides Pm (x), for 0 ≤ m < n.
151
8.5 The interpolation formula
We proceed by induction on m. For m = 0, it follows that
P0 (x) =
n
#
k=0
k0
n k
x = (1 + x)n ,
k
then the statement is true.
Suppose that for m, with 0 < m < n, Pm−1 (x) has −1 as a root of multiplicity
′
n − (m − 1). By the previous theorem, Pm−1
(x) has one root of multiplicity n − m
′
n−m
in −1. Then Pm−1 (x) = (x + 1)
Q(x), for some polynomial Q(x). But
′
n
n
#
#
n k−1
1
′
k
m−1 n
Pm−1 (x) =
k
= Pm (x).
=
km
x
x
x
k
k
k=0
k=0
′
Hence, Pm (x) = xPm−1
(x) = x(x + 1)n−m Q(x), therefore (x + 1)n−m divides
Pm (x).
Proposition 8.4.3. If P (x) is a polynomial such that P (a) = P ′ (a) = · · · =
P (m−1) (a) = 0 and P (m) (a) = 0, then a is a zero of P (x) of multiplicity m.
That is, P (x) = (x − a)m Q(x) for some polynomial Q(x), with Q(a) = 0. Here
P (j+1) (x) is the derivative of P (j) (x) and P (1) (x) = P ′ (x).
Proof. Since P (a) = 0, then P (x) = (x − a)Q1 (x) for some polynomial Q1 (x).
Since P ′ (x) = Q1 (x) + (x − a)Q′1 (x), it follows that P ′ (a) = Q1 (a) = 0. Then
Q1 (x) = (x − a)Q2 (x) for some polynomial Q2 (x), hence P (x) = (x − a)2 Q2 (x).
Proceeding in the same way we get that P (x) = (x − a)m Qm (x), where Qm (x) is
the polynomial Q(x) we are looking for.
Exercise 8.20. Determine if the polynomial P (x) = x3 − x2 − 8x + 12 has multiple
roots.
Exercise 8.21. Find all the triplets of real numbers (a, b, c) such that the polynomial
P (x) = x3 + ax2 + bx + c is divisible by (x + 1)2 .
8.5 The interpolation formula
Given two points in the Cartesian plane, there is a unique straight line that joins
these two points. Then, for two pairs of real numbers (α0 , β0 ), (α1 , β1 ), with α0 =
α1 , there is a unique polynomial P (x) of degree at most 1, such that P (α0 ) = β0
and P (α1 ) = β1 . This can be generalized as follows.
Theorem 8.5.1 (Lagrange interpolation formula). Let α0 , α1 , . . . , αn be different
real numbers and let β0 , β1 , . . . , βn be another n + 1 set of real numbers. Then
there exists a unique polynomial P (x) of degree at most n such that P (αi ) = βi ,
for 0 ≤ i ≤ n.
152
Chapter 8. Polynomials
Proof. First, let us find a polynomial that satisfies the conditions. Consider the
polynomials
Dk (x) =
(x − α0 )(x − α1 ) · · · (x − αk−1 )(x − αk+1 ) · · · (x − αn )
,
(αk − α0 )(αk − α1 ) · · · (αk − αk−1 )(αk − αk+1 ) · · · (αk − αn )
with 0 ≤ k ≤ n, where the numerator and the denominator have n factors. It is
clear that Dk (x) has degree n, that Dk (αk ) = 1 and that Dk (αi ) = 0, if i = k.
Then, the polynomial that satisfies the conditions is
P (x) =
n
#
βk Dk (x).
k=0
To show the uniqueness of the polynomial, suppose that there are two polynomials P1 (x) and P2 (x) of degree at most n, such that
Pj (αi ) = βi , for 0 ≤ i ≤ n, j = 1, 2.
Then the polynomial P (x) = P1 (x) − P2 (x) has degree at most n and has n + 1
distinct roots α0 , α1 , . . . , αn . By Theorem 8.2.4, it follows that P (x) is the zero
polynomial, then P1 (x) = P2 (x).
Example 8.5.2. For 1 < m < n, the following identity holds:
n
#
(−1)k k m
k=1
n
k
= 0.
Construct a polynomial of degree at most n which takes the values βk =
k m in the points αk = k, with k = 0, 1, 2, . . . , n, respectively. By the Lagrange
interpolation formula it follows that the polynomial we are looking for is given by
P (x) =
n
#
k m Dk (x),
k=0
where
Dk (x) =
x(x − 1)(x − 2) · · · (x − k + 1)(x − k − 1) · · · (x − n)
.
k(k − 1) · · · 1(−1)(−2) · · · (k − n)
The coefficient of xn , in the polynomial Dk (x), is equal to
n
1
(−1)n
1
= (−1)k−n
=
(−1)k
k
k(k − 1) · · · 1(−1)(−2) · · · (k − n)
k!(n − k)!
n!
and therefore the coefficient of xn , in the polynomial P (x), turns out to be
n
n
(−1)n #
(−1)k k m
.
k
n!
k=0
8.6 Other tools to find roots
153
On the other hand, the polynomial Q(x) = xm satisfies that Q(k) = k m for every
0 ≤ k ≤ n. Then, by Theorem 8.5.1, P (x) and Q(x) are equal. Since m < n, the
coefficient of xn in the polynomial P (x) is equal to 0.
Example 8.5.3. Find a polynomial P (x) such that xP (x − 1) = (x + 1)P (x), for
all x ∈ R.
For x = 0, the condition is equivalent to P (0) = 0. If P (n) = 0, then
P (n + 1) = 0 since (n + 1)P (n) = (n + 2)P (n + 1). Therefore P (x) has an infinite
number of zeros. Then, P (x) ≡ 0 is the only polynomial that satisfies the equation.
Exercise 8.22. Let P (x) be a polynomial of degree n such that P (k) = 2k , for
k = 0, 1, . . . , n. Find P (n + 1).
8.6 Other tools to find roots
8.6.1 Parameters
In order to study the roots of a polynomial of degree greater than two, sometimes
it is useful to consider the independent terms as variables. These independent
terms are called parameters. To illustrate this let us see the next example.
Example 8.6.1. Find the solutions of the equation
x3 (x + 1) = 2(x + a)(x + 2a),
where a is a real parameter.
Solving the equation is equivalent to solving the quartic equation
x4 + x3 − 2x2 − 6ax − 4a2 = 0.
This equation is difficult to solve. In some cases it is possible to factorize without
any trouble the left-hand side, but in many cases it is not easy. However we can use
some algebraic tricks; for instance, we can consider the number a as the variable
and x as the parameter or constant. Then, we get the quadratic equation in a,
4a2 + 6ax − x4 − x3 + 2x2 = 0.
Using the formula to solve second-degree equations, the discriminant is given by
36x2 + 16(x4 + x3 − 2x2 ) = 4x2 (2x + 1)2 ,
which is a square number. Solving the equation for a, we obtain the two roots
of the equation: a1 = − 21 x2 − x and a2 = 12 x2 − 21 x. Then, we can factorize the
154
Chapter 8. Polynomials
equation as
1
1
a − x2 + x
2
2
1
x4 + x3 − 2x2 − 6ax − 4a2 = −4 a + x2 + x
2
= (x2 + 2x + 2a)(x2 − x − 2a).
Finally,
solving these second-degree
equations we obtain
! the "solutions x1,2 = −1 ±
√
√
1 − 2a, x3,4 = 12 ± 12 1 + 8a, which are real if a ∈ − 81 , 12 .
Exercise 8.23.
√
(i) Solve the equation 5 − x = 5 − x2 .
√
(ii) Solve the equation a − x = a − x2 , with a > 0.
Exercise 8.24. Solve the equation x =
a−
√
a + x, where a > 0 is a parameter.
8.6.2 Conjugate
The idea we are about to consider is very simple: when in some algebraic expression
there is a square root in the denominator of a fraction, sometimes it is useful to
multiply the expression by some factor that cancels out the square root. Let us
show this procedure with some simple examples.
Example 8.6.2. If we want to eliminate the square root in the denominator of the
expression
1
√ ,
a+ b
we should multiply it by
The number a −
√
a−√b
a− b
in order to obtain
1√
a+ b
=
√
a− b
a2 −b .
√
√
b is known as the conjugate of a + b.
Example 8.6.3. Solve the equation
√
√
1 + mx = x + 1 − mx,
where m is a real parameter.
√
√
The equation is equivalent to 1 + mx − 1 − mx = x.
√
√
and dividing by 1 + mx + 1 − mx, which is the conjugate of
√ Multiplying
√
1 + mx − 1 − mx, it follows that
2mx
√
√
= x.
1 + mx + 1 − mx
155
8.6 Other tools to find roots
√
√
A solution is x = 0 and, if x = 0, then 2m = 1 + mx + 1 − mx, hence m is
positive. Squaring and simplifying, it follows that
2m2 − 1 =
1 − m2 x2 ,
hence 2m2 − 1 ≥ 0. Squaring again and solving for x we get
x = ±2 1 − m2
'
&
and, since 1 − m2 ≥ 0, it follows that m ∈ √12 , 1 .
Exercise 8.25. Solve the equation
√
√
x+ x− x− x=m
x
√ ,
x+ x
where m is a real parameter.
Exercise 8.26 (Short list OIM, 2009). Find all triplets (x, y, z) of positive real
numbers that satisfy the system of equations:
x+
y + 11 = y + 76
√
√
y + z + 11 = z + 76
√
√
z + x + 11 = x + 76.
Exercise 8.27. Let a, b, c > 0, solve the system:
1
=c
xy
1
=a
bz − cx +
zx
1
cy − az +
= b.
yz
ax − by +
8.6.3 Descartes’ rule of signs ⋆
The estimation of the number of positive roots of a polynomial P (x), with real
coefficients, can be achieved counting the number of changes of sign C(P ), in the
sequence of non-zero coefficients of P (x).
Example 8.6.4. The polynomial
P (x) = 3x12 + 4x10 − 2x9 − 4x8 − x6 + 3x5 − 2x4 − 6x2 + 11x,
has 4 changes of sign, that is, C(P ) = 4.
Note that zero coefficients are neglected.
156
Chapter 8. Polynomials
Theorem 8.6.5 (Descartes’ rule of signs). The number of positive real roots of
a polynomial P (x), with real coefficients, is less than or equal to the number of
changes of sign, C(P ), that are produced among its coefficients (neglecting the
zero coefficients and counting multiplicities of the roots). Similarly, the number of
negative real roots of the polynomial is less than or equal to the changes of sign
that appear among the coefficients of P (−x).
Moreover, if the number of positive roots is less than the number of changes of
sign, then the number of positive roots differs from the number of changes of sign
by an even number.
Proof. Suppose that the polynomial P (x) has degree n, is monic and that the
constant term is not zero, that is, P (0) = 0. Otherwise, we can factor one term
of the form xk , which does not contribute to the positive roots. Let us show first
that the number of changes of sign and the number of positive roots have the same
parity. The proof is by induction on n.
For n = 1, the polynomial has degree 1 and the result is clear, since P (x) = x − a,
with a > 0, has one change of sign and the only positive root is x = a. If P (x) =
x + a, there is no change of sign and the only solution is x = −a, which is negative.
Now suppose that P (x) is a monic polynomial of degree n > 1, with P (0) = 0.
There are two cases:
Case 1. If P (0) < 0, then the number of changes of sign must be odd since it
starts with a positive number because the polynomial is monic, and it finishes
with a negative number P (0). Let us see that the number of positive roots of the
polynomial is also odd.
Since the degree of P (x) is n, the term xn dominates for large values of x. Then
for some large and positive value of x, say x0 , it follows that P (x0 ) is positive,
then P (x) must have a root24 in the interval (0, x0 ), which is clearly positive.
Let k be such a root. Then P (x) = (x − k)Q(x), with Q(x) a polynomial of degree
(0)
positive. Hence, applying the induction hypothesis to Q(x),
n − 1 and Q(0) = P−k
we obtain that this polynomial has an even number of positive roots, therefore
P (x) has an odd number of positive zeros, the zeros of Q(x) and k.
Case 2. If P (0) > 0 and the equation has no positive solutions, we are done,
since zero is an even number. When the equation has some positive solution, we
consider one of them, say k. As before, we have that P (x) = (x − k)Q(x), with
(0)
negative. Then Q(x) has an
Q(x) a polynomial of degree n − 1 and Q(0) = P−k
odd number of changes of sign. Applying the induction hypothesis to Q(x), we
obtain that Q(x) has an odd number of positive roots. Therefore, P (x) has an
even number of positive roots.
Until now we have shown that the number of changes of sign and the number of
positive roots of a polynomial have the same parity. It is only left to prove that the
number of sign changes is greater than or equal to the number of positive roots,
24 This
claim is guaranteed by the Intermediate Value Theorem, see [21].
8.7 Polynomials that commute
157
that is, the number of sign changes is an upper bound for the number of positive
roots. If there were more positive roots than sign changes in the coefficients of
P (x), then there must be at least two more positive roots than the number of sign
changes, that is, there must be at least C(P ) + 2 positive roots.
On the other hand, P ′ (x) has at least one change of sign25 between every
two roots of P (x), hence there would be at least C(P ) + 1 roots of P ′ (x).
But P ′ (x) has at least as many sign changes as P (x), that is, C(P ), and moreover
its degree is n − 1. Under these conditions, the induction hypothesis tells us that
such polynomial satisfies the rule of signs, that is, it has more sign changes than
positive roots, which is a contradiction. Therefore, there are more changes of sign
than positive roots.
Example 8.6.6 (Poland, 2001). Let n ≥ 3 be an integer. Prove that a polynomial
of the form an xn + an−1 xn−1 + · · · + a1 x + a0 , with an−1 = an−2 = 0 and at least
one ak = 0, cannot have all its roots real.
Suppose that the polynomial satisfies that a0 = 0 in order to assure that if
it has a real root, it will be positive or negative. Applying Descartes’ rule of signs,
it follows that the number of real roots of the polynomial is at most n − 1, hence
at least one of them is a complex root.
If a0 = 0, we can factorize the highest power of x that divides the polynomial
and, in this case, work with the quotient polynomial as in the previous case.
8.7 Polynomials that commute
Two monic polynomials P (x) and Q(x), with real coefficients in one variable,
commute if, for every real number x, it follows that
P (Q(x)) = Q(P (x)).
This means that they commute as polynomial functions. In this section we try to
characterize all monic polynomials Q(x) that commute with a given polynomial
P (x). Let us see a first example.
Example 8.7.1. Find all monic polynomials of degree 3 that commute with the
polynomial P (x) = x2 − α, for some α.
Let Q(x) = x3 + ax2 + bx + c. The equality P (Q(x)) = Q(P (x)) can be
written as
(x3 + ax2 + bx + c)2 − α = (x2 − α)3 + a(x2 − α)2 + b(x2 − α) + c.
Expanding these last expressions, we get that the right-hand side of the equality
has only even powers of x, whereas on the left-hand side there is x5 with coefficient
25 See
Rolle’s theorem [21].
158
Chapter 8. Polynomials
2a, forcing a to be zero. Hence the coefficient of x3 in the left-hand side is 2c, so
that c is also zero. Therefore, Q(x) = x3 + bx.
Canceling the parenthesis and equating coefficients of the corresponding powers
of x, we get
2b = −3α, b2 = 3α2 + b, α = α3 + bα.
Letting α = 2γ, it follows that b = −3γ. The second equation implies that 9γ 2 =
12γ 2 − 3γ, that is, γ 2 − γ = 0, then γ = 0 or γ = 1. It is easy to verify that each
of these values represents a solution of the system.
Summarizing, a polynomial Q(x) of degree 3 that commutes with P (x) = x2 − α
exists only when α = 0 or α = 2. If α = 0, Q(x) = x3 and if α = 2, Q(x) = x3 −3x.
Similarly, it can be shown that the only polynomial of degree 2 that commutes
with P (x) = x2 − α is P (x) itself and that the only polynomial Q(x) of degree 1
that commutes with P (x) is Q(x) = x.
An important example of polynomials that commute are the so-called
Tchebyshev polynomials, defined in the following way.
Let k be a non-negative integer, the Tchebyshev polynomial Tk (x) is defined, for
−1 ≤ x ≤ 1, in a recursive way as follows: T0 (x) = 1, T1 (x) = x and, for k ≥ 2,
Tk (x) = 2xTk−1 (x) − Tk−2 (x).
(8.7)
The first Tchebyshev polynomials are T2 (x) = 2x2 − 1, T3 (x) = 4x3 − 3x, T4 (x) =
8x4 − 8x2 + 1. The identity (8.7) and the induction principle guarantee us that Tk
is a polynomial of degree k.
Lemma 8.7.2. The Tchebyshev polynomials satisfy Tk (cos t) = cos (kt).
Proof. We proceed by induction. We have that T0 (cos t) = 1 = cos 0 = cos (0 · t),
moreover, T1 (x) = x implies that T1 (cos t) = cos t = cos (1 · t).
Suppose that Tk−1 (cos t) = cos [(k − 1)t] and Tk−2 (cos t) = cos [(k − 2)t].
Let us show the result for k. Since the equation (8.7) holds, then
Tk (cos t) = 2 cos t Tk−1 (cos t) − Tk−2 (cos t)
= 2 cos t cos[(k − 1)t] − cos[(k − 2)t],
(8.8)
where we have used the induction hypothesis.
Now, cos(kt) = cos[(k−1)t+t] = cos[(k−1)t]·cos t−sin[(k−1)t]·sin t. On the
other hand, since sin[(k−1)t] = sin[(k−2)t+t] = sin[(k−2)t] cos t+sin t cos[(k−2)t],
159
8.7 Polynomials that commute
it follows that,
sin[(k − 1)t] sin t = sin[(k − 2)t] cos t sin t + cos[(k − 2)t] sin2 t
= sin[(k − 2)t] cos t sin t + cos[(k − 2)t](1 − cos2 t)
= sin[(k − 2)t] cos t sin t + cos[(k − 2)t] − cos[(k − 2)t] cos2 t
= cos t (sin[(k − 2)t] sin t − cos[(k − 2)t] cos t) + cos[(k − 2)t]
= cos t (− cos[(k − 1)t]) + cos[(k − 2)t].
Therefore,
cos kt = cos[(k − 1)t] cos t − cos[(k − 2)t] + cos[(k − 1)t] cos t
= 2 cos t cos[(k − 1)t] − cos[(k − 2)t],
as desired.
Also, it follows that the Tchebyshev polynomials commute,
Tk (Tm (x)) = Tkm (x) = Tm (Tk (x)).
This follows from the simple fact that cos [k(mt)] = cos [kmt] = cos [m(kt)]. However, Tk is not a monic polynomial, its principal coefficient is 2k−1 , which is an
inconvenience. But this can be easily fixed if we define Pk (x) = 2Tk ( x2 ). This
procedure keeps the commuting property.
Example 8.7.3 (IMO, 1976). Let P1 (x) = x2 − 2 and let Pj (x) = P1 (Pj−1 (x)),
for j = 2, 3, . . . . Prove that for every positive integer n, the roots of the equation
Pn (x) = x are real and distinct.
Let us write x(t) = 2 cos t. This function sends the interval [0, π] to the interval
[−2, 2]. Now, observe that
P1 (x) = P1 (2 cos t) = 4(cos t)2 − 2 = 2 cos 2t,
P2 (x) = P1 (P1 (x)) = 4(cos 2t)2 − 2 = 2 cos 4t,
..
..
.
.
Pn (x) = 2 cos 2n t.
Equation Pn (x) = x is equivalent to 2 cos 2n t = 2 cos t, with solutions 2n t =
±t + 2πk, with k = 0, 1, . . . . That is, the following 2n values of t,
t=
2kπ
2n − 1
and
t=
2kπ
,
2n + 1
give 2n real distinct values of x(t) = 2 cos t that satisfy Pn (x) = x.
Let us see another example.
160
Chapter 8. Polynomials
Example 8.7.4. There exists an infinite sequence of polynomials P1 (x), P2 (x), . . . ,
Pk (x), . . . , such that any two of them commute, the degree of Pk (x) is k, and
P1 (x) = x and P2 (x) = x2 − 2.
One solution is immediate, considering the Tchebyshev polynomials that, as
we already saw, commute. However, we will give a constructive proof. There is
only one polynomial Pk (x), of degree k, that commutes with P2 (x) = x2 − 2 (see
Exercise 8.28).
Let us write the first terms of the sequence we are looking for:
P1 (x) = x,
P2 (x) = x2 − 2,
P3 (x) = x3 − 3x,
P4 (x) = x4 − 4x2 + 2,
P5 (x) = x5 − 5x3 + 5x,
P6 (x) = x6 − 6x4 + 9x2 − 2.
Here, P4 (x) = P2 (P2 (x)), P6 (x) = P2 (P3 (x)). Observing the previous polynomials, we note they satisfy the relation Pk+1 (x) = xPk (x) − Pk−1 (x). This makes it
natural to define the polynomials using this last recursion, for k > 1, and with
P1 (x) = x, P2 (x) = x2 − 2. We can prove, using induction, that these polynomials commute with P2 (x) and, by Exercise 8.30, we obtain that any two of them
commute.
Exercise 8.28. Prove that there exists at most one polynomial of a given degree
that commutes with a given polynomial of degree 2.
Exercise 8.29. Find all polynomials of degrees 4 and 8 that commute with a given
polynomial of degree 2.
Exercise 8.30. Prove that if the polynomials Q(x) and R(x) both commute with
the polynomial P (x) of degree 2, then they commute with each other.
Exercise 8.31. Prove that two polynomials P (x) and Q(x), of degree 1, commute
if and only if either both are monic or both have a common fixed point.
Exercise 8.32. Given a polynomial R(x), define the polynomial Ra (x) = R(x −
a) + a. Prove that if two polynomials P (x) and Q(x) commute, then Pa (x) and
Qa (x) commute as well.
8.8 Polynomials of several variables
161
8.8 Polynomials of several variables
If x and y are the solutions of the quadratic equation at2 + bt + c = 0, then
c
−b
a = x + y and a = xy. The expressions x + y and xy are examples of polynomials
in two variables x and y. In general, a polynomial in two variables x and y, is a
sum of terms of the form cxk y m , where c is a constant, k and m are non-negative
integers and we denote it by P (x, y). The number k + m is called the degree of the
term, and the degree of the polynomial P (x, y) is equal to the degree of the term
with the largest degree. We can add, subtract and multiply polynomials of several
variables in the same way as the polynomials in one variable. To simplify the
polynomials, the terms of the same degree are grouped together. We will consider
two types of polynomials with two variables: symmetric polynomials, that is, the
ones that satisfy P (x, y) = P (y, x), and homogeneous polynomials where all the
terms have the same degree.
Similar definitions can be given for polynomials in three variables x, y and z. A
polynomial, in three variables, is any finite sum of terms of the form cxk y n z m ,
where k, n and m are non-negative integers. The degree of the polynomial is
given by the maximum sum of the powers k + m + n. If all the terms have the
same degree, we say that the polynomial is homogeneous and if it satisfies that
P (x, y, z) = P (z, x, y) = P (y, z, x) = P (x, z, y) = P (y, x, z) = P (z, y, x), then we
say that P (x, y, z) is symmetric.
Example 8.8.1.
(a) The elementary symmetric polynomials σ1 = x + y and σ2 = xy are also
homogeneous, the first one of degree 1 and the second one of degree 2.
(b) The polynomial x2 + x + y + y 2 is symmetric but it is not homogeneous,
meanwhile x2 y + 2y 3 is homogeneous but not symmetric.
(c) The sum of powers si = xi + y i , with i = 0, 1, 2, . . . , are symmetric.
Theorem 8.8.2. Any symmetric polynomial in x and y can be represented as a
polynomial in σ1 = x + y and σ2 = xy.
Proof. In fact,
sn = xn + y n = (x + y)(xn−1 + y n−1 ) − xy(xn−2 + y n−2 ) = σ1 sn−1 − σ2 sn−2 ,
where si = xi + y i . Then, we have the recursion
s0 = 2,
s1 = σ1 ,
sn = σ1 sn−1 − σ2 sn−2 , for n ≥ 2.
Now, the proof for any symmetric polynomial is simple. The terms of the form
axk y k do not cause any problem, since axk y k = aσ2k . If the term bxi y k , with i < k,
appears in the polynomial, then, by symmetry, the term bxk y i also has to be part
of the polynomial. Grouping terms, it follows that
bxi y k + bxk y i = bxi y i (xk−i + y k−i ) = bσ2i sk−i .
But sk−i can be expressed as a polynomial in terms of σ1 and σ2 .
162
Chapter 8. Polynomials
The non-linear systems of symmetric equations in two variables x and y can be
simplified using the substitution σ1 = x + y and σ2 = xy. The degree of these
equations will be reduced, since σ2 = xy is a second-degree term in x and y. As
soon as we find σ1 and σ2 , we can find the solutions x, y of the system of symmetric
equations, either solving the quadratic equation z 2 − σ1 z + σ2 = 0 or solving the
system x + y = σ1 , xy = σ2 .
Example 8.8.3. Solve the system
x5 + y 5 = 31,
x + y = 1.
Take σ1 = x + y, σ2 = xy. Then
x5 + y 5 = σ1 s4 − σ2 s3 = (x + y)(x4 + y 4 ) − xy(x3 + y 3 ) = 31.
(8.9)
Since x4 + y 4 = σ1 s3 − σ2 s2 , making the substitutions recursively, we obtain that
equation (8.9) is transformed into σ15 − 5σ13 σ2 + 5σ1 σ22 = 31. Then, the system we
need to solve is
σ15 − 5σ13 σ2 + 5σ1 σ22 = 31, σ1 = 1.
Substituting the value of σ1 in the first equation, we get σ22 − σ2 − 6 = 0. This last
equation has as solutions σ2 = 3, −2. Hence, we need to solve x + y = 1, xy = 3,
−2. Therefore, the solutions are:
√
√
√
√
1 + i 11 1 − i 11
1 − i 11 1 + i 11
,
, (2, −1), (−1, 2).
,
,
2
2
2
2
The symmetric polynomials in three variables can be expressed in terms of
the symmetric polynomials
σ1 = x + y + z,
σ2 = xy + yz + zx and σ3 = xyz.
The sum of the powers si = xi + y i + z i , with i = 0, 1, 2, . . . , can be expressed
with σ1 , σ2 and σ3 , in the following way:
s0 = x0 + y 0 + z 0 = 3,
s1 = x + y + z = σ1 ,
s2 = x2 + y 2 + z 2 = σ12 − 2σ2 ,
s3 = x3 + y 3 + z 3 = σ13 − 3σ1 σ2 + 3σ3 ,
and, for n ≥ 3, we have the recursive equation
sn = σ1 sn−1 − σ2 sn−2 + σ3 sn−3 .
The non-linear systems of symmetric equations in three variables x, y and z
can be simplified using σ1 , σ2 and σ3 . Once we have the equations in σ1 , σ2 and
163
8.8 Polynomials of several variables
σ3 , we write the cubic equation u3 − σ1 u2 + σ2 u − σ3 = 0, where σ1 , σ2 and σ3
are its coefficients. Then, the solutions (x1 , x2 , x3 ) of this cubic equation are the
solutions of the system. The other solutions can be obtained by permutation of
the variables.
Example 8.8.4. Solve the system
x + y + z = 4,
x2 + y 2 + z 2 = 14,
x3 + y 3 + z 3 = 34.
Note that x, y, z are roots of the polynomial P (u) = u3 − (x + y + z)u2 +
(xy + yz + zx)u − xyz. This becomes
P (u) = u3 − σ1 u2 + σ2 u − σ3 ,
where σ1 = x + y + z, σ2 = xy + yz + zx and σ3 = xyz. Then
4 = s1 = σ1 ,
14 = s2 = σ12 − 2σ2 = 16 − 2σ2 , then σ2 = 1.
The numbers x, y and z are roots of P (u), then
x3 − 4x2 + x − σ3 = 0,
y 3 − 4y 2 + y − σ3 = 0,
z 3 − 4z 2 + z − σ3 = 0.
Adding the equations, we get that σ3 = −6, and so the roots we need to find are
the roots of the polynomial P (u) = u3 −4u2 +u+6. Observe that u1 = −1 is a root,
then we can factorize P (u) as P (u) = (u + 1)(u2 − 5u + 6). Solving u2 − 5u + 6 = 0,
we obtain the other roots, which in this case are u2 = 2 and u3 = 3. Hence, the
solution of the system is (x, y, z) = (−1, 2, 3) and all its permutations.
A technique used to generate integer roots for a family of quadratic polynomials or for a polynomial of several variables, is known as Vieta’s jumping. This
tool is used to find the roots of a quadratic polynomial in a recursive way. In general, when this technique is applied to polynomials of several variables only one
variable is taken into account while the other variables are considered as constants.
Let us see the following example.
Example 8.8.5. If x, y are positive integers such that
such integer is equal to 3.
x2 +y 2 +1
xy
is an integer, then
Suppose that (x0 , y0 ) is a pair of positive integers such that
x20 + y02 + 1
= k,
x0 y0
with k an integer. Using this solution, we will find another solution (x1 , y1 ) of
positive integers such that x1 + y1 < x0 + y0 . Suppose that x0 > y0 and let
164
Chapter 8. Polynomials
P (x) = x2 − ky0 x + y02 + 1, then x0 is a root of P (x) and if x1 is the other root, by
Vieta’s formulas, it follows that x0 + x1 = ky0 and x0 x1 = y02 + 1. From the first
equality, it follows that x1 is an integer, and from the second equality, x1 is positive
y 2 +1
since x0 and y02 + 1 are positive. Then x1 = 0x0 and, since x0 , y0 are integers
with x0 > y0 , it follows that x0 ≥ y0 + 1, so that x20 ≥ (y0 + 1)2 > y02 + 1, and
therefore x1 < x0 . In this way, we have found another solution (x1 , y1 ) = (x1 , y0 )
such that x1 + y1 < x0 + y0 .
This process can be continued and we can find another solution of positive
integers (x1 , y1 ) such that the sum x1 + y1 is less than the sum of the elements of
the previous solution. Since this process cannot be done infinitely, we must get a
solution (x0 , y0 ) such that x0 = y0 . Then
k=
1
y02 + y02 + 1
= 2 + 2,
y02
y0
hence 2 < k ≤ 3, from where we deduce k = 3.
Exercise 8.33. Solve the system
x5 + y 5 = 33,
x + y = 3.
√
√
Exercise 8.34. Find the solutions of the equation 4 97 − x + 4 x = 5.
Exercise 8.35. What is the relation between a, b and c if
x + y = a,
x2 + y 2 = b,
x3 + y 3 = c?
Exercise 8.36 (IMO, 1961). What conditions must satisfy a and b in order that x,
y, z are different positive real numbers and such that
x + y + z = a,
x2 + y 2 + z 2 = b2 ,
xy = z 2 ?
Exercise 8.37. Solve the system
x + y + z = a,
x2 + y 2 + z 2 = b2 ,
x3 + y 3 + z 3 = a3 .
Exercise 8.38. Find the integer solutions of the equation
(x + y 2 )(x2 + y) = (x + y)3 .
Exercise 8.39 (China, 2010). Find all integers k for which there are positive integers a, b, such that
a+1 b+1
+
= k.
b
a
Exercise 8.40. Let x, y, z be non-zero integers, and such that
x2 + y 2 + z 2
xyz
is an integer. Prove that this integer is either 1 or 3.
Chapter 9
Problems
Problem 9.1. Find the irrational numbers a such that a2 + 2a and a3 − 6a are
rational numbers.
Problem 9.2 (OMCC, 2010). Let p, q, r be rational numbers different from zero,
such that
3
pq 2 + 3 qr2 + 3 rp2
is a rational number different from zero. Prove that
1
3
pq 2
+
1
3
qr2
+
1
3
rp2
,
is also a rational number.
Problem 9.3 (APMO, 2005). Prove that for every irrational number a there exist
irrational numbers b and b′ such that a + b and ab′ are rational numbers and such
that a + b′ and ab are irrational numbers.
Problem 9.4. Let x1 , x2 , . . . , xn be positive integers less or equal than an integer
number m. Prove that if the least common multiple of each pair of integers is
greater than m, then the sum of the reciprocals of the n numbers is less than 32 .
Problem 9.5 (APMO, 2013). Find all positive integers n such that
integer number.
2
√n +1
⌊ n⌋2 +2
is an
Problem 9.6 (IMO shortlist, 1998). Determine all the pairs (a, b) of real numbers
such that a ⌊nb⌋ = b ⌊na⌋, for every positive integer n.
Problem 9.7. If n and m are positive integers without common factors, prove that
4
n 2n 4 3n 4
(m − 1)n
(m − 1)
.
+
+
+ ··· +
=
m
m
m
m
2
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5_9
165
166
Chapter 9. Problems
Problem 9.8. Find the real solutions of the system
1 1
+
a b
1
1
+ √
18 = √
3
3
a
b
9=
1
1+ √
3
a
1
1+ √
3
b
.
Problem 9.9. The different real numbers a, b, c satisfy the identities a +
b + 1c = c + a1 . Find the values that abc can get.
1
b
=
Problem 9.10. The positive real numbers a, b, c satisfy the identity abc(a+b+c) =
1. Find the minimum value of (a + b)(a + c).
Problem 9.11. The real numbers a, b, c different from zero, satisfy the identity
$
1+
a%
b
1+
bc
ca
Find the value of a + b + c.
$
c%
1+
=
ab
1 1 1
+ + −1
a b
c
2
.
Problem 9.12. Let a, b, c be real numbers different from zero and such that a +
b + c = 0. Find the value of
b−c c−a a−b
+
+
a
b
c
a
b
c
+
+
b−c c−a a−b
.
Problem 9.13. Let a, b, c be non-zero and distinct real numbers, such that
b
c
a
+
+
= 0.
b−c c−a a−b
Find the value of
a
b
c
+
+
.
2
2
(b − c)
(c − a)
(a − b)2
Problem 9.14. Let a, b, c be integers that satisfy the equality ab + bc + ca = 1.
Prove that the number (a2 + 1)(b2 + 1)(c2 + 1) is a perfect square.
Problem 9.15. Let a, b, c be integers that satisfy the equality
that the number a2 + b2 + c2 is a perfect square.
1
a
+ 1b + 1c = 0. Prove
Problem 9.16. The real numbers a, b, c satisfy the identity (a+b+c)2 = a2 +b2 +c2 .
Prove that:
b2
c2
a2
+ 2
+ 2
= 1.
(i) 2
a + 2bc b + 2ca c + 2ab
bc
ca
ab
(ii) 2
+
+
= 1.
a + 2bc b2 + 2ca c2 + 2ab
167
Chapter 9. Problems
Problem 9.17 (Slovenia, 2005). The real numbers a, b, c satisfy the identity abc =
1. Find the value of the expression
b+1
c+1
a+1
+
+
.
ab + a + 1 bc + b + 1 ca + c + 1
Problem 9.18. Let a, b, c be distinct, non-zero integers, such that
What is the value of a − b?
a−c
b
+
b+c
a
= 2.
Problem 9.19 (Czech-Slovak-Polish, 2005). Let n be a positive integer. Find the
real non-negative numbers x1 , . . . , xn that solve the following system of equations:
x1 + x22 + x33 + · · · + xnn = n
x1 + 2x2 + 3x3 + · · · + nxn =
n(n + 1)
.
2
Problem 9.20 (Canada,
1971).
Let x, y be positive real numbers with x + y = 1.
%
$
Prove that 1 + x1 1 + y1 ≥ 9.
Problem 9.21 (Romania, 2007). For real non-negative numbers x, y, z, prove that
x3 + y 3 + z 3
3
≥ xyz + |(x − y)(y − z)(z − x)|.
3
4
Problem 9.22 (Great Britain, 2010). Let a, b, c be the lengths of the sides of a
triangle, that satisfy ab + bc + ca = 1. Prove that (a + 1)(b + 1)(c + 1) < 4.
Problem 9.23 (OMCC, 2012). Let a, b, c be real numbers such that
1
a+c = 1 and ab + bc + ca > 0. Prove that
a+b+c−
1
a+b
+
1
b+c
+
abc
≥ 4.
ab + bc + ca
Problem 9.24 (IMO, 1964). Let a, b, c be the lengths of the sides of a triangle.
Prove that
a2 (b + c − a) + b2 (a + c − b) + c2 (a + b − c) ≤ 3abc.
Problem 9.25 (IMO, 1975). Consider two collections of numbers x1 ≤ x2 ≤ · · · ≤
xn , y1 ≤ y2 ≤ · · · ≤ yn and a permutation (z1 , z2 , . . . , zn ) of (y1 , y2 , . . . , yn ).
Prove that
(x1 − y1 )2 + · · · + (xn − yn )2 ≤ (x1 − z1 )2 + · · · + (xn − zn )2 .
Problem 9.26 (IMO, 1978). Let x1 , x2 , . . . , xn be distinct positive integers. Prove
that
x2
xn
1 1
1
x1
+ 2 + ···+ 2 ≥ + + ··· + .
12
2
n
1 2
n
168
Chapter 9. Problems
Problem 9.27 (IMO shortlist, 2010). Let x1 , . . . , x100 be non-negative real numbers
such that xi + xi+1 + xi+2 ≤ 1, for every i = 1, . . . , 100 (where x101 = x1 , x102 =
x2 ). Find the maximum possible value of the sum
S=
100
#
xi xi+2 .
i=1
Problem 9.28 (Czech-Slovak-Polish, 2010). Let a, b, x, y be positive real numbers,
with a ≥ b and ab ≥ ax + by. Prove that:
(i) x + y ≤ a,
√
√
√
(ii) a + b ≥ x + y.
Problem 9.29 (Thailand, 2005). Let a, b, c be real numbers, prove that
2a − b
a−b
2
+
2b − c
b−c
2
+
2c − a
c−a
2
≥ 5.
Problem 9.30. Find all triplets of positive real numbers a, b, c such that they
satisfy the following inequalities
ab + 1 ≤ 2c,
bc + 1 ≤ 2a,
ca + 1 ≤ 2b.
Problem 9.31 (Czech-Slovak-Polish, 2010). Determine all triplets of positive integers (a, b, c), that satisfy the identities:
√
a b−c=a
√
b c−a=b
√
c a − b = c.
Problem 9.32 (IMO, 1968). Let a, b, c be real numbers. Prove that the system of
equations
ax21 + bx1 + c = x2
ax22 + bx2 + c = x3
..
..
.
.
ax2n−1 + bxn−1 + c = xn
ax2n + bxn + c = x1 ,
has a unique real solution if and only if (b − 1)2 − 4ac = 0.
169
Chapter 9. Problems
Problem 9.33 (IMO shorlist, 1967). Solve the following system:
x2 + x − 1 = y
y2 + y − 1 = z
z 2 + z − 1 = x.
Problem 9.34 (OMM, 2011). Solve the system,
x21 + x1 − 1 = x2
x22 + x2 − 1 = x3
..
..
.
.
x2n−1 + xn−1 − 1 = xn
x2n + xn − 1 = x1 .
Problem 9.35. For n ≥ 3, find all the positive solutions of the system
x21 = x2 + x3
x22 = x3 + x4
..
..
. =
.
x2n−1 = xn + x1
x2n = x1 + x2 .
Problem 9.36. Let x, y, z be real numbers such that x+y+z = x−1 +y −1 +z −1 = 0.
Prove that
x6 + y 6 + z 6
= xyz.
x3 + y 3 + z 3
Problem 9.37 (Peru, 2009). Let a, b, c, d be integers such that a + b + c + d = 0.
Prove that (bc − ad)(ac − bd)(ab − cd) is a perfect square.
Problem 9.38. Let a, b and c be real numbers different from zero, with a+b+c = 0.
Prove that
a2 + b 2
b 2 + c2
c2 + a2
a3
b3
c3
+
+
=
+
+ .
a+b
b+c
c+a
bc
ca ab
Problem 9.39. Find all triplets of positive integers (a, b, c), such that a3 + b3 +
c3 − 3abc = p, where p > 3 is a prime number.
Problem 9.40. Find all positive integers that are solutions of the equation x3 −y 3 =
xy + 61.
170
Chapter 9. Problems
Problem 9.41 (Great Britain, 2008). Find the minimum of x2 + y 2 + z 2 , where x,
y, z are real numbers that satisfy x3 + y 3 + z 3 − 3xyz = 1.
Problem 9.42 (Shortlist OMCC, 2011). The positive real numbers x, y, z, satisfy
that
z
x
y
x + = y + = z + = 2.
z
x
y
Find the value of x + y + z.
Problem 9.43. If {an } ⊂ R+ is an arithmetic progression, prove that
1
1
n
1
√ +√
√ + ···+ √
√ = √
√ .
√
a0 + a1
a1 + a2
an−1 + an
a0 + an
Problem 9.44. The lengths of the sides of a right triangle are a < b < c and they
are in arithmetic progression. Prove that their difference d is equal to the radius
of the incircle of the triangle.
Problem 9.45. Prove that in any partition of the set {1, 2, . . . , 9} into two subsets,
it is possible to find, in one of them, an arithmetic progression with three terms.
Problem 9.46. Let a, b, c be real numbers in arithmetic progression; prove that
2
(a + b + c)3 = a2 (b + c) + b2 (c + a) + c2 (a + b).
9
Problem 9.47. Prove that in the arithmetic progression {3, 7, 11, . . . , 4k − 1, . . . },
there are an infinite number of primes.
Problem 9.48. Is it possible to divide the set of natural numbers in two subsets,
such that none of them contains a non-constant arithmetic progression?
Problem 9.49. Is there a non-constant arithmetic progression where the terms of
the progression are all prime numbers?
Problem 9.50. Prove that if an arithmetic progression of positive integers contains
a perfect square then it has an infinite number of perfect squares.
Problem 9.51 (OMM, 1999). Prove that there do not exist 1999 prime numbers
in arithmetic progression, all of them smaller than 12345.
Problem 9.52 (OMM, 2005). Let us say that a list of numbers a1 , a2 , . . . , am
contains an arithmetic triplet ai , aj , ak , if i < j < k and 2aj = ai + ak . For
example, 8, 1, 5, 2, 7 has the following arithmetic triplet (8, 5, 2), but 8, 1, 2, 5,
7 does not. Let n be a positive integer. Prove that the numbers 1, 2, . . . , n can be
rearranged in a list, such that this list does not contain an arithmetic triplet.
171
Chapter 9. Problems
Problem 9.53 (Czech-Slovak, 2010). The four real solutions of the equation ax4 +
bx2 + a = 1, form an increasing arithmetic progression. One of the solutions is
also a solution of the equation bx2 + ax + a = 1. Find all possible real values of a
and b.
Problem 9.54 (APMO, 2013). For 2k real numbers a1 , a2 , . . . , ak , b1 , b2 , . . . , bk ,
define the sequence of numbers Xn by
Xn =
k
#
i=1
⌊ai n + bi ⌋ , (n = 1, 2, . . . ).
*k If the sequence Xn forms an arithmetic progression, prove that the sum
i=1 ai has to be an integer number.
(
)
Problem 9.55. Prove that in any partition of 1, 2, 22 , . . . , 256 into two subsets,
it is possible to find in one of them three terms that are in geometric progression.
Problem 9.56. Let n > 1 and consider the collection of numbers a0 , a1 , . . . , an
defined by
a0 =
1
2
and
ak+1 = ak +
ak
,
n2
with
k = 0, 1, . . . , n − 1.
Prove that an < 1.
Problem 9.57. Let a1 < a2 < · · · < an < · · · be an increasing sequence of positive
integers such that:
(i) For every n ≥ 1, a2n = an + n.
(ii) If ap is a prime number, then p is prime.
Prove that an = n, for all n ≥ 1.
Problem 9.58. An arbitraty set of m + n numbers is divided into two arbitrary sets
a1 , a2 , . . . , am , b1 , b2 , . . . , bn . Order the numbers, in each set, in increasing form
a1 < a2 < · · · < am ,
b1 < b2 < · · · < bn .
Then, the numbers in each set are divided again into two subsets c1 , c2 , . . . , cm
and d1 , d2 , . . . , dn and let us arrange them in increasing order
c1 < c2 < · · · < cm ,
d1 < d2 < · · · < dn .
Prove the equality
|a1 − c1 | + |a2 − c2 | + · · · + |am − cm | = |b1 − d1 | + |b2 − d2 | + · · · + |bn − dn |.
172
Chapter 9. Problems
Problem 9.59 (Poland, 1986). Prove that, for every integer n ≥ 3, the number n!
can be represented as the sum of n different divisors of n!.
Problem 9.60. For each prime number p > 3, prove that the number
divisible by p3 .
2p−1
p−1
− 1 is
Problem 9.61 (UK, 2004). Let S be a set of rational numbers that satisfy:
(i) 12 ∈ S.
(ii) If x ∈ S, then
1
x+1
∈ S and
x
x+1
∈ S.
Prove that S contains all rational numbers in the interval 0 < x < 1.
Problem 9.62 (IMO, 1988). Prove that, if a, b are positive integers such that ab+1
2
+b2
divides a2 + b2 , then aab+1
is a perfect square.
Problem 9.63 (Ireland, 2007). If a, b, c are roots of the polynomial P (x) = x3 −
2007x + 2002, determine the value of
a−1
a+1
b−1
b+1
c−1
c+1
.
Problem 9.64. Determine all positive rational numbers a, b, c such that a + b + c,
abc and a1 + 1b + 1c are integers.
Problem 9.65. Let a, b, c be real numbers, not all zero. Prove that one of the
equations ax2 + 2bx + c = 0, bx2 + 2cx + a = 0, cx2 + 2ax + b = 0 has a real root.
Problem 9.66 (Estonia, 2005). Find all pairs of real numbers (a, b) such that the
roots of the polynomials
6x2 − 24x − 4a
and
x3 + ax2 + bx − 8
are all non-negative real numbers.
Problem 9.67. Let P (x) be a polynomial such that |P (x)| ≤ 1, for |x| ≤ 1.
(i) (Short list IMO, 1986) If P (x) = ax2 + bx + c, find the maximum value of
|a| + |b| + |c|.
(ii) (Short list IMO, 1996) If P (x) = ax3 + bx2 + cx + d, find the maximum value
of |a| + |b| + |c| + |d|.
Problem 9.68. Let a, b, c, d be real numbers, with a and d distinct from zero.
Prove that if the roots of the polynomials ax3 + bx2 + cx + d and dx3 + cx2 + bx + a
bc
are positive, then ad
≥ 9.
173
Chapter 9. Problems
Problem
9.69 (IMO, 1963). Find all real solutions of the equation
√
2 x2 − 1 = x, where p is a real number.
x2 − p +
Problem 9.70 (China, 2008). Let P (x) = ax3 + bx2 + cx + d be a polynomial with
real coefficients. If P (x) has three positive real roots and P (0) < 0, prove that
2b3 + 9a2 d − 7abc ≤ 0.
Problem 9.71 (India, 2010). Let a, b, c be integers with b even and c odd. Suppose
that the equation x3 + ax2 + bx + c = 0 has roots α, β, γ, with α2 = β + γ. Prove
that α is an integer and that β = γ.
Problem 9.72. Consider all quadratic equations x2 + px + q = 0, where the coefficients p, q belong to the interval [−1, 1]. Find all possible values of the solutions
of those equations.
Problem 9.73 (Hermite’s identity). Given a positive integer n and a real number
x, prove that
1
2
n−1
⌊nx⌋ = ⌊x⌋ + x +
+ x+
+ ··· + x +
.
n
n
n
Problem 9.74 (IMO, 1968). For every positive integer n, prove that
n + 2k
n+1
n+2
+ ···+
+ · · · = n.
+
2
22
2k+1
Problem 9.75 (USA, 1981). Prove that if x is a positive real number and n is a
positive integer, then
⌊nx⌋ ≥
⌊x⌋ ⌊2x⌋
⌊nx⌋
+
+ ··· +
.
1
2
n
Problem 9.76. The function f assigns to each non-negative integer n, the nonnegative integer f (n), such that:
(i) f (nm) = f (n) + f (m), for n, m ≥ 0.
(ii) f (n) = 0 if the units digit of n is 3.
(iii) f (10) = 0.
Find the value of f (1985).
Problem 9.77 (Czech-Slovak, 2010). Find all functions f : R+ → R+ that satisfy
f (x)f (y) = f (y)f (xf (y)) +
1
,
xy
with x, y ∈ R+ .
174
Chapter 9. Problems
Problem 9.78 (Thailand, 2004). Let f : [0, 1] → R be a function such that:
(i) f (0) = f (1) = 0.
(ii) |f (x) − f (y)| < |x − y|, for x, y ∈ [0, 1] with x = y.
Prove that |f (x) − f (y)| < 12 , for all x, y ∈ [0, 1].
Problem 9.79 (IMO, 1978). The set of all positive integers is the union of two
disjoint subsets {f (1), f (2), . . . , f (n), . . . }, {g(1), g(2), . . . , g(n), . . . }, where
f (1) < f (2) < · · · < f (n) < . . . ,
g(1) < g(2) < · · · < g(n) < . . . ,
and
g(n) = f (f (n)) + 1,
for all
n ≥ 1.
Determine f (240).
Problem 9.80 (IMO, 1981). The function f (x, y) satisfies
f (0, y) = y + 1
f (x + 1, 0) = f (x, 1)
f (x + 1, y + 1) = f (x, f (x + 1, y)),
(9.1)
(9.2)
(9.3)
for all non-negative integers x, y. Determine the value of f (4, 1981).
Problem 9.81 (IMO, 1982). The function f (n) is defined for all positive integers
n and takes non-negative values. Also, for all m and n,
f (n + m) − f (m) − f (n) = 0 or 1,
f (2) = 0, f (3) > 0 and f (9999) = 3333.
Find f (1982).
Problem 9.82 (IMO, 1983). Find all functions f defined in the set of positive real
numbers that satisfy the conditions:
(i) f (xf (y)) = yf (x), for all positive numbers x, y.
(ii) f (x) → 0, when x → ∞.
Problem 9.83 (IMO, 2010). Find all functions f : R → R such that, for all x,
y ∈ R, the equality
f (⌊x⌋y) = f (x)⌊f (y)⌋
holds, where ⌊x⌋ denotes the greatest integer less than or equal to x.
175
Chapter 9. Problems
Problem 9.84 (APMO, 2011). Find all upper bounded functions f : R → R that
satisfy
f (xf (y)) + yf (x) = xf (y) + f (xy), for x, y ∈ R.
(9.4)
Problem 9.85 (OIM, 2009). Let {an } be a sequence defined by
a1 = 1,
a2n = an + 1,
a2n+1 =
1
,
a2n
for n ≥ 1.
Prove that, for every rational number r, there is a unique positive integer n with
an = r.
Problem 9.86. Find the term a1000 of the sequence defined by
a0 = 1
and
an+1 =
an
,
1 + nan
for
n ≥ 0.
Problem 9.87 (Short list IMO, 2010). A sequence x1 , x2 , . . . , is defined by x1 = 1
and x2k = −xk , x2k−1 = (−1)k+1 xk , for all k ≥ 1. Prove that x1 +x2 +· · ·+xn ≥ 0,
for all n ≥ 1.
Problem 9.88 (IMO, 2010). Let a1 , a2 , . . . , an , . . . be a sequence of positive real
numbers. Suppose that for some positive integer s, we have that
an = max{ak + an−k : 1 ≤ k ≤ n − 1},
for all n > s. Prove that there exist positive integers l and N , with l ≤ s, such
that an = an−l + al , for all n ≥ N .
Problem 9.89 (Bulgaria, 1987). Let k be an integer greater than 1. Prove that
there exist a prime number p and an increasing sequence of positive integers
a1 , a2 , . . . , an , . . . , such that the terms of the sequence
p + ka1 , p + ka2 , . . . , p + kan , . . .
are all prime numbers.
Problem 9.90 (Austria, 2005). For real numbers a, b, c, set sn = an + bn + cn , for
n ≥ 0. Suppose that s1 = 2, s2 = 6 and s3 = 14. Prove that s2n − sn−1 sn+1 = 8,
for all n ≥ 2.
Problem 9.91 (IMO, 1982). Consider an infinite sequence of positive real numbers
{xn }, such that x0 = 1 and xi+1 ≤ xi , for all i ≥ 0.
(i) Prove that, for any sequence that satisfies the given conditions, there exists
an integer n ≥ 1 such that
x2
x2
x20
+ 1 + · · · + n−1 ≥ 3.999.
x1
x2
xn
176
Chapter 9. Problems
(ii) Find a sequence with the given conditions such that
x2
x2
x20
+ 1 + · · · + n−1 < 4,
x1
x2
xn
for all n ≥ 1.
Problem 9.92 (Great Britain, 2009). Find all sequences {an } that satisfy the
following conditions:
(i) an+1 = 2a2n − 1, for all n ≥ 1.
(ii) a1 is a rational number.
(iii) ai = aj , for some i, j with i = j.
Problem 9.93. If a0 = 0, a1 = 1 and an = 2an−1 + an−2 , prove that 2k |an if and
only if 2k |n.
Problem 9.94 (Russia, 1989). The sequence {an } is such that
|am + an − am+n | ≤
1
, for all m, n ≥ 1.
m+n
Prove that {an } is an arithmetic progression.
Problem 9.95 (Bulgaria, 1996). The sequence {an } is defined by
a1 = 1 and an+1 =
n
an
+
, for n ≥ 1.
n
an
Prove that ⌊a2n ⌋ = n, for all n ≥ 4.
Problem 9.96. Let bn be the units digit of the number 11 + 22 + 33 + · · · + nn .
Prove that the sequence {bn } is periodic with period 100.
√
Problem 9.97. Prove that ⌊(1 + 3)2n+1 ⌋ is an even integer, for n ≥ 0.
Problem 9.98. The sequence of integers {an } is defined by a1 = 3, a2 = 5 and
an+2 = 3an+1 − 2an , for n ≥ 1. Prove that an = 2n + 1, for all integers n ≥ 1.
Problem 9.99. The sequence of integers {an } is defined by a1 = 1, a2 = 2 and
an+2 = an+1 − an , for n ≥ 1. Prove that an+6 = an , for all integers n ≥ 1.
Problem 9.100 (Short list IMO, 1986). Let a0 = a1 = 1 and, for n ≥ 0, an+2 =
7an+1 − an − 2. Prove that an is a perfect square, for every integer number n ≥ 0.
Problem 9.101 (IMO, 1976). The sequence {un } is defined by u0 = 2, u1 =
2n −(−1)n
3
5
2
and
, where ⌊x⌋ is
un+1 = un (u2n−1 − 2) − u1 , for n ≥ 1. Prove that ⌊un ⌋ = 2
the integer part of x.
√
Problem 9.102. If an = ⌊ 2n⌋, for n ≥ 1, what is the value of a2020 ?
177
Chapter 9. Problems
Problem 9.103. Let P (x) be a polynomial of degree n with P (j) =
0, 1, . . . , n. Find P (m), for m > n.
j
j+1 ,
for j =
1
j,
for j =
Problem 9.104. Let P (x) be a polynomial of degree n with P (j) =
20 , 21 , . . . , 2n . Find P (0).
Problem 9.105 (Short list IMO, 1981). Let P (x) be a polynomial of degree n with
−1
, for j = 0, 1, . . . , n. Find P (n + 1).
P (j) = n+1
j
Problem 9.106 (IMO, 1993). Let n > 1 be an integer and let P (x) = xn +5xn−1 +3.
Prove that P (x) is irreducible over Z[x].
Problem 9.107 (Short list, 1997). Let p be a prime number and let Q(x) be a
polynomial of degree n with integer coefficients such that:
(i) Q(0) = 0, Q(1) = 1,
(ii) For any integer n, Q(n) is congruent to 0 or 1 module p.
Prove that n ≥ p − 1.
Problem 9.108 (Short list IMO, 1997). Let P (x) be a polynomial with real coefficients and P (x) > 0, for x ≥ 0. Prove that there is a positive integer n such that
the coefficients of the polynomial (1 + x)n P (x) are all positive.
Problem 9.109 (Short list IMO, 2002). Let P (x) = ax3 + bx2 + cx + d be a cubic
polynomial with integer coefficients a, b, c, d and a = 0. If xP (x) = yP (y), for an
infinite number of integers x, y, with x = y, prove that P (x) has an integer root.
Problem 9.110. Consider the positive real numbers a, b, c. Solve the system of
equations:
xy = a,
yz = b,
zx = c.
Problem 9.111. Consider the real numbers a, b, c. Solve the system of equations:
x(y + z) = a,
y(z + x) = b,
z(x + y) = c.
Problem 9.112. Consider the positive real numbers a, b, c. Solve the system of
equations:
xyz
xyz
xyz
= c,
= a,
= b.
x+y
y+z
z+x
Problem 9.113 (Balkanic, 2002). Solve the system of equations:
a3 + 3ab2 + 3ac2 − 6abc = 1,
b3 + 3ba2 + 3bc2 − 6abc = 1,
c3 + 3cb2 + 3ca2 − 6abc = 1.
178
Chapter 9. Problems
Problem 9.114. Let P (x) be a polynomial that takes integer values in the integers.
Prove that there are integers c0 , c1 , . . . , cn such that
P (x) = cn
where
x
j
=
x
x
x
,
+ · · · + c0
+ cn−1
0
n−1
n
x(x−1)(x−2)...(x−j+1)
.
j!
Problem 9.115. Let p be an odd prime number and let P (x) = xp − x + p. Prove
that P (x) is irreducible over Z[x].
Problem 9.116 (IMO, 2004). Find all polynomials P (x) with real coefficients that
satisfy the equality
P (a − b) + P (b − c) + P (c − a) = 2P (a + b + c),
for all real numbers a, b, c, with ab + bc + ca = 0.
Problem 9.117 (IMO, 2006). Let P (x) be a polynomial of degree n > 1, with
integer coefficients and let k be a positive integer. Consider the polynomial Q(x) =
P (P (. . . P (P (x)) . . . )), with k pairs of parenthesis. Prove that Q(x) has at most n
integer fixed points, that is, integers that satisfy the equation Q(x) = x.
Problem 9.118. Find all polynomials P (x) such that P (0) = 0 and P (x2 + 1) =
P (x)2 + 1, for all real numbers x.
Problem 9.119. Find all polynomials P (x) such that P (x)2 − 2 = 2P (2x2 − 1), for
all real numbers x.
Problem 9.120. Let P (x) and Q(x) be monic polynomials such that P (P (x)) =
Q(Q(x)), for all real numbers x. Prove that P (x) = Q(x).
Chapter 10
Solutions to Exercises and Problems
The first eight sections of this chapter contain all solutions of the exercises in the
first eight chapters. In Section 9, you can find the solutions to the problems of
Chapter 9. The difficulty of the problems in Chapter 9 is usually greater than the
difficulty of the exercises that you find in the first eight chapters. However, solving
the problems of this last chapter would be an excellent training in preparation for
international mathematical competitions.
We recommend that the reader consult this last chapter just in case he or
she cannot solve the exercises and problems alone.
10.1 Solutions to exercises of Chapter 1
Solution 1.1. (i) If a < 0, then −a > 0. Also use (−a)(−b) = ab. (ii) (−a)b > 0. (iii)
a < b ⇔ b − a > 0, now use property (a). (iv) Use property (a). (v) aa−1 = 1 > 0.
(vi) If a < 0, then −a > 0.
Solution 1.2. Observe that if a2 + b − (a + b2 ) ∈ Q, then (a − b)(a + b − 1) ∈ Q and,
since a + b − 1 ∈ Q \ {0}, then (a − b) ∈ Q. Therefore, if a + b ∈ Q and a − b ∈ Q,
then 2a and 2b are in Q. Therefore, a and b are rational numbers.
Solution 1.3. If a = 0 or b = 0, then the result is clear. Now suppose ab = 0.
Since (a2 + b2 )2 − (a4 + b4 ) = 2a2 b2 , we have that a2 b2 ∈ Q. Note that a6 + b6 =
(a2 + b2 )3 − 3a2b2 (a2 + b2 ) ∈ Q, then (a3 + b3 )2 − (a6 + b6 ) = 2a3 b3 ∈ Q. Therefore,
a3 + b 3
a3 b 3
∈ Q.
∈
Q
and
a
+
b
=
a2 b 2
a2 + b2 − ab
√
√
Solution 1.4. (i) Suppose p is not an irrational number, that is, p = m
n , where
m, n are integers with (m, n) = 1, that is, m and n are relatively prime numbers.
Squaring both sides of the equality, we get pn2 = m2 , that is, p divides m2 , then
p divides m. Therefore, m = ps and pn2 = p2 s2 imply n2 = ps2 , this guarantees
that p divides n2 and also divides n. Therefore, p divides m and n contradicting
the fact that m and n are relatively prime.
ab =
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5_10
179
180
Chapter 10. Solutions to Exercises and Problems
√
√
(ii) Suppose m is not an irrational number, that is, m = rs , where r, s
are integers with (r, s) = 1. Squaring we have ms2 = r2 . Since m is not a perfect
square, it has a factor of the form pα , where p is a prime number and α is a positive
odd integer. Then pα divides r2 , which means that the prime p appears an even
number of times in the factor decomposition of r2 . Since r and s are relatively
prime numbers, p does not divide s, hence p appears an odd number of times as
a factor of ms2 , which is a contradiction.
Solution 1.5. If a + b = ab = n, then b = n − a and n = a(n − a). The last equation
is equivalent to a2 − na + n = 0; solving the equation we have
√
√
n ± n2 − 4n
n ∓ n2 − 4n
a=
, from where b =
.
2
2
√
For n ≥ 5, we have (n − 3)2 < n2 − 4n < (n − 2)2 , therefore n2 − 4n is an
irrational number, and then a and b are irrational numbers.
Solution 1.6. Suppose
m
n
is a root, with (m, n) = 1, then m and n cannot both be
2
2
+b m
even. On the other hand, since a m
n
n + c = 0, we have that am + bmn +
2
cn = 0. The right-hand side of the last equation is even and the left-hand side
is odd. If m and n are odd numbers, the three terms of the left-hand side of the
equation are odd. Now, if one term is even and the other is odd then two terms
are even, the third odd and the sum is odd again. This contradiction implies that
the equation cannot have rational roots.
Second Solution. The discriminant b2 − 4ac has to be a perfect square. But, since
a, b and c are odd numbers, we can prove that b2 − 4ac ≡ 5 mod 8. However, the
square of an odd number has remainder 1 modulo 8.
√
√
Solution 1.7. Let u = a + b and v = a − b, then
√
√
√
√
√
√
u+ v
u− v
a+ b= u=
+
2
2
√
√ 2
√
√
( u + v)
( u − v)2
+
=
4
4
√
√
u+v
u+v
uv
uv
2 +
2 −
+
=
2
2
√
√
√
√
√
√
a+ b+a− b
a+ b+a− b
+ a2 − b
− a2 − b
2
2
+
=
2
2
√
√
a + a2 − b
a − a2 − b
+
,
=
2
2
as we wanted to prove.
181
10.1 Solutions of Chapter 1
Solution 1.8. (i) Let x =
a
a
√
a a . . ., then x2 = a
a
a
√
a a . . ., therefore,
x2 = ax. Factorizing, x(x − a) = 0. Therefore, since a is positive the solution is
x = a.
Second Solution. We can give another solution using series. We have
1
1
1
1
1
1
x = a 2 a 4 a 8 · · · = a 2 + 4 + 8 +··· = a,
since
*∞
1
j=1 2j
= 1, see Section 7.3.1.
√
b a b . . ., then x2 = a
√
3
a2 bx. Since x = 0, x3 = a2 b, then x = a2 b.
a
(ii) Let x =
b
a
√
b a . . ., therefore x4 =
Second Solution. We can give another solution using series. We have
1
since
*∞
1
j=1 22j
Solution 1.9.
=
1
3
1
1
1
1
1
2
1
x = a 2 + 8 + 32 +··· b 4 + 16 + 64 +··· = a 3 b 3 ,
*
2
1
and ∞
j=0 22j+1 = 3 , see Section 7.3.1.
(i) If xy, yz and zx are in Q, then (xy)(zx)
= x2 ∈ Q. Similarly, y 2 , z 2 ∈ Q.
yz
Therefore, x2 + y 2 + z 2 ∈ Q.
(ii) By (i) we have (x2 )2 + (xy)y 2 + (xz)z 2 = x(x3 + y 3 + z 3 ) ∈ Q, then x ∈ Q.
Similarly, y, z ∈ Q.
$
√ %
√
√
Solution 1.10. Since a − ab = a 1 − √ab , it is sufficient to prove that 1 − √ab
is a rational
number
from
zero to claim that a is a rational number.
√ √ different
√
√
√
b( b− a)
b
b− √ab
√
√
But a− ab = √a(√a− b) = − a is a rational number different from −1 (since
a = b), therefore 1 −
rational number.
√
√b
a
is a rational number different from 0. Similarly, b is a
Solution 1.11. To solve (i), define x = 0.111 . . . , then 10x = 1.11 . . . . Subtracting
the first equation from the second, we get 9x = 1, therefore x = 19 .
(ii) Let x = 1.141414 . . ., then 100x = 114.141414 . . .. Subtracting the first equation from the second, we get 99x = 113, therefore x = 113
99 .
Solution 1.12.
(i) First observe that 121b = (1 × b2 ) + (2 × b) + 1 = (b + 1)2 , then 121b is a
perfect square in any base b ≥ 2.
(ii) Since 232b = 2b2 + 3b + 2 has to be a square and since 3 is one of its digits,
b ≥ 4.
182
Chapter 10. Solutions to Exercises and Problems
For b = 4, 2324 = 46, for b = 5, 2325 = 67, for b = 6, 2326 = 92 and for b = 7,
2327 = 121. Then, b = 7 is the smallest positive integer such that 232b is a perfect
square.
Solution 1.13. Suppose that a > b. Then for all integers 0 ≤ k ≤ n, xn xk an bk ≥
xn xk bn ak , where the equality holds only when k = n or xk = 0. In particular, we
have a strict inequality for k = n − 1. Adding, this becomes
xn an
n
#
xk bk > xn bn
k=0
or
n
#
xk ak
k=0
xn bn
xn an
>
.
An
Bn
This implies that
An−1
xn an
xn bn
Bn−1
=1−
<1−
=
.
An
An
Bn
Bn
Bn−1
An−1
An = Bn , and if a < b, using what
> BBn−1
. Therefore, AAn−1
< BBn−1
if and
n
n
n
On the other hand, if a = b, then clearly
we proved before, it follows that
only if a > b.
An−1
An
Solution 1.14. Note that |a| = |a − b + b| ≤ |a − b| + |b|, therefore we have
|a|−|b| ≤ |a−b|. Similarly, following the same procedure, we have |b|−|a| ≤ |b−a|.
From these two inequalities, we get ||a| − |b|| ≤ |a − b|.
Solution 1.15.
(i) |x − 1| − |x + 1| = 0 is equivalent to |x − 1| = |x + 1|. Squaring both sides
of the previous equation and solving (x − 1)2 = (x + 1)2 , we have 4x = 0,
therefore the only solution is x = 0.
(ii) |x − 1||x + 1| = 1 is equivalent to |x2 − 1| = 1, hence
x2 − 1 = 1
x2 = 2
or −(x2 − 1) = 1,
or
x2 = 0,
√
x = 0,
x = ± 2 or
√
therefore the solutions are x = ± 2 and x = 0.
(iii) If x > 1 we get |x + 1| = x + 1 > 2, therefore there are no solutions.
If x < −1 we get |x − 1| = −x + 1 > 2, therefore there are no solutions.
If −1 ≤ x ≤ 1, then x − 1 ≤ 0 ≤ x + 1, therefore
|x − 1| + |x + 1| = (1 − x) + (x + 1) = 2.
Thus, the only values of x that satisfy the equality are −1 ≤ x ≤ 1.
183
10.1 Solutions of Chapter 1
Solution 1.16. From the first and third inequalities we have z ≥ |x + y| − 1 ≥ 0.
Therefore, z 2 ≥ (|x + y| − 1)2 . Now, 2xy ≥ z 2 + 1 ≥ (|x + y| − 1)2 + 1 ≥ 0, then
2xy ≥ x2 + 2xy + y 2 − 2|x + y| + 2 ≥ |x|2 + 2xy + |y|2 − 2|x| − 2|y| + 2,
cancelling out, 0 ≥ |x|2 + |y|2 − 2|x| − 2|y| + 2 = (|x| − 1)2 + (|y| − 1)2 . Therefore
|x| = 1 and |y| = 1. Since x and y have to be −1 or 1, but since xy ≥ 0, both
numbers have the same sign. For x = y = 1 or x = y = −1 we get, substituting
in the original equations, that 2 − z 2 ≥ 1 and z − 2 ≥ −1. Therefore, z 2 ≤ 1 and
z ≥ 1. The only value of z that satisfies both inequalities is z = 1. Therefore, there
are two solutions to the problem x = y = z = 1 and x = y = −1, z = 1.
Solution 1.17. Suppose that a1 < a2 < · · · < an is a collection with the largest
quantity of integers that satisfy the property. It is clear that ai ≥ i, for all i =
1, . . . , n.
ab
,
If a and b are two integers from the collection a > b, since |a − b| = a − b ≥ 100
b
100b
we get a 1 − 100 ≥ b; therefore, if 100 − b > 0, then a ≥ 100−b .
Note that there are no two numbers a and b in the collection greater than 100; in
ab
> a, which is false.
fact if a > b > 100, then a − b = |a − b| ≥ 100
100b
100a
≥ 100−b
if
We also have that for integers a and b smaller than 100, we have 100−a
and only if 100a − ab ≥ 100b − ab if and only if a ≥ b.
It is clear that {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is a collection whose elements satisfy the
property.
100a10
100·10
= 100
≥ 100−10
Now, a11 ≥ 100−a
9 > 11, which implies that a11 ≥ 12.
10
a12 ≥
100a11
100−a11
≥
100·12
100−12
=
1200
88
> 13, hence a12 ≥ 14.
a13 ≥
100a12
100−a12
≥
100·14
100−14
=
1400
86
> 16, hence a13 ≥ 17.
a14 ≥
100a13
100−a13
≥
100·17
100−17
=
1700
83
> 20, hence a14 ≥ 21.
a15 ≥
100a14
100−a14
≥
100·21
100−21
=
2100
79
> 26, hence a15 ≥ 27.
a16 ≥
100a15
100−a15
≥
100·27
100−27
=
2700
73
> 36, hence a16 ≥ 37.
a17 ≥
100a16
100−a16
≥
100·37
100−37
=
3700
63
> 58, hence a17 ≥ 59.
a18 ≥
100a17
100−a17
≥
100·59
100−59
=
5900
41
> 143, hence a18 ≥ 144.
Moreover, as we have already observed, there are no two integers of the
collection greater than 100, so the largest quantity is 18. A collection with 18
integers that satisfies the conditions is
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 17, 21, 27, 37, 59, 144} .
184
Chapter 10. Solutions to Exercises and Problems
Solution 1.18. By Example 1.3.2, ⌊2a⌋ = ⌊a⌋ + ⌊a + 12 ⌋ and ⌊2b⌋ = ⌊b⌋ + ⌊b + 12 ⌋,
then the inequality that we have to prove is equivalent to
1
1
+ ⌊b⌋ + b +
≥ ⌊a⌋ + ⌊b⌋ + ⌊a + b⌋,
⌊a⌋ + a +
2
2
then we only have to prove that a + 21 + b + 21 ≥ ⌊a + b⌋.
Let a = n+ y, b = m+ x, with n, m ∈ Z and 0 ≤ x, y < 1. Then 0 ≤ x+ y < 2
and a + b = n + m + x + y. We have two cases:
(i) If 1 ≤ x + y < 2, then ⌊a + b⌋ = n + m + 1, and at least one of the
numbers x or y is greater than or equal to 21 . Suppose that x ≥ 12 , then
⌊b+ 21 ⌋ = ⌊m+x+ 21 ⌋ = m+1, therefore ⌊a+ 21 ⌋+⌊b+ 21 ⌋ ≥ m+n+1 = ⌊a+b⌋.
(ii) If 0 ≤ x+ y < 1, then ⌊a+ b⌋ = n+ m and ⌊a+ 21 ⌋+ ⌊b + 12 ⌋ ≥ m+ n = ⌊a+ b⌋.
Solution 1.19. (i) We have that ⌊x⌊x⌋⌋ = 1 if and only if 1 ≤ x⌊x⌋ < 2. If
x = m + y, with m ∈ Z and 0 ≤ y < 1, then 1 ≤ m2 + my < 2. Observe that
m = 0 is impossible, as well as m ≥ 2 or m ≤ −2. Therefore, it only remains to
be proved for m = 1 or m = −1.
If m = 1, then 1 ≤ 1 + y < 2, from which 0 ≤ y < 1 and then any x in
the interval [1, 2) satisfies the equation. If m = −1, then since 1 ≤ m2 + my < 2,
we have 1 ≤ 1 − y < 2, from which 0 ≤ −y < 1 and then y = 0 and x = −1.
Therefore, the numbers that satisfy the equation are x = −1 and x ∈ [1, 2).
(ii) Since ⌊x⌋ ≤ x ≤ |x|, it follows that |x| − ⌊x⌋ ≥ 0, therefore ||x| − ⌊x⌋| =
|x| − ⌊x⌋. On the other hand, by property (c) in 1.3.1 we obtain ⌊|x| − ⌊x⌋⌋ =
⌊|x|⌋ − ⌊x⌋. Using the last equalities, the equation becomes |x| − ⌊x⌋ = ⌊|x|⌋ − ⌊x⌋,
which is equivalent to |x| = ⌊|x|⌋; then |x| is an integer number and the values of
x that satisfy the equation are all the integers.
Solution 1.20. Add the three equations to obtain 2x + 2y + 2z = 6.6, so that
x + y + z = 3.3. If you subtract this equation from the original ones, we obtain
{y} + ⌊z⌋ = 2.2,
{x} + ⌊y⌋ = 1.1,
{z} + ⌊x⌋ = 0.
For the first equation, we obtain ⌊z⌋ = 2 and {y} = 0.2; the second equation
becomes ⌊y⌋ = 1, {x} = 0.1, and the third ⌊x⌋ = 0 and {z} = 0. Therefore, the
solution is x = 0.1, y = 1.2 and z = 2.
√
√
√ √
2
Solution 1.21. We have n+ n +
√ 1 < 4n + 2 if and only if 2n+1+ 4n + 4n <
2
again,√the last
4n + 2, which is equivalent to 4n + 4n < 2n + 1. Squaring √
2
2
inequality
is
equivalent
to
4n
+4n
<
4n
+4n+1.
This
proves
that
n+ n + 1 <
√
√
√
√
4n + 2, then ⌊ n + n + 1⌋ ≤ ⌊ 4n + 2⌋.
10.1 Solutions of Chapter 1
185
√
√
√
⌊ 4n + 2⌋.
Suppose
√ integer n, ⌊√ n + n + 1⌋ =
√ that, for some
√ positive
Let q =√ ⌊ 4n + 2⌋, then n + n + 1 < q ≤√ 4n + 2. Squaring, we obtain
2n+1+ 4n2 + 4n < q 2 ≤ 4n+2, or equivalently, 4n2 + 4n < q 2 −2n−1 ≤ 2n+1.
Squaring again we find that 4n2 + 4n < (q 2 − 2n − 1)2 ≤ 4n2 + 4n + 1 = (2n + 1)2 .
Since there does not exist a square between two consecutive integers, we have
q 2 − 2n − 1 = 2n + 1 or q 2 = 4n + 2, which is equivalent to saying that q 2 ≡ 2
mod 4. But this is a contradiction, since any square number is congruent to 0 or
1 mod 4. Therefore, we get the equality.
√
√
√
We now prove that ⌊ 4n + 1⌋ = ⌊ 4n + 2⌋ = ⌊ 4n + 3⌋.
√
For the
that there exists n such
√ first equality, suppose √
√ that m = ⌊2 4n + 1⌋ <
m + 1 = ⌊ 4n + 2⌋, therefore m ≤ 4n + 1 < m + 1 ≤ 4n + 2, or m ≤ 4n + 1 <
(m + 1)2 ≤ 4n + 2. Therefore, since 4n + 1 and 4n + 2 are two consecutive integers,
and since (m + 1)2 > 4n + 1, therefore (m + 1)2 = 4n + 2, and again we found a
square number which has remainder 2 when we divide the number by 4, which is
impossible. For the second equality, proceed in the same way.
Solution 1.22. For the first five parts use equations (1.2), (1.3) and (1.5). To prove
(vi), use (iv) and (v).
Solution 1.23. For the first two parts (i) and (ii), use equations (1.2) and (1.3).
To prove (iii), use (i) and (ii).
Solution 1.24. To prove (i) and (ii) just expand the left-hand side of the equations
and rearrange the terms.
To prove (iii), (iv), (v) and (vi) make the operations on both sides of the equality
and observe that they are equal.
Solution 1.25. To prove (i) and (ii) expand the right-hand side of the equations
and simplify.
Solution 1.26. Use equations (1.2) and (1.3), and perform the operations on both
sides of the equation.
Solution 1.27. Let x =
3
√
√
3
5 + 2 − 5, then
√
√
3
3
x − 2 + 5 − 2 − 5 = 0.
2+
By equation (1.7), if a + b + c = 0, then a3 + b3 + c3 = 3abc, therefore
$
$
√ %
√ %
√ %$
√ % $
x3 − 2 + 5 − 2 − 5 = 3x 3 2 + 5 2 − 5 ,
186
Chapter 10. Solutions to Exercises and Problems
simplifying we have that x3 + 3x − 4 = 0. Clearly a root of the equation is x = 1
and the other roots satisfy the equation x2 + x + 4 = 0 which does not have real
√
√
3
3
solutions. Since 2 + 5 + 2 − 5 is a real number, it follows that
√
√
3
3
2 + 5 + 2 − 5 = 1,
which is a rational number.
Solution 1.28. Observe that, if x + y + z = 0, then it follows from equation (1.7)
that x3 + y 3 + z 3 = 3xyz. Since (x − y) + (y − z) + (z − x) = 0, we obtain the
factorization
(x − y)3 + (y − z)3 + (z − x)3 = 3(x − y)(y − z)(z − x).
Solution 1.29. Observe that (x+ 2y − 3z)+ (y + 2z − 3x)+ (z + 2x− 3y) = 0, then it
follows, from equation (1.7), that (x + 2y − 3z)3 + (y + 2z − 3x)3 + (z + 2x − 3y)3 =
3(x + 2y − 3z)(y + 2z − 3x)(z + 2x − 3y).
√
√
√
Solution 1.30. Let a = 3 x − y, b = 3 y − z, c = 3 z − x, and suppose that
a+ b + c = 0, then it follows, from equation (1.7), that a3 + b3 + c3 = 3abc,
√ but then
√
√
0 = (x − y) + (y − z) + (z − x) = a3 + b3 + c3 = 3abc = 3 3 x − y 3 y − z 3 z − x = 0,
which is absurd.
√
1
Solution 1.31. If we define a = 3 r, b = − √
3 r and c = −1, we have a + b + c = 0,
%
$
√
1
(−1) = 3, therefore r− r1 = 4. Similarly, r3 − r13 −43 =
then r− r1 −1 = 3 3 r − √
3 r
1
3r − r (−4) = 12, therefore r3 − r13 = 76.
Solution 1.32. It follows from
a b
a3 + b3 + c3 − 3abc = c a
b c
c 100b + 10c + a
b = 100a + 10b + c
a 100c + 10a + b
b
a
c
c bca
b = abc
a cab
b
a
c
c
b .
a
Solution 1.33. If we rewrite the equation as m3 + n3 + (−33)3 − 3mn(−33) = 0,
and using equation (1.9), we get
!
"
(m + n − 33) (m − n)2 + (m + 33)2 + (n + 33)2 = 0.
The equation m + n = 33 has 34 solutions with mn ≥ 0 which are (k, 33 − k), with
k = 0, 1, . . . , 33, and the second factor is 0 only when m = n = −33, therefore
there are 35 solutions.
Solution 1.34. If we rewrite the equation as x3 + y 3 + (−1)3 − 3xy(−1) = 0, and
using equation (1.9), we have
"
!
(x + y − 1) (x − y)2 + (y + 1)2 + (−1 − x)2 = 0.
Therefore, the points (x, y) satisfy x + y = 1 or x = y = −1.
187
10.1 Solutions of Chapter 1
Solution 1.35. Substituting the equation of the hypothesis in equation (1.7), we get
(x + y + z)3 − 3xyz = x3 + y 3 + z 3 − 3xyz
= (x + y + z)(x2 + y 2 + z 2 − xy − yz − zx)
= (x + y + z)((x + y + z)2 − 3xy − 3yz − 3zx)
= (x + y + z)3 − 3(x + y + z)(xy + yz + zx),
from where it is clear that xyz = (x + y + z)(xy + yz + zx), therefore (x + y)(y +
z)(z + x) = 0. Or use that (x + y + z)3 = x3 + y 3 + z 3 + 3(x + y)(y + z)(z + x).
Therefore, the solutions are (x, −x, z), (x, y, −y), (x, y, −x), with x, y, z any real
numbers.
Solution 1.36.
(i) Since 0 ≤ b ≤ 1 and 1 + a > 0, it follows that b(1 + a) ≤ 1 + a, then
b−a
≤ 1.
0 ≤ b − a ≤ 1 − ab, therefore 0 ≤
1 − ab
1
1
≤ 1+a
, then
(ii) The inequality is clear. Since 1 + a ≤ 1 + b, we have 1+b
a
b
a
b
a+b
+
≤
+
=
≤ 1.
1+b 1+a
1+a 1+a
1+a
Solution 1.37. If we define X =
times 1, leads to
X=
a
b+c
+
b
a+c
+
c
a+b ,
adding and substracting three
b+c
b
a+c
c
a+b
a
+
+
+
+
+
−3
b+c b+c a+c a+c a+b a+b
a+b+c a+b+c a+b+c
+
+
−3
b+c
a+c
a+b
1
1
1
+
+
−3
= (a + b + c)
b+c a+c a+b
1
1
1
1
= ((a + b) + (b + c) + (a + c))
+
+
2
b+c a+c a+b
=
− 3.
√
Now, from the arithmetic and geometric mean inequality, we get x+y +z ≥ 3 3 xyz
and x1 + y1 + z1 ≥ 3 3 x1 y1 1z . Therefore, X ≥ 12 · 3 · 3 − 3 = 23 .
Solution 1.38. Without loss of generality we can assume that a ≥ b ≥ c; the
inequality is equivalent to −a3 + b3 +! c3 + 3abc ≥ 0. But, by equation
(1.9),
"
−a3 + b3 + c3 + 3abc = 12 (−a + b + c) (a + b)2 + (a + c)2 + (b − c)2 ≥ 0, since,
by the triangle inequality, a < b + c.
Solution 1.39. Observe that
implies s ≥ 4.
1
p
+ 1q = 1 implies p + q = pq = s. Now, (p + q)2 ≥ 4pq
188
Chapter 10. Solutions to Exercises and Problems
To prove (i), observe that
1
1
1
1
1
p+q+2
1
+
= −
+ −
= 1−
p(p + 1) q(q + 1)
p p+1 q
q+1
(p + 1)(q + 1)
s+2
s−1
=1−
=
.
2s + 1
2s + 1
Therefore, we have to prove
s−1
1
1
≤
≤ ,
3
2s + 1
2
but 2s + 1 ≤ 3s − 3 ⇔ 4 ≤ s and 2s − 2 ≤ 2s + 1 ⇔ −2 ≤ 1.
To prove (ii), show that
1
1
1
1
1
1
p+q−2
+
=
− +
− =
−1
p(p − 1) q(q − 1)
p−1 p q−1 q
(p − 1)(q − 1)
s−2
− 1 = s − 3 ≥ 1.
=
s−s+1
Solution 1.40. First, note that
a
a
a
+
=
≥ 4a,
b
1−b
b(1 − b)
since
b(1 − b) ≤
b + (1 − b)
2
2
=
1
.
4
Moreover, the equality holds if and only if b = 21 . Similarly,
Therefore,
b
b
+
≥ 4b.
a 1−a
√
√
a
b
a
b
+ +
+
≥ 4a + 4b ≥ 2 42 ab = 8 k.
b
a 1−b 1−a
With equality if and only if a = b. Then,
√
a
b
a
b
+ +
+
≥8 k≥4
b
a 1−b 1−a
if and only if k ≥ 14 , then the smallest number k is 14 .
Solution 1.41. Prove that (a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) −
abc = 98 (a + b + c)(ab + bc + ca) + 19 (a + b + c)(ab + bc + ca) − abc, and from the
arithmetic
inequality, we have that (a + b + c)(ab + bcb + ca) ≥
$
% $ and geometric mean
%
√
3
3
3 abc 3 (ab)(bc)(ca) = 9abc.
189
10.1 Solutions of Chapter 1
Solution 1.42. Using the arithmetic and geometric mean inequality, and the condition (a + b)(b + c)(c + a) = 1, leads to
a+b
b+c
c+a
3
= ,
2
2
2
2
√ √ √
a+b
b+c
c+a
abc = ab bc ca ≤
2
2
2
a+b+c≥33
=
1
.
8
Now, 1 = (a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) − abc ≥ 32 (ab + bc + ca) − 18 ,
see Exercise 1.24 (iii).
3
≥ 4.
Solution 1.43. By Exercise 1.24 (iii), it is enough to prove ab + bc+ ca+ a+b+c
But
ab + bc + ca +
ab + bc + ca
3
3
=3
a+b+c
≥4
4
ab + bc + ca
3
+
3
3
a+b+c
3
.
a+b+c
Now use that (ab√+ bc + ca)2 ≥ 3(ab · bc + bc · ca + ca · ab) = 3(a + b + c), and that
3
ab + bc + ca ≥ 3 a2 b2 c2 = 3.
Solution 1.44. Without loss of generality we can assume that a ≤ b ≤ c. Therefore
c2 < c2 + a + b ≤ c2 + 2c < (c + 1)2 ; this proves that c2 + a + b cannot be a perfect
square.
Solution 1.45. To prove all the equalities of the exercise, just perform the operations and simplify.
Solution 1.46. To prove all the equalities of the exercise, just use the identity
(1.7).
Solution 1.47. Expand both sides of the identities and compare.
Solution 1.48. We have
0 = x2 (y + z) − y 2 (x + z) = xy(x − y) + (x2 − y 2 )z
= (x − y)(xy + xz + yz).
Since x = y, we have xy + xz + yz = 0. Multiplying by x − z we obtain
0 = (x − z)(xy + xz + yz) = xz(x − z) + (x2 − z 2 )y
= x2 (y + z) − z 2 (x + y),
then z 2 (x + y) = x2 (y + z) = 2.
190
Chapter 10. Solutions to Exercises and Problems
Solution 1.49. See that (x + y + z)(xy + yz + zx) = xyz and by equation (1.23) we
get (x + y)(y + z)(z + x) = 0. Therefore, the solutions (x, y, z, w) are of the form
(x, −x, z, z), (x, y, −y, x) and (x, y, −x, y), with x, y and z real numbers different
from zero.
Solution 1.50. By equation (1.23) the condition is equivalent to (x + y)(y + z)(z +
x) = 0. Therefore, one factor is zero, say x+ y = 0. Then, since n is odd, xn + y n =
0, and also x1n + y1n = 0.
10.2 Solutions to exercises of Chapter 2
Solution 2.1. Call cn the sum of the first n even numbers, then we have
cn
cn
2cn
=
=
=
2
+
4
+
2n
+ 2n − 2 +
(2n + 2) + (2n + 2) +
··· +
2n
··· +
2
· · · + (2n + 2)
(10.1)
therefore, 2cn = n(2n + 2) = 2n(n + 1), then cn = n(n + 1). We can represent this
sum as arrangements of points forming rectangles such as those below:
Solution 2.2.
(i) Let d and d′ be the differences of the progressions {an } and {bn }, respectively.
Then, (an+1 − an ) ± (bn+1 − bn ) = d ± d′ . Rearranging the terms we have
(an+1 ± bn+1 ) − (an ± bn ) = d ± d′ .
(ii) If d is the difference of the progression {an }, we obtain
bn+1 − bn = (a2n+2 − a2n+1 ) − (a2n+1 − a2n )
= (an+2 − an+1 )(an+2 + an+1 ) − (an+1 − an )(an+1 + an )
= d(an+2 + an+1 − an+1 − an )
= d(an+2 − an ) = 2d2 .
Solution 2.3. If d is the difference of the progression {an }, we have
n−1
#
j=0
n−1
1# 1
1
1
1
=
−
=
aj aj+1
d j=0 aj
aj+1
d
=
1
d
an − a0
a0 an
=
1
d
1
1
−
a0
an
a0 + nd − a0
a0 an
=
n
.
a0 an
191
10.2 Solutions of Chapter 2
Solution 2.4. If {an } is an arithmetic progression,
a0 + an−1
·n
2
d
d
2a0 + (n − 1)d
· n = n2 + a0 −
n,
=
2
2
2
Sn = a0 + a1 + · · · + an−1 =
and with A =
d
2
and B = a0 −
d
2
we get the result.
Suppose now that Sn = a0 + a1 + · · · + an−1 = An2 + Bn, then Sn+1 = a0 + a1 +
· · · + an−1 + an = A(n + 1)2 + B(n + 1). Subtracting the first equation from the
second, we have
an = A(n + 1)2 + B(n + 1) − An2 − Bn
= A(2n + 1) + B = 2An + (A + B)
and using Proposition 2.1.3, we get the expected result.
Solution 2.5. Suppose that {an } is an arithmetic progression of order 2. Consider
Sn = (a1 − a0 ) + (a2 − a1 ) + · · · + (an − an−1 ) = an − a0 . By Exercise 2.4, we have
an − a0 = An2 + Bn. Therefore, an = An2 + Bn + a0 = P (n), where P (x) is the
polynomial of degree 2, given by Ax2 + Bx + a0 .
Suppose now that each term of the progression an is equal to P (n), where P (x)
is a polynomial of degree 2, that is, an = An2 + Bn + C. It follows that
an+1 − an = A(n + 1)2 + B(n + 1) + C − (An2 + Bn + C)
= 2An + (A + B).
Therefore, by Proposition 2.1.3, {an+1 − an } is an arithmetic progression and thus
{an } is an arithmetic progression of order 2.
Solution 2.6. A consequence of the inequality between the geometric mean and
√
n
. By Proposition 2.1.1, we get
the arithmetic mean, is n a1 a2 · · · an ≤ a1 +a2 +···+a
n
√
a1 + a2 + · · · + an
a1 + an
1
a1 + an
n
=
·n· =
.
a1 a2 · · · an ≤
n
2
n
2
To prove the left-hand side inequality, we use a similar version of the equality of
Exercise 2.3
1
1
1
n−1
+
+ ···+
=
.
a1 a2
a2 a3
an−1 an
a1 an
By the inequality between the harmonic mean and the geometric mean, we get
a1 an =
1
a1 a2
n−1
≤
1
+ · · · + an−1
an
n−1
(a1 a2 )(a2 a3 ) · · · (an−1 an ).
Then,
(a1 an )n−1 ≤ a1 (a2 · · · an−1 )2 an .
√
√
Therefore, (a1 an )n ≤ (a1 a2 · · · an )2 , that is, a1 an ≤ n a1 a2 · · · an .
192
Chapter 10. Solutions to Exercises and Problems
Solution 2.7. If p, p + 6, p + 12, p + 18, p + 24 are the 5 prime numbers, when
we consider these numbers modulo 5, we have that they are congruent to p, p + 1,
p+ 2, p+ 3, p+ 4, but between five consecutive numbers we have always one that is
divisible by 5, and since p is prime, then p = 5 and hence the numbers are forced
to be 5, 11, 17, 23 and 29.
Solution 2.8. The numbers m between 2n and 2n+1 are part of an arithmetic
progression with difference 1 starting in 2n + 1 and ending in 2n+1 − 1. Then, by
Proposition 2.1.1,
Sn =
3 · 2n n+1
2n + 1 + 2n+1 − 1 n+1
(2
(2
− 1 − 2n ) =
− 1 − 2n ),
2
2
and from this it is clear that 3 divides Sn .
Solution 2.9. Since a, b and c are in harmonic progression, in that order, we can
1
= A(A+s)
,
suppose that a1 = A − s, 1b = A and 1c = A + s, with s = 0. We have b−c
s
4
c−a
=
−2(A2 −s2 )
1
, a−b
s
=
A(A−s)
s
and
1
c
−
1
a
= 2s. Therefore,
4
1
A2 + As − 2A2 + 2s2 + A2 − As
1 1
1
+
+
=
= 2s = − .
b−c c−a a−b
s
c a
Solution 2.10. If a, b, c and d are in harmonic progression, then their inverses are
in arithmetic progression. Suppose that 1b = a1 + s, 1c = a1 + 2s and d1 = a1 + 3s,
1
1+2as
and d1 = 1+3as
with s = 0, that is, we have 1b = 1+as
a , c =
a
a . Then we have
a
a
a
that, b = 1+as , c = 1+2as and d = 1+3as .
Then,
a+d=
2a + 3a2 s
1 + 3as
and b + c =
2a + 3a2 s
.
1 + 3as + 2(as)2
Since 1 + 3as < 1 + 3as + 2(as)2 , we have that a + d > b + c, as desired.
Solution 2.11. Note that
1
1
b−a
b 2 − a2
−
=
=
,
c+a b+c
(b + c)(c + a)
(b + a)(b + c)(c + a)
1
c−b
c2 − b 2
1
−
=
=
,
a+b c+a
(a + b)(c + a)
(b + a)(b + c)(c + a)
1
1
1
1
− b+c
= a+b
− c+a
if and only if b2 − a2 = c2 − b2 . Then, b + c, c + a
then c+a
and a + b are in harmonic progression if and only if a2 , b2 and c2 are in arithmetic
progression.
Solution 2.12. Suppose that a0 , a1 , . . . is the progression and that d is the difference of the progression, that is, d = an+1 − an for all n ≥ 0, then an = a0 + nd.
193
10.2 Solutions of Chapter 2
By hypothesis, a0 (a0 + d) and a0 (a0 + 2d) also belong to the progression, therefore
a0 (a0 + 2d) − a0 (a0 + d) = n0 d for some integer number n0 ≥ 1, then a0 d = n0 d.
Since the progression is increasing, it follows that d > 0 and therefore a0 = n0 .
Change a0 for any am in the previous argument and conclude that am is an
integer number.
Solution 2.13.
(i) Observe that in the array, the left-hand side number of each row is as follows:
1◦
2◦
3◦
4◦
5◦
row
row
row
row
row
..
.
1
1+2= 3
1+2+4 = 7
1 + 2 + 4 + 6 = 13
1 + 2 + 4 + 6 + 8 = 21
..
.
100◦ row 1 + 2 + · · · + 2 · 99 = 1 + 2(1 + · · · + 99) = 9901.
(ii) The sum of the numbers on the 100th row is,
9901 + 9903 + · · · + 10099 = (9900 + 1) + (9900 + 3) + · · ·
+ (9900 + 2 · 100 − 1)
= 9900 · 100 + (1 + 3 + · · · + 2 · 100 − 1)
= 990000 + 1002 = 106 .
Solution 2.14. The array will be a square of size 100 × 100. Let S1 be the sum
of the numbers in the diagonal which goes from the top left corner to the bottom
right corner and let S2 be the sum of the numbers of the other main diagonal.
When we move one column to the right in the same row, the number increases by
1; if we move one column to the left in the same row, the number decreases by 1.
When we move down in the same column, the number increases by 100.
From the top left corner to the bottom right corner through the diagonal, each
number is one column to the right and one row below, that is, it is 1 + 100 = 101
greater than the previous number of the diagonal. That is, the sum we want to
calculate is the sum of the progression 1, 1+101, . . . , 1+99·101 that, by Proposition
2.1.1 (b), is
2 · 1 + 99 · 101
100 = 500050.
S1 =
2
From the top right corner, through the diagonal, each number is in the previous
column, that is, one column to the left and one row below, that is, it is −1 + 100 =
99 greater than the previous number in the diagonal. That is, the sum we are
194
Chapter 10. Solutions to Exercises and Problems
looking for is the sum of the progression 100, 100 + 1 · 99, . . . , 100 + 99 · 99 that,
by Proposition 2.1.1 (b), is
S2 =
2 · 100 + 99 · 99
2
100 = 500050.
Therefore, S1 = S2 = 500050.
Solution 2.15. Call aij , where i denotes the number of the row and j denotes the
number of the column, the corresponding position in the table.
Let x0 and x1 be the two numbers neighboring 0, x0 on the right-hand side of 0
and x1 on top of 0, then the last row can be filled up with 0, x0 , 2x0 , 3x0 and 4x0 ,
and the first column with 4x1 , 3x1 , 2x1 , x1 and 0.
If x is the number in the position a32 , the number that occupies the a42 position will
be 12 (x + x0 ), but we also know that the number in the a42 position is 21 (x1 + 103).
Therefore, 12 (x + x0 ) = 21 (x1 + 103), solving for x we have that x = x1 + 103 − x0 .
Now, let y be the number occupying the a44 position, then we have 103 = 12 (y +
1
1
2 (x1 + 103)), solving for y we obtain y = 2 (309 − x1 ).
1
1
The number 2 (309−x1)−103 = 2 (103−x1 ) is the difference of the progression
in the fourth row, but this difference, added to the number in the a44 position,
gives the number in position a45 , that is, 21 (309 − x1 ) + 12 (103 − x1 ) = 206 − x1 .
However, we know that 206 − x1 = 12 (186 + 4x0 ), then
2x0 + x1 = 113.
(10.2)
Observe also that 21 74 + 12 (x1 + 103) = x1 + 103 − x0 . Simplifying we obtain
4x0 − 3x1 = 161.
(10.3)
Solving the system of equations (10.2) and (10.3), we get x0 = 50 and x1 = 13.
With these values, now it is easy to complete the table, so the filled board is
52
39
26
13
0
82
74
66
58
50
112
109
106
103
100
142
144
146
148
150
172
179
186
193
200
Solution 2.16. If {an } is a geometric progression with ratio r, we have that an =
n
a0 rn . Similarly, if {bn } is a geometric progressionwith
ratio s, then bn = b0 s .
an
a0 r n
Therefore, since bn = 0 for all n, we have bn = b0 s .
Solution 2.17. Let {an } be a geometric progression with ratio r and having the
property that an+2 = an+1 + an . Since an = a0 rn , this property is equivalent to
a0 rn+2 = a0 rn+1 +a0 rn . Since a0 = 0 and r = 0, we certainly have that r2 = r +1,
195
10.2 Solutions of Chapter 2
which has as solutions r =
$ √ %n
.
an = a0 1−2 5
√
1± 5
2 .
$ √ %n
Then, the solutions are an = a0 1+2 5
and
Solution 2.18.
(i) Since Pn = a0 · a1 · · · · · an−1 and an = a0 rn , for all n, we have
Pn = a0 · a1 · · · · · an−1 = a0 (a0 r)(a0 r2 ) · · · (a0 rn−1 )
= an0 r1+2+···+(n−1) = an0 r
(ii) Since Pn = an0 r
n(n−1)
2
n(n−1)
2
.
, it is clear that
$
%2
n
n(n−1)
(Pn )2 = an0 r 2
= an0 an0 rn(n−1) = an0 a0 rn−1 = an0 ann−1 .
Solution 2.19. Since an+1 = an · r, then bn+1 = log an+1 = log (an · r) = log an +
log r = bn + log r, and the result follows.
Solution 2.20. Factorizing we have
a3 b3 + b3 c3 + c3 a3 − abc(a3 + b3 + c3 ) = −bc + a2 −ca + b2 −ab + c2 .
Solution 2.21. If the arithmetic progression is a, a + d, a + 2d, . . . , it is clear
that a + ad = a(1 + d) is an element of the arithmetic progression, and also the
integers a(1 + d)n , with n ≥ 1, are part of the progression. But these terms form
a geometric progression, then it is clear that these terms have to remain in the
sequence and we have to eliminate the remaining terms of the original progression.
Solution 2.22. Consider a < b < c, then since the lengths of the sides are in
geometric progression, we have b = ar and c = ar2 , with r positive. Since the
triangle is a right triangle, it follows that a2 + (ar)2 = (ar2 )2 . Simplifying √
the
1+ 5
2
4
2
2
equation we get 1 + r = r , which can be solved for r . That is, r = 2 .
√
Therefore, r = 1+2 5 .
Solution 2.23. Let d be the common difference of the progression. Then a2 = 1+d,
a5 = 1 + 4d and a11 = 1 + 10d. Since a2 , a5 , a11 form a geometric progression, we
have (1 + 4d)2 = (1 + d)(1 + 10d) or 6d2 = 3d. Since the arithmetic progression
is not constant, we conclude that d = 21 and the sum of the first 2009 terms is
· 21 = 2009 · 503.
2009 + 2009·2008
2
Solution 2.24. By the similarity of the triangles, we have that
Again, using the similarity of the triangles, we have ay = yz .
b
a
= ya , then y =
a2
b .
196
Chapter 10. Solutions to Exercises and Problems
A
y
z
B
Then z =
3
= ab2 .
z=
a4
b2 a
2
y
a
b
a
C
, and substituting the value of y in this equation results in
(i) Carry on with this process and find that the sides of the polygonal measure
2 3
2
3
4
b, a, ab , ab2 , ab3 , . . ., which can be written as b, a, a ab , a ab , a ab , . . ..
n−2
Therefore, the nth segment measures a ab
.
(ii) The length of the n-sided polygonal line is then
b+a
$ a %0
b
+a
$ a %1
b
+ ···+ a
$ a %n−2
b
n−1
1 − ab
=b+a
1 − ab
n−1
1 − ab
.
= b + ab
b−a
(iii) Since the number ab is less than one, then raising this number to the nth
power and making n go to infinite leads to 0 as its limit. Then, the length of
the polygonal line with an infinite number of sides is
n−1
1 − ab
ab
b2
=b+
=
.
lim b + ab
n→∞
b−a
b−a
b−a
,
Solution 2.25. The sum of each row is $
Rn = 12 + 22%+ · · · + n2 = n(n+1)(2n+1)
6
n(n+1)(2n+1)
then the sum of all the array is ST = n
. See now that the sum of
6
each corridor is
Ck = 12 + 22 + · · · + (k − 1)2 + k 3 =
k2
k
8 k3
−
+ .
6
2
6
Then
ST = C1 + C2 + · · · + Cn
1
1
8
= (13 + 23 + · · · + n3 ) − (12 + 22 + · · · + n2 ) + (1 + 2 + · · · + n)
6
2
6
1
n(n
+
1)(2n
+
1)
8 3
1 n(n + 1)
= (1 + 23 + · · · + n3 ) −
+
.
6
2
6
6
2
197
10.2 Solutions of Chapter 2
Therefore
8 3
n2 (n + 1)(2n + 1) n(n + 1)(2n + 1) 1
(1 + 23 + · · · + n3 ) =
+
−
6
6
12
6
=
2
n(n + 1)
2
8
6
n(n + 1)
2
.
From this the sought for equality follows immediately.
Solution 2.26. Observe that the numbers {1, 4, 7, . . . , 2998, 3001} form an arithmetic progression {an } with difference 3 and a1 = 1. Each term of the sum can
be seen as
1
1
1
1
1
1
=
=
−
−
.
ai ai+1
ai+1 − ai ai
ai+1
3ai 3ai+1
Therefore, we can calculate the sum as
1
1
−
3a1
3a2
1
1
−
+ ···+
3a2
3a3
1
1000
1
=
.
= −
3 3 · 3001
3001
+
Solution 2.27.
(i) We have
n
#
k=1
1
k(k+2)
=
1
2
$
1
k
−
1
k+2
n
%
1
1
−
3 · 3000 3 · 3001
, then
1
1
−
k k+2
#1
1
=
k(k + 2)
2
k=1
1
1
1 1 1 1 1 1 1
− + − + − + ···+ −
2 1 3 2 4 3 5
n n+2
1
1
1
1
−
=
1+ −
2
2 n+1 n+2
2n + 3
3
.
= −
4 2(n + 1)(n + 2)
=
(ii) We have
2k+1
k2 (k+1)2
n
#
k=1
=
1
k2
−
1
(k+1)2 ,
n
#
2k + 1
=
k 2 (k + 1)2
k=1
then
1
1
−
k2
(k + 1)2
1
1
1
1
1
1
− 2 + 2 − 2 + ···+ 2 −
2
1
2
2
3
n
(n + 1)2
1
.
=1−
(n + 1)2
=
198
Chapter 10. Solutions to Exercises and Problems
Solution 2.28.
(i) Since
k
(k + 1) − 1
k+1
1
1
1
=
=
−
=
−
(k + 1)!
(k + 1)!
(k + 1)! (k + 1)!
k! (k + 1)!
the sum we have to calculate becomes
1
1
−
1! 2!
+
1
1
−
2! 3!
+ ···+
1
1
−
n! (n + 1)!
=1−
1
.
(n + 1)!
(ii) The general term can be written as
k+1
k+1
=
(k − 1)! + k! + (k + 1)!
(k − 1)![1 + k + k(k + 1)]
k+1
=
(k − 1)!(k + 1)2
k
1
=
.
=
(k − 1)!(k + 1)
(k + 1)!
By the first part of the exercise, the sum is equal to 1 −
1
(n+1)! .
Solution 2.29. For any positive integer n, we have
1+
1
1
n2 (n + 1)2 + (n + 1)2 + n2
(n2 + n + 1)2
+
=
=
.
2
2
2
2
n
(n + 1)
n (n + 1)
n2 (n + 1)2
Then
1+
1
1
1
n2 + n + 1
=1+
.
+
=
n2
(n + 1)2
n2 + n
n(n + 1)
Therefore, the sum is equal to
2011
#
n=1
1+
1
n(n + 1)
=
2011
#
1+
n=1
1
1
−
n n+1
= 2012 −
1
.
2012
Solution 2.30. Define S as the product we have to calculate, that is,
1
2
1
1
· · · 1 + 2n .
22
2
Multiplying
both sides of the equality by 1 − 21 and using that 1 − 21 1 + 12
= 1 − 212 , we get
S=
1−
1
2
1+
S=
1−
1+
1
22
1+
1
22
··· 1 +
1
22n
.
199
10.3 Solutions of Chapter 3
Proceeding in this way, we arrive at
1
1−
2
Therefore,
S=
1−
1
22n
2 %
1 − 221n
=2 1−
S=
1 − 12
$
2
1
22n
.
2
.
10.3 Solutions to exercises of Chapter 3
Solution 3.1.
(i) If n = 1, then 1 =
1−q
1−q
= 1.
Suppose that 1 + q + · · · + q n−1 =
1−qn
1−q .
1 + q + · · · + q n−1 + q n =
Then,
1 − q n+1
1 − qn
+ qn =
,
1−q
1−q
as we wanted to prove.
Second Solution. Let S = 1+q+· · ·+q n−1 , then Sq = q+q 2 +· · ·+q n . Substracting
n
the first equality from the second leads to Sq − S = q n − 1, then S = 1−q
1−q .
(ii) The proof is immediate from (i).
Solution 3.2. The result is valid for n = 1, since a3 = a2 + a1 = 2 and a3 = 1 + a1 = 2.
Suppose that the result is valid for n, that is, an+2 = 1 + a1 + a2 + · · · + an .
Then the formula is valid for n + 1, since
an+3 = an+2 + an+1 = 1 + a1 + a2 + · · · + an + an+1 .
Solution 3.3. For n = 0 the statement is valid, because 3 divides 2 + 1 = 3. The
n−1
+ 1.
induction hypothesis for n − 1 tells us that 3n divides 23
We prove now the result for n. We start with
+$
,
%
n
n−1
n−1 2
n−1
+ 1) 23
23 + 1 = (23
− 23
+1 .
By the induction hypothesis, the first factor is divisible by 3n . The second factor
n−1
is divisible by 3, since 23
≡ −1 mod 3. This proves the statement.
Solution 3.4. If n = 1 we have three coins. Place in each plate of the weighing
scale one coin, if the plates balance, the false coin is the third coin, which we did
not place. If not, the plate that lifts is the one that has the false coin.
200
Chapter 10. Solutions to Exercises and Problems
Suppose that it is true for 3n coins. Consider 3n+1 coins. Divide the coins in three
groups with 3n coins in each. Put one of the groups in one plate of the balance
and another group in the other plate. If the plates balance, the false coin is in the
third group. If not, the false coin is in the group of the plate that raises. In both
cases the problem will reduce to finding the false coin in a group with 3n coins,
but this can be done in n weighings and with the weighing already done we have
n + 1 weighings.
Solution 3.5. First note that for n ≥ 1, it follows that 2n+3 ± 1 = 2n (23 − 1) + 2n ±
1 = 7 · 2n + (2n ± 1), then 7 | 2n ± 1 if and only if 7 | 2n+3 ± 1. This equivalence
shows an inductive step of the form 7 | 2n ± 1 ⇒ 7 | 2n+3 ± 1, and of the form
7 ∤ 2n ± 1 ⇒ 7 ∤ 2n+3 ± 1.
Now let us see the induction basis.
(i) For n = 1, 2 it follows that 7 does not divide 2n − 1. For n = 3, it follows
that 7 divides 23 − 1 = 7. Therefore, the integers n we are looking for are the
multiples of 3.
(ii) For n = 1, 2, 3, it follows that 7 does not divide 2n + 1 (which are 3, 5 and
9). Thus 7 does not divide 2n + 1 with n ≥ 1.
Solution 3.6. Observe that a1 + a2 = 22 . Solving with respect to a2 , we show that
a2 = 4 − a1 = 4 − 3 = 3. This suggests to us that an = 2n − 1. Suppose that aj =
2j−1, for all j < n. It follows that a1 +a2 +· · ·+an = 1+3+· · ·+(2n−3)+an = n2 .
Since 1 + 3 + · · · + (2n − 3) = (n − 1)2 , it follows that (n − 1)2 + an = n2 .
From this we conclude that an = n2 − (n − 1)2 = n2 − (n2 − 2n + 1) = 2n − 1.
Solution 3.7. For n = 1, the identity is valid, the left-hand side is ⌊ 12 ⌋ = 0 and
the right-hand side is ⌊ 12 ⌋⌊ 22 ⌋ = 0.
For n = 2, the identity is also valid; on the one hand we have ⌊ 12 ⌋ + ⌊ 22 ⌋ = 1
and on the other hand we have ⌊ 22 ⌋⌊ 23 ⌋ = 1 · 1 = 1.
Now we suppose valid the identity for n and we prove that it is true for n + 2.
The left-hand side is
n n + 1 n + 2
2
1
+
+
+ ···+
+
2
2
2
2
2
n n + 1
n+1
n+2
+
+
=
2
2
2
2
n n + 1 n + 1 n
=
+1
+
+
2
2
2
2
n n + 1 n + 1 n
+1
+
+
=
2
2
2
2
$ n
% n + 1
n+2
n+3
=
+1
+1 =
,
2
2
2
2
which is what we expected from the right-hand side.
201
10.3 Solutions of Chapter 3
√
√
1· a
Solution 3.8. Observe that a1 = 2 2 . Solving this equation gives as a result
√
2 = a2 , from where a2 = 22 . This suggests an = n2 . Suppose that aj = j 2 , for
all j < n. We then have
√
(n − 1) an
√
√
(n − 1)n
√
=
,
a1 + a2 + · · · + an−1 = 1 + 2 + · · · + n − 1 =
2
2
√
and therefore n = an , that is, an = n2 .
Solution 3.9.
(i) For n = 1, we have x2 − 2x + 1 = (x − 1)2 .
For n = 2 we get x3 − 3x + 2 = (x − 1)2 (x + 2), since (x − 1)2 (x + 2) =
(x2 − 2x + 1)(x + 2) = x3 − 3x + 2.
Suppose that xn+1 − (n + 1)x + n = (x − 1)2 (xn−1 + 2xn−2 + · · · + n), then
(x−1)2 (xn + 2xn−1 + · · · + (n − 1)x2 + nx + (n + 1))
= (x − 1)2 x(xn−1 + 2xn−2 + · · · + (n − 1)x + n) + (n + 1)
= x (x − 1)2 (xn−1 + 2xn−2 + · · · + (n − 1)x + n + (x − 1)2 (n + 1)
= x xn+1 − (n + 1)x + n + (x − 1)2 (n + 1)
= xn+2 − (n + 1)x2 + nx + (n + 1)x2 − 2x(n + 1) + (n + 1)
= xn+2 − (n + 2)x + (n + 1).
Therefore we have proved that xn+1 − (n + 1)x + n = (x − 1)2 (xn−1 + 2xn−2 +
· · · + n). Now, if x > 0 the right-hand side of the above equality is greater
tan or equal to zero, then xn+1 − (n + 1)x + n ≥ 0.
(ii) If x = ab , with a =
x1 +x2 +···+xn+1
n+1
and b =
x1 +x2 +···+xn
,
n
we have that
an+1
a
− (n + 1) + n ≥ 0
bn+1
b
x1 +···+xn+1
an+1
n+1
≥ (n + 1)
−n
x1 +···+xn
bn+1
n
then
n(x1 + · · · + xn+1 )
an+1
≥
−n
bn+1
x1 + · · · + xn
an+1
n(x1 + · · · + xn )
nxn+1
≥
+
−n
n+1
b
x1 + · · · + xn
x1 + · · · + xn
nxn+1
an+1
≥
,
n+1
b
x1 + · · · + xn
an+1 ≥ xn+1
which is what we wanted to prove.
bn+1
x1 +···+xn
n
= xn+1 bn ,
202
Chapter 10. Solutions to Exercises and Problems
(iii) Suppose that
x1 +x2 +···+xn
n
x1 + · · · + xn+1
n+1
≥
√
n
x1 x2 · · · xn , then
n+1
≥ xn+1
x1 + · · · + xn
n
n
≥ xn+1 (x1 · · · xn ),
where the first inequality is due to the (ii) part and the second by the induc√
n+1
tive step. Therefore, x1 +···+x
≥ n+1 x1 · · · xn+1 , which is what we wanted
n+1
to prove.
Solution 3.10.*For n = 1 the result follows.
Suppose that the result is true for
n−1
n
n
−
1.
Then
different
values and the same holds for
e
a
has
at
least
2
$*
% i=1 i i
n−1
i=1 ei ai − an . The greater value of these sums is a1 + a2 + · · · + an−1 − an ,
but it is clear that
a1 + a2 + · · · + an−1 − an < a1 + a2 + · · · + an−2 − an−1 + an
< a1 + · · · + an−3 − an−2 + an−1 + an
..
..
.
.
< a1 − a2 + a3 + · · · + an
< −a1 + a2 + a3 + · · · + an
< a1 + a2 + a3 + · · · + an ,
then we have n additional different sums. Since
proved.
n
2
+n =
n+1
2 , the result is
Solution 3.11. For n = 1 the statement is true, because in a sequence of three
numbers (chosen from the set of numbers {0, 1}) there are two that are equal.
Suppose the statement to be valid for 2n + 1 numbers and consider a sequence
with 2n + 3 elements.
If for any i we have that ai = ai+1 , then by the induction hypothesis, a1 ,
. . ., ai−1 , ai+2 ,. . ., a2n+3 has a subsequence even-balanced with 2n elements. To
this subsequence add two equal elements and you will have a subsequence evenbalanced with 2(n + 1) elements.
On the contrary, we have that ai = ai+1 for all i ∈ {1, 2, . . . , n + 1}, therefore ai = ai+2 for all i. Since the sequence has an odd number of elements, the
initial and last numbers have to be the same. Then, if we take out the element
that occupies the central place of the sequence, we will have a subsequence with
2n elements that clearly is even-balanced, since the sequence is symmetric with
respect to the central element.
Solution 3.12. Prove, by induction, that ak = k. For k = 1 the statement is true,
since a31 = a21 and a1 > 0, hence a1 = 1. Suppose the result is true for 1, 2, . . . , k
and prove that it is valid for k + 1.
10.3 Solutions of Chapter 3
203
We then have that the equality
a31 + a32 + · · · + a3n = (a1 + a2 + · · · + an )2 ,
is satisfied by n = 1, 2, . . . k, k + 1. By the induction hypothesis, we have that
a1 = 1, a2 = 2, . . . , ak = k and the condition for n = k + 1 is
13 + 23 + · · · + k 3 + a3k+1 = (1 + 2 + · · · + k + ak+1 )2 .
Expanding the right-hand side of the equality, we obtain
(13 + 23 + · · · + k 3 ) + a3k+1 = (1 + 2 + · · · + k)2 + 2(1 + 2 + · · · + k)ak+1 + a2k+1 .
The first terms on both sides of the equality can be canceled out, and from there
we get a3k+1 = 2(1 + 2 + · · · + k)ak+1 + a2k+1 ; but this is equivalent to a3k+1 =
ak+1 + a2k+1 . Dividing by ak+1 = 0, we obtain a2k+1 − ak+1 − k(k + 1) = 0.
2 k(k+1)
2
This is a quadratic equation in ak+1 with roots −k and k+1. Since ak+1 is positive,
we end up with ak+1 = k + 1, which ends the proof.
Solution 3.13. We will do the proof using induction. For n = 1, since a1 ≥ 1, we
have that a31 ≥ a21 , moreover, a31 = a21 if and only if a1 = 1. Suppose the inequality
is true for n = k and consider k + 1 positive integers such that a1 < a2 < · · · <
ak < ak+1 . Then, we have ak+1 ≥ ak + 1, therefore
(ak+1 − 1)ak+1
ak (ak + 1)
≥
= 1 + 2 + · · · + ak .
2
2
Note that the sum 1 + 2 + · · · + ak contains all positive integers less than or equal
k+1
≥
to ak , then it is greater than or equal to a1 + a2 + · · · + ak , hence (ak+1 −1)a
2
2
a1 + a2 + · · · + ak , which multiplied by 2ak+1 , is equivalent to (ak+1 − ak+1 )ak+1 ≥
2(a1 + a2 + · · · + ak )ak+1 , that is, a3k+1 ≥ 2(a1 + a2 + · · · + ak )ak+1 + a2k+1 .
On the other hand, the induction hypothesis implies that
a31 + a32 + · · · + a3k ≥ (a1 + a2 + · · · + ak )2 .
Adding the last two inequalities, we obtain
a31 + a32 + · · · + a3k + a3k+1 ≥ (a1 + a2 + · · · + ak + ak+1 )2 ,
from where the inequality is true for n = k + 1.
It is not difficult to deduce, from the previous proof, that the equality follows if
and only if ak+1 = ak = 1 and a31 + a32 + · · · + a3k = (a1 + a2 + · · · + ak )2 . In the last
identity, the induction hypothesis implies that if the equality holds, then a1 = 1,
a2 = 2, . . . , ak = k. Therefore, ak+1 = ak + 1 implies that ak+1 = k + 1. That is,
the sequence is ai = i, for i = 1, 2, . . . , k + 1. Reciprocally, we have equality for
a1 = 1, a2 = 2, . . . , ak = k, ak+1 = k + 1 thanks to the classical formula.
204
Chapter 10. Solutions to Exercises and Problems
Solution 3.14. (i) The original inequality for n = 1 can be verified directly. For
n ≥ 2, it is enough to prove by induction that
1+
1
2
1+
1
22
··· 1 +
1
2n
5
2
1−
1−
2n+1
≤
1
2n
.
For n = 2 we have the equality.
The inductive step is reduced to verifying that
5
2
1−
1
2n
1
1+
2n+1
≤
5
2
1
.
To see this, observe that the previous inequality is equivalent to
1−
1−
1
2n
1+
1
2n+1
≤1−
1
2n+1
1
1
1
1
+ n+1 − 2n+1 ≤ 1 − n+1
2n
2
2
2
2
1
1
≤ n + 2n+1
2n+1
2
2
1
1
1
≤ n 1 + n+1
n
2
2
2
,
which is clearly true.
(ii) As in the previous part, it will be easier to see that
1+
1
13
1+
1
23
··· 1 +
1
n3
<3−
1
.
n
For the inductive step, it is necessary to verify that
3−
1
n
1+
1
(n + 1)3
≤3−
1
.
n+1
Performing all the operations and simplifying leads to
3−
1
3
1
1
+
−
≤3−
n (n + 1)3
n(n + 1)3
n+1
3
1
1
1
≤
+
1+
(n + 1)3
n+1
n
(n + 1)3
1
3
1
1
+1 ≤
1+
n + 1 (n + 1)2
n
(n + 1)3
n(n2 + 2n + 4) ≤ n3 + 3n2 + 3n + 2
0 ≤ n2 − n + 2,
and the last inequality follows.
205
10.3 Solutions of Chapter 3
Solution 3.15. Define P (n) as the statement we want to prove. If n = 1 the equality
follows and then P (1) is true. To see that P (2) is valid, consider the following set
of equivalences:
1
2
1
+
≥
√
1 + a1
1 + a2
1 + a1 a2
√
⇔ (2 + a1 + a2 )(1 + a1 a2 ) ≥ 2(1 + a1 )(1 + a2 )
√
√
⇔ 2 a1 a2 + (a1 + a2 ) a1 a2 ≥ a1 + a2 + 2a1 a2
√
√
√
⇔ 2 a1 a2 (1 − a1 a2 ) + (a1 + a2 )( a1 a2 − 1) ≥ 0
√
√
⇔ ( a1 a2 − 1)(a1 − 2 a1 a2 + a2 ) ≥ 0.
√
√
√
But the last inequality is true, since a1 a2 ≥ 1 and ( a1 − a2 )2 ≥ 0.
We now see that P (2n ) ⇒ P (2n+1 ).
n+1
2#
i=1
n
n+1
2
2#
#
1
1
1
=
+
1 + ai
1
+
a
1
+
ai
i
i=1
i=2n +1
2n
2n
+
√
√
n
2
1+
a1 · · · a2n
1+
a2n +1 · · · a2n+1
2
n
≥2
√
√
1 + 2n a1 · · · a2n 2n a2n +1 · · · a2n+1
≥
=
2n
2n+1
.
√
1 + 2n+1 a1 · · · a2n+1
For the first inequality we used P (2n ) twice and for the second we applied P (2)
√
√
to the numbers 2n a1 · · · a2n and 2n a2n +1 · · · a2n+1 .
that
Now, let us see that P (n + 1) ⇒ P (n).
√
If we apply P (n + 1) to the numbers a1 , . . . , an , an+1 = n a1 · · · an , we have
1
1
1
+ ··· +
+
≥
1 + a1
1 + an
1 + an+1
1+
n+1
√
a1 · · · an n a1 · · · an
n+1
=
n+1
1+ 1
1+
(a1 · · · an ) n
n+1
.
=
1 + an+1
Hence
n+1
1
1
n
n
+ ··· +
≥
=
.
√
n
1 + a1
1 + an
1 + an+1
1 + a1 · · · an
Solution 3.16. To prove (i) and (ii) apply the binomial Theorem 3.2.3 to (1 + 1)n
and (1 + 2)n , respectively.
206
Chapter 10. Solutions to Exercises and Problems
Solution 3.17. To prove (i) just apply formula (3.5) and for (ii) take r = 1 in (i).
Solution 3.18. (i) In equation (1 + x)n (1 + x)n =(1 + x)2n , the coefficient of the
term xn on the right-hand side of the equation is 2n
n . If we expand the left-hand
side of the equation we find that
+
,
n
n
n j
n n
+
x + ···+
x + ··· +
x
0
1
j
n
+
,
n
n
n k
n n
+
x + ···+
x + ···+
×
x .
0
1
k
n
Therefore the term xn will appear when we multiply nj xj and nk xk , with j +k =
n, and then
#
n n
2n
=
.
j
k
n
0≤j,k≤n
j+k=n
But
#
0≤j,k≤n
j+k=n
since
n
j
=
n
n−j
n
j
n
k
n
#
n
=
j
j=0
n
n−j
n
#
n
=
j
j=0
2
,
.
(ii) In the same way as in (i), it is enough to compare the coefficient of xr
on both sides of the identity
(1 + x)n (1 + x)m = (1 + x)n+m
which turns out to be
#
0≤j,k≤r
j+k=r
n
k
m
j
=
n+m
.
r
Then,
r
#
n
m
n+m
=
.
k
r−k
r
k=0
m+1
on the left-hand side of the equality and using
(iii) Changing m
0 by
0
Pascal’s formula (3.6), we have that
m+1
m+1
m+2
m+n
+
+
+ ··· +
0
1
2
n
m+2
m+2
m+n
=
+
+ ···+
1
2
n
207
10.3 Solutions of Chapter 3
m+3
m+3
m+n
+
+ ···+
2
3
n
..
.
=
=
m+n
m+n
+
n−1
n
m+n+1
.
n
=
=
m+1
(iv) Changing m
m by m+1 on the left-hand side of the equality and using
Pascal’s formula (3.6), we obtain
m+1
m+1
m+2
n
+
+
+ ···+
m+1
m
m
m
m+2
m+2
=
+
+ ···+
m+1
m
m+3
m+3
=
+
+ ···+
m+1
m
..
=
.
n
n
+
m+1
m
=
=
n
m
n
m
n+1
.
m+1
Solution 3.19. (i) Using part (ii) of Exercise 3.17, it follows that
j
n
j
=n
n−1
,
j−1
then
n
#
j=1
j
n
j
=
n
#
n
j=1
=n
n−1
#
j=0
n
#
n−1
j−1
j=1
n−1
j−1
=n
n−1
j
= n · 2n−1 .
(ii) By part (i) of Exercise 3.17, it follows that
from where
n+1
j+1
=
n+1 n
,
j+1 j
1
n
j+1 j
=
n+1
1
n+1 j+1
208
Chapter 10. Solutions to Exercises and Problems
hence,
n
#
j=0
n
1
j+1 j
=
n
#
j=0
=
1
n+1
n+1 j+1
n
1 # n+1
n + 1 j=0 j + 1
n+1
1 # n+1
j
n + 1 j=1
⎡⎛
⎞
⎤
n+1
#
n+1 ⎠
n+1 ⎦
1 ⎣⎝
=
−
j
0
n+1
j=0
=
=
1
[2n+1 − 1].
n+1
Solution 3.20. (i) First, using repeatedly Pascal’s formula (3.6) and part (ii) of
Exercise 3.17, it follows that
+
,
n−1
n−1
1 n
1
1 n
1 n−1
=
+
+
=
j j
j
j
n j
j
j
j−1
+
,
1
n−2
n−2
1 n
=
+
+
j
n j
j
j−1
n−1
1
1 n
1 n−2
+
+
,
=
j
n−1
n j
j
j
and carrying on in this way, we obtain
1 n
j j
=
n−1
1 n
1
1 j
+
+ ···+
.
n j
n−1
j j
j
Therefore
n
#
j=1
n
j j
j+1 1
(−1)
=
n
#
j+1
(−1)
j=1
-n−j
#
i=0
n−i
1
n−i
j
.
.
If we change the sum’s order, our previous identity changes to
⎡
⎤
n−1
n−i
# #
n−i ⎦
⎣ (−1)j+1 1
.
n
−
i
j
i=0 j=1
Set k = n − i, then the right-hand side of the identity becomes
⎤
⎡
n #
k
n
k
n
#
#
#
1 k
k ⎦ #1
1⎣
(−1)j+1
.
(−1)j+1
=
=
j
k j
k j=1
k
j=1
k=1
k=1
k=1
10.3 Solutions of Chapter 3
209
(ii) Observe that the left-hand side of the equality is
1 n
(−1)n+1 n
1 n
−
+ ···+
1·2 1
2·3 2
n(n + 1) n
1
1
n
n
n
1
1 1
= 1−
−
−
−
+ · · · + (−1)n+1
2
2 3
n n+1
1
2
n
, +
+
1 n
1 n
n
1 n
1 n
+ · · · + (−1)n+1
−
−
+ ···
=
−
2 2
n n
2 1
3 2
1
,
(−1)n+1 n
+
.
(10.4)
n+1
n
Let U be equal to the terms inside the first square bracket. By the previous
statement we have that U = 1 + 12 + · · · + n1 , and for the terms inside the second
n
n+1
1
1
square bracket we use j+1
j = n+1 j+1 . Then the above identity (10.4) takes
the form
1 n
1 n
(−1)n+1 n
−
+ ···+
1·2 1
2·3 2
n(n + 1) n
,
+
1
n+1
n+1
n+1 n + 1
=U−
−
+ · · · + (−1)
n+1
n+1
2
3
+
n+1
n+1
n+1
1
−
=U−
+
+ ···
n+1
0
1
2
,
n+1
− (1 − (n + 1))
+(−1)n+1
n+1
n
1
[n] = U −
=U−
n+1
n+1
n+1−1
1
1
=U−
= + ··· +
.
n+1
2
n+1
(iii) Call T the left-hand side of the equality, then
T =
1 n
1 n
(−1)n n
−
+
·
·
·
+
.
12 0
22 1
(n + 1)2 n
Multiplying by n + 1, we have
n+1 n
n+1 n
n+1 n
−
+ · · · + (−1)n
.
0
1
12
22
(n + 1)2 n
n n+1
Using the fact that n+1
j+1 j = j+1 , our identity becomes
(n + 1)T =
1
n+1
1 n+1
1 n+1
−
+ · · · + (−1)n
1
2
n+1 n+1
1
2
1
1
.
= 1 + + ···+
2
n+1
1
1
Therefore, the sum we are looking for is T = n+1
1 + 12 + · · · + n+1
.
(n + 1)T =
210
Chapter 10. Solutions to Exercises and Problems
Solution 3.21. (i) Using the equality
n
#
(−1)j j
j=1
n
j
=
n
#
n
m
(−1)j n
j=1
=n
n−1
j−1
n−1
#
n n−1
m m−1
=
=n
leads to the set of identities
n
#
(−1)j
j=1
n−1
j−1
n−1
j
(−1)j+1
j=0
and using Example 3.2.4, we reach the result
n
n−1
#
n−1
j
(−1)j+1
j=0
(ii) Again, using the equality
n
#
j=1
(−1)j j 2
n
j
=
n
#
n
m
(−1)j n · j
j=1
n
#
=
= n · 0 = 0.
, it follows that
n n−1
m m−1
n−1
j−1
=n
n
#
(−1)j j
j=1
n−1
j−1
n
#
n−1
n−1
=n
(−1)j
(−1) (j − 1)
+n
,
j−1
j
−
1
j=1
j=1
=n
n
#
j=2
j
(−1)j (j − 1)
n
#
n−1
n−1
,
(−1)j
+n
j−1
j−1
j=1
since in the above identity the first term of the first sum is zero.
Finally, by Example 3.2.4, and using the previous part of this exercise, we
find
n
n
n
#
#
n−1
n−1
(−1)j
(−1)j (j − 1)
+n
j−1
j−1
j=1
j=2
= n · 0 + n · 0 = 0.
Solution 3.22. Consider the following
array, where in the mth row, for m ≥ 0,
we have the binomial coefficients m
j modulo 2. On the left column is shown the
number of the row written in base 2 and in the right column the number of odd
binomial coefficients written in base 2.
0
1
2
3 = 2 + 20
4 = 22
5 = 22 + 20
6 = 22 + 21
7 = 22 + 21 + 20
1
1
1
1
1
1
1
1
1
0
1
0
1
0
1
1
0
0
1
1
1
0
0
1
1
0
1
1
1
1
1
0
1
1
1
1
20
21
21
22
21
22
22
23
211
10.3 Solutions of Chapter 3
This array suggests to us that the number of odd binomial coefficients will
be 2k , with k the number of non-zero digits when we write n in base 2.
Note that if n = 2α1 + 2α2 + · · · + 2αr , with α1 > α2 > · · · > αr ≥ 0, then
α1
(1 + x)n = (1 + x)2
α1
αr
· · · (1 + x)2
αr
≡ (1 + x2 ) · · · (1 + x2 ) mod 2.
The previous identity follows because if we expand the binomial of the form (1 +
x)2a the first and the last binomial coefficients are 1 and the rest are even.
α1
αr
It is clear that if we expand (1 + x2 ) · · · (1 + x2 ), there are 2r terms.
*n 2
Solution 3.23. The proof is based on the identity j=0 pj = 2p
p , as seen in
Exercise
3.18
part
(i).
Since pj is divisible by p for all j = 1, 2, . . . , p−1, each term of the sum is divisible
2
by p2 , with
2p exception of the first and the last, which are equal to 1. Therefore, p
divides p − 2. To finish the proof, observe that
2p − 1
1
−1=
2
p−1
2p
−2 .
p
Solution 3.24. For p = 2, we have that 21 = 13 + 13 and for p = 3, we get
32 = 13 + 23 .
Let us see now that there is no prime number p > 3, for which there exist a, b and
n such that a3 + b3 = pn .
Suppose that we can find such numbers and that n is the smallest integer
number that fulfills the conditions of the problem. Since p ≥ 5, one of the numbers
a or b is greater than 1, then a3 + b3 ≥ 5. Since
a3 + b3 = (a + b)(a2 − ab + b2 )
and a2 − ab + b2 = (a − b)2 + ab ≥ 2, then p must divide a + b and a2 − ab + b2 .
But then, p divides
(a + b)2 − (a2 − ab + b2 ) = 3ab.
Since p ≥ 5, p has to divide a or b, but since p|a + b, it follows that p divides a
and it also divides b. Then, a3 + b3 ≥ 2p3 , hence n > 3. Since
pn−3 =
a3 + b 3
pn
=
=
p3
p3
a
p
3
+
b
p
3
,
it follows that n − 3 also satisfies the condition; then n is not a minimum.
Solution 3.25. Consider the equation x2 + y 2 + z 2 = 2xyz. The left-hand side of
the equation has exactly one even term or all three terms are even. If exactly one
212
Chapter 10. Solutions to Exercises and Problems
term is even, then the right-hand side of the equation is divisible by 4 and the
left-hand side is divisible only by 2, so we have a contradiction. Then all terms are
even, that is, x = 2x1 , y = 2y1 , z = 2z1 and
x21 + y12 + z12 = 4x1 y1 z1 .
(10.5)
From equation (10.5), following the same reasoning leads to x1 = 2x2 , y1 = 2y2 ,
z1 = 2z2 and
(10.6)
x22 + y22 + z22 = 8x2 y2 z2 .
Again from equation (10.6), it follows that x2 , y2 , z2 are even, and so on and so
forth. Then
x = 2x1 = 22 x2 = 23 x3 = · · · = 2n xn = · · · ,
y = 2y1 = 22 y2 = 23 y3 = · · · = 2n yn = · · · ,
z = 2z1 = 22 z2 = 23 z3 = · · · = 2n zn = · · · ,
that is, if (x, y, z) is a solution, then x, y and z are divisible by 2n for all n. This
is impossible, unless x = y = z = 0.
Solution 3.26. The solutions are a = b = 1 or a and b consecutive square numbers.
We can write the divisibility condition as
k(ab + a + b) = a2 + b2 + 1,
(10.7)
for some integer number k. If k = 1, then the equation (10.7), is equivalent to
(a − b)2 + (a − 1)2 + (b − 1)2 = 0, from where a = b = 1. If k = 2, then equation
(10.7) can be written as 4a = (b − a − 1)2 , from where we deduce that a has to be
a square number, that is, a = d2 . Then b − d2 − 1 = ±2d, that is, b = ±(d ± 1)2
and, a and b are consecutive square numbers.
Suppose now that k ≥ 3, and let (a, b) be a solution with a the minimum
and a ≤ b. Write equation (10.7) as a quadratic equation in b,
b2 − k(a + 1)b + (a2 − ka + 1) = 0.
Since root b is an integer, the other root r satisfies b + r = k(a + 1) and it is also
an integer. Since equation (10.7) has to be true if we substitute b by r, note that
k(ar + a + r) = a2 + r2 + 1 > 0 implies ar + a + r > 0, and then we can conclude
that r > 0. And since a ≤ b and the product of the roots a2 − ka + 1 is less than
a2 , we have r < a. But (r, a) is also a solution of (10.7), which contradicts a being
a minimum.
Solution 3.27. First we prove that we cannot find an equilateral triangle such
that the vertices are points with integer coordinates. Suppose we can find such
a triangle. Let a be the length of the sides of the triangle such that the vertices
213
10.3 Solutions of Chapter 3
√
are points with integer coordinates. The area of the triangle is a2 43 which is an
irrational number, since a2 is an integer number. On the other hand, the area
of any polygon whose vertices are points with integer coordinates is a rational
number26 .
The vertices of a regular hexagon P1 P2 P3 P4 P5 P6 cannot be points with integer coordinates, since P1 P3 P5 would be an equilateral triangle whose vertices
have integer coordinates.
Let n = 3, 4, 6. Suppose that P1 P2 P3 . . . Pn is a regular n-gon with vertices
having integer coordinates. Through the points P1 , P2 , . . . , Pn draw the parallels
−−−→ −−−→
−−−→
to P2 P3 , P3 P4 , . . . , P1 P2 , respectively, as shown in the figure.
P1
P5
P2
P4
P3
The points of intersection of the parallels are also points with integer coordinates and form a regular n-gon inside the first one. With the new n-gon we can
proceed in the same form. This process can continue to generate an infinite number of n-gons. The square of the length of the sides of these polygons are integers
that decrease in each step, but this is impossible.
Solution 3.28. Error. The given arguments assume that the set has at least 3
elements, and we use that a1 , an , an+1 are different. We can say that the statement
P1 ⇒ P2 is not valid.
Solution 3.29. Error. Statement P(n) is: for n coins among which one is false, it
is enough to weigh 4 times to identify the false coin. When we take out one coin
there are two cases: (a) the coin that we took out is genuine, (b) the coin we took
out is false.
In the first case, the inductive step works, but in the second case it does not,
because among the coins that remain we do not have a false coin.
Solution 3.30. Error. Statement P(0) implies P(1) is false.
26 See
[13].
214
Chapter 10. Solutions to Exercises and Problems
10.4 Solutions to exercises of Chapter 4
2
2
Solution 4.1. The equation can be written as m − 23 − 54 = n + 12 − 45 or
2
2
m − 32 − n + 12 = 0, that is, (m + n − 1)(m − n − 2) = 0. Since m and n
are positive integers, then m + n − 1 > 0, therefore m and n are solutions of the
equation if and only if m − n − 2 = 0, that is, the solutions are (m, n) = (a + 2, a),
where a is any positive integer.
Solution 4.2. Since P (−1) = a − b + c, P (0) = c and P (1) = a + b + c are integers,
we have that 2a, 2b y c are integers.
For n = 2m, with m an integer number, we have that P (n) = P (2m) =
a(4m2 )+b(2m)+c = (2m2 )(2a)+m(2b)+c is an integer, and for n = 2m+1, with m
an integer number, we have that P (n) = P (2m+ 1) = a(2m+ 1)2 + b(2m+ 1)+ c =
(2m2 + 2m)(2a) + (2a) + m(2b) + (a + b + c) is also an integer.
Solution 4.3. If pq is a solution of ax2 + bx + c = 0, it follows, multiplying by q 2 ,
that ap2 + bpq + cq 2 = 0. Then p | cq 2 and q | ap2 , but since (p, q) = 1, it follows
that p | c and q | a.
Solution 4.4. Note that cx2 + bx + a = x2 (c + b x1 + a x12 ), then the roots are the
inverse of the roots of ax2 + bx + c, therefore
(α + β)(α′ + β ′ ) = (α + β)
1
1
+
α β
≥ 4.
Another way, by Vieta’s formulas (4.1), α + β = − ab and α′ + β ′ = − bc , then
b2
(α + β)(α′ + β ′ ) = ac
. But b2 − 4ac ≥ 0, as they are real roots and since a = αβ,
′ ′
c = α β are positive.
Solution 4.5. Let x1 , x2 be the zeros of P (x). Then, by Vieta’s formulas (4.1), we
have x1 + x2 = a − 2 and x1 x2 = −a − 1. Substitute in the identity x21 + x22 =
(x1 + x2 )2 − 2x1 x2 the values of the sum and the product of x1 and x2 to obtain
(a − 2)2 + 2(a + 1) = a2 − 2a + 6 = (a − 1)2 + 5 ≥ 5, with equality for a = 1. Then,
a = 1 is the only number.
Solution 4.6. Observe that p1 + 1q +
qr + pr + pq = 3 and pqr = −1. Then,
1
r
1
p
= qr+pr+pq
. By Vieta’s formula (4.2),
pqr
+ q1 + r1 = −3.
Solution 4.7. Since p, q and r are roots of the given cubic equation, by Vieta’s
formula (4.2), it follows that p + q + r = −b and pq + qr + rp = c.
Since (p+q +r)2 = p2 +q 2 +r2 +2(pq +qr+rp), we obtain (−b)2 = p2 +q 2 +r2 +2c,
and rearranging terms b2 − 2c = p2 + q 2 + r2 .
Therefore, a quadratic equation with the desired roots is
(x − (−b))(x − (b2 − 2c)) = x2 + (b − b2 + 2c)x + (2bc − b3 ) = 0.
10.4 Solutions of Chapter 4
215
2
x + m 2−3 = 0,
Solution 4.8. Since x1 and x2 are solutions to equation x2 − 2m−1
2
2
it follows that x1 + x2 =2m−1
and x1 x2 = m 2−3 . We need x1 = x2 − 12 , then it
2
2m−1
1
1
is necessary that x2 − 2 + x2 = 2x2 − 2 = 2 , that is, x2 = m
2 . Similarly, we
x2
m2
3
m2 −3
1
2
want to have x2 − 2 x2 = 2 , then x2 − 2 = 2 − 2 . Hence, substituting
m2
m
m2
3
x2 = m
2 , we get 4 − 4 = 2 − 2 , that is (m + 3)(m − 2) = 0. Then, it follows
that m = −3 or m = 2. Since m has to be positive, then m = 2. Therefore, x1 = 12
and x2 = 1.
Solution 4.9. Since P (x2 + 1) = x4 + 4x2 , the polynomial P (x) has to be of degree
2 and it is monic, that is, P (x) = x2 + bx + c, then
(x2 + 1)2 + b(x2 + 1) + c = x4 + 4x2 .
Expand the equation to obtain x4 + 2x2 + 1 + bx2 + b + c = x4 + 4x2 , hence 2 + b = 4
and 1 + b + c = 0, then b = 2, c = −3. Substitution leads to P (x) = x2 + 2x − 3,
therefore P (x2 − 1) = (x2 − 1)2 + 2(x2 − 1) − 3 = x4 − 2x2 + 1 + 2x2 − 2 − 3 = x4 − 4.
Solution 4.10. Note that a3 + b3 = c3 + d3 if and only if (a + b)(a2 − ab + b2 ) =
(c + d)(c2 − cd + d2 ), but since a + b = c + d = 0, we can cancel these factors and
obtain a2 − ab + b2 = c2 − cd + d2 . On the other hand, from a + b = c + d we
get, squaring this equality, that a2 + 2ab + b2 = c2 + 2cd + d2 . Subtracting these
two last identities, it follows that ab = cd. Then, the two quadratic polynomials
x2 − (a + b)x + ab and x2 − (c + d)x + cd coincide, in particular their roots are the
same. But, by Vieta’s formula (4.1) we know that the roots of the polynomials are
{a, b} and {c, d}, respectively. Therefore, {a, b} = {c, d}.
Solution
has
equal roots if the discriminant is zero, that is,
4.11. The polynomial
4 − 4λ 1 − λ1 = 0, then λ 1 − λ1 = 1. Thus λ = 2 is the only possibility.
Solution 4.12. It is not possible. Otherwise, if the three polynomials had two real
roots, the discriminants, b2 − 4ac and c2 − 4ab, a2 − 4bc would be positive. Hence,
b2 > 4ac, c2 > 4ab, a2 > 4bc, and multiplying the inequalities we would have
a2 b2 c2 > 64a2 b2 c2 , which is false.
Solution 4.13. The solutions of the equation are given by
x=
1 − 2k ±
√
(1 − 2k)2 − 4k(k − 2)
1 − 2k ± 1 + 4k
=
.
2k
2k
The number x will be rational if 1 + 4k is a perfect square, that is, k has to be
2
an integer of the form k = n 4−1 , with n a positive integer. Since we want k to
be an integer, n2 − 1 has to be divisible by 4, but n2 − 1 = (n + 1)(n − 1) is
2
divisible by 4 if and only if n is odd. Then, for k = n 4−1 , with n odd, the roots of
kx2 − (1 − 2k)x + k − 2 = 0 are rational numbers.
216
Chapter 10. Solutions to Exercises and Problems
Solution 4.14. If a + b is a root of the polynomial P (x) = x2 + ax + b, then
0 = (a + b)2 + a(a + b) + b = b2 + (3a + 1)b + 2a2 , and so b has to be a root of
the polynomial Q(x) = x2 + (3a + 1)x + 2a2 . But Q(x) will have an integer root
if its discriminant (3a + 1)2 − 4 · 2a2 = (a + 3)2 − 8 is a perfect square. But two
square numbers have difference 8 if and only if they are 1 and 9, then (a + 3)2 = 9,
therefore a = −6 or a = 0. If a = −6, then b = 8 or 9 and if a = 0 then b = 0 or −1.
Hence, the only possible pairs (a, b) are: (−6, 8), (−6, 9), (0, 0) and (0, −1).
Solution 4.15. We want to solve equation x2 − 5x − 1 = n2 , that is, x2 − 5x −
(n2 + 1) = 0. The solutions of the equation are given by
√
5 ± 25 + 4n2 + 4
.
(10.8)
x1,2 =
2
For x to be an integer number it is necessary that 4n2 + 29 = t2 . Then, t2 − 4n2 =
(t − 2n)(t + 2n) = 29, that is, t + 2n = ±29 and t − 2n = ±1 or t + 2n = ±1
and t − 2n = ±29. Solving these equations, it follows that 4n = ±28, then n = 7
and t = 15 or n = −7 and t = −15. Substituting n in equation (10.8), we obtain
x1 = 10 and x2 = −5.
Solution 4.16. We present the solution due to R. Descartes. Using Vieta’s formulas
(4.1), we find that if α and β are the polynomial roots, then α + β = −b and
αβ = −c2 , then there is a negative root.
R
S
c
O
b
2
Q
T
Consider the triangles RQS and RT Q, which are similar, since they share
the angle of the vertex R and ∠RT Q = ∠RQS. Then, we have that RT · RS =
RQ2 = c2 , therefore, RS · (−RT ) = −c2 . On the other hand, since RT = RS + b,
we have RS +(−RT ) = −b. Then, −RT and RS satisfy Vieta’s relations, therefore
those numbers are the roots of the equation.
Solution 4.17. If P (x) = x does not have real solutions, then P (x) > x, for all x
or P (x) < x, for all x. Hence, P (P (x)) > P (x) > x or P (P (x)) < P (x) < x, for
all x, therefore it is impossible to have P (P (x)) = x.
10.4 Solutions of Chapter 4
217
Solution 4.18. If P (P (P (x0 ))) = P (x0 ) = 0 for some x0 , then
P (P (0)) = P (P (P (x0 ))) = 0.
Hence, P (0) is a zero of P (x) and it is an integer because P (x) is a polynomial
with integer coefficients. Moreover, P (P (P (P (0)))) = P (P (0)) = 0, then P (0) is
also a root of P (P (P (x))).
Solution 4.19. Note that ax2 +bx+c > cx is equivalent to P (x) = ax2 +(b−c)x+c >
0; this guarantees that a > 0, c = P (0) > 0 and (b − c)2 − 4ac < 0. On the
other hand, to prove that cx2 − bx + a > cx − b is equivalent to proving that
cx2 − (b + c)x + a + b > 0, but since c > 0, it is enough that the discriminant be
negative; such discriminant is (b + c)2 − 4c(a + b) = (b − c)2 − 4ac, but we have
already proved that it is negative.
Solution 4.20. Suppose that 20(b − a) is an integer number. By symmetry we
can also suppose that b > a, and then 20(b − a) ≥ 1. Since there are no real
solutions, the discriminant of the polynomial x2 + 20bx + 10a is negative, therefore
1
1
. Hence 0 < b − a < b < 10
10b2 − a < 0. Then, we have 10b2 < a < b and b < 10
and, then 20(b − a) < 2, but if 20(b − a) is an integer number we have 20(b − a) = 1
1
1 2
1
and then b = a + 20
. Thus, 10b2 − a = 10(a + 20
) − a = 10a2 + 40
> 0, which is
a contradiction. Therefore, 20(b − a) can never be an integer number.
Solution 4.21. If P (x) = x2 + bx + c satisfies P (P (P (x0 ))) = P (x0 ) = 0, for some
x0 , then P (P (0)) = P (P (P (x0 ))) = 0. Therefore,
0 = P (P (0)) = P (c) = c2 + bc + c = c(c + b + 1) = P (0)P (1).
Solution 4.22. Expand the following polynomial and use the relation ab+ac+bc =
de + df + ef , to get that
(x + a)(x+b)(x + c) − (x − d)(x − e)(x − f )
= x3 + (a + b + c)x2 + (ab + ac + bc)x + abc
− x3 + (d + e + f )x2 − (de + df + ef )x + def
= N x2 + abc + def.
Then, if we let x = d, the above expression becomes
(d + a)(d + b)(d + c) = N d 2 + (abc + def ).
Then, if N divides abc + def , then N divides (d + a)(d + b)(d + c). Let p be a prime
number such that p divides N , then p divides at least one of the factors d + a,
d + b or d + c. Then, p ≤ max(d + a, d + b, d + c) < N , that is, p is a factor of N
and N is a composite number.
218
Chapter 10. Solutions to Exercises and Problems
Solution 4.23. Since P (x) and Q(x) have integer coefficients, we can divide by the
main coefficient and assume that P (x) = (x − α)(x − r) and Q(x) = (x − α)(x − s),
where α ∈ R\Q, that is, P (x) and Q(x) are monic polynomials with rational
coefficients. In a quadratic polynomial, if one root is an irrational number the
other root is also irrational, since the sum of both roots has to be a rational
number. Then, r, s ∈ R\Q.
Note that α + r, αr, α + s and αs are all rational numbers since they are the
coefficients of P (x) and Q(x). Then
(α + r) − (α + s) = r − s ∈ Q
r
αr
= ∈ Q.
αs
s
Let r − s = pq and rs = m
equation, it follows that
n . Solving for r from the second
m
p
p
m
m
·
s,
from
where
we
get
·
s
−
s
=
,
that
is,
s
r= m
n
n
q
n − 1 = q . If n − 1 = 0,
it follows that
p
s=
which is a contradiction, then
m
n
q
m
n
−1
∈Q
= 1. Thus, r = s.
Solution 4.24. Multiplying both equations, it follows that
x4 − (b1 + b2 )x3 + (c1 + c2 + b1 b2 )x2 − (b1 c2 + b2 c1 )x + c1 c2 = 0.
On the other hand, since b1 , c1 , b2 and c2 are roots, we have by, Vieta’s formulas
(4.1.1) and equating the coefficients, that
b 1 + b 2 + c1 + c2 = b 1 + b 2
b 1 b 2 + b 1 c1 + b 1 c2 + b 2 c1 + b 2 c2 + c1 c2 = c1 + c2 + b 1 b 2
b 1 b 2 c1 + b 1 b 2 c2 + b 1 c1 c2 + b 2 c1 c2 = b 1 c2 + c1 b 2
b 1 b 2 c1 c2 = c1 c2 .
From the first equation, we obtain c1 = −c2 , then from the second equation it
follows that c1 c2 = c1 + c2 = 0, from where we get c1 = c2 = 0, which contradicts
the fact that c1 and c2 are different numbers. Hence, those polynomials do not
exist.
Solution 4.25. Since (a − b) + (b − c) + (c − a) = 0, then some of the terms of
the sum a − b, b − c or c − a are less than or equal to zero. Suppose, without
loss of generality, that a − b ≤ 0. Then, the discriminant of the third equation is
(c − a)2 − 4(a − b) ≥ 0, that is, the third equation has a zero that is a real number.
Solution 4.26. Let P (x) = x3 − (a + b + c)x2 + (ab + bc + ca)x − abc be the
monic polynomial with zeros a, b and c. Let A = ab + bc + ca, B = abc and
219
10.4 Solutions of Chapter 4
Tn = an + bn + cn . Then, T0 = 3, T1 = 0, T2 = (a + b + c)2 − 2(ab + bc + ca) = −2A.
The equation P (x) = 0 is equivalent to x3 = −Ax + B, from where xn+3 =
−Axn+1 + Bxn . Then, it follows that Tn+3 = −ATn+1 + BTn . Now find T3 , T4
and T5 .
Solution 4.27. Using (a+b+c)2 = a2 +b2 +c2 +2(ab+bc+ca), we get ab+bc+ca = 2.
Using the identity a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca),
we get abc = − 32 . Then, the cubic polynomial with roots a, b and c is P (x) =
x3 −3x2 +2x+ 32 ; now, from aP (a)+bP (b)+cP (c) = 0, it follows that a4 +b4 +c4 = 9.
Solution 4.28. Each of the equations ax2 +bx+c = 0 and cx2 +bx+a = 0 have two
different real solutions, a = 0 and c = 0. Moreover, r is the root of ax2 + bx +
c = 0
if and only if r1 is a root of cx2 + bx + a = 0. Therefore, {q1 , q2 } = p11 , p12 .
If p1 , q1 , p2 , q2 are in arithmetic progression,
1
1 |p1 − p2 |
,
|p1 − p2 | = |q1 − q2 | = − =
p1
p2
|p1 p2 |
from where |p1 p2 | = 1.
Using Vieta’s formula (4.1), we have p1 p2 = ac , so |c| = |a| and then a = ±c.
If a = c, the two given quadratic equations are equal, and then p1 = q1 , p2 = q2 ,
which tell us that the difference of the progression is 0. Then, p1 = q1 = p2 = q2
which is a contradiction. Therefore, a = −c or a + c = 0.
Solution 4.29. Let Tn = an + bn + cn for each integer number n, then T0 = 3,
T1 = 0 and T2 = (a + b + c)2 − 2(ab + bc + ca) = −2(ab + bc + ca). Now, define
A = ab + bc + ca and B = abc; then, by Vieta’s formulas (4.2), it follows that a, b
and c are the roots of the equation x3 + Ax − B = 0 and T2 = −2A.
For n ≥ 0, we substitute a, b and c in xn+3 = −Axn+1 + Bxn and adding we
obtain Tn+3 = −ATn+1 + BTn . Then
T3 = −AT1 + BT0 = 3B,
T4 = −AT2 + BT1 = 2A2 ,
T5 = −AT3 + BT2 = −5AB.
Hence,
T5
5
= −AB =
T3
3
·
T2
2 .
Since T3 = T5 , the last equality implies that T2 = 65 .
Solution 4.30. Let Q(x) = P (x) − 2, since a, b and c are the roots of Q(x), it
is clear that Q(x) = α(x − a)(x − b)(x − c), for some integer number α. If for
some integer number d we have P (d) = 3; then, since 1 = P (d) − 2 = Q(d) =
α(d − a)(d − b)(d − c), the factors on the right-hand side of the equation have to be
−1 or 1, then two of d − a, d − b, d − c are equal, so a, b, c are not different, which
is a contradiction. This guarantees that there does not exist an integer number d
with P (d) = 3.
220
Chapter 10. Solutions to Exercises and Problems
10.5 Solutions to exercises of Chapter 5
Solution 5.1. For (i) and (ii) apply directly the definition. The proof of (iii) and
(iv) is straightforward once you do the operations. To prove (v), after squaring
both sides, observe that z w̄ + wz̄ ≤ 2|z w̄|. (vi) Do the operations using |z|2 = z z̄.
x−iy
1
= (x+iy)(x−iy)
= xx−iy
Solution 5.2. Let z = x + iy, then 1z = x+iy
2 +y 2 . Hence,
−y
1
2
2
Im z + z = y + x2 +y2 = 0 is equivalent to y(x + y − 1) = 0, with solutions
y = 0 or x2 + y 2 = 1. The solution y = 0 means that the real axis satisfies the
original equation and the solution x2 + y 2 = 1 is the unit circle.
Second Solution. Write z in its polar form, that is, z = r(cos
θ). Then,
θ + i sin
1
1
1
1
1
=
(cos
(−θ)+i
sin
(−θ))
=
(cos
θ−i
sin
θ).
Thus,
Im
z
+
)
=
(r−
z
r
r
z
r sin θ = 0,
and then r − 1r = 0 or sin θ = 0. Equation r − 1r = 0 implies that r = 1, that
is, the complex numbers that satisfy the equation are the ones on the unit circle;
meanwhile the complex numbers z that satisfy sin θ = 0 are the complex numbers
with argument 0 or π, that is, all the real numbers.
Third Solution. Observe that
Im z +
1
z
1
1
= z̄ +
z
z̄
⇔ (z z̄ − 1)(z − z̄) = 0
⇔ |z| = 1 or z = z̄
=0 ⇔ z+
⇔ z is on the unit circle or the real axis.
Solution 5.3. Use the fact that z z̄ = |z|2 for every complex number z, and see
that
2
|z + w| = (z + w)(z + w) = z z̄ + ww̄ + z w̄ + z̄w,
2
|z − w| = (z − w)(z − w) = z z̄ + ww̄ − z w̄ − z̄w.
Then, |z + w| = |z − w| if and only if z w̄+ z̄w = −z w̄− z̄w, that is, 2(z w̄+ z̄w) = 0.
Using the fact that Re z = z+z̄
2 , we obtain 2(z w̄ + z̄w) = 4 Re w̄z = 0. Since
2
2
|w|
z
z
w̄ = |w|
w , then w̄z = w z, hence Re w̄z = Re w = 0. Therefore, w is purely
iz
z
imaginary, that is, w = ir for some r ∈ R. Thus, w = −r, which is a real number.
Second Solution. A geometric proof is the following. Since w = 0, we can divide
the original equation by w, to obtain
z
z
+ 1 = − 1 ,
w
w
which means that wz is in the perpendicular bisector of the segment that joins 1
and −1, that is, the imaginary axis, then wz is purely imaginary and now we can
conclude as above.
221
10.5 Solutions of Chapter 5
Solution 5.4. (i) Use the fact that z z̄ = |z|2 . The left-hand side of the identity is
|1 − z̄w|2 − |z − w|2 = (1 − z̄w)(1 − z w̄) − (z − w)(z̄ − w̄)
2
2
2
2
2
2
= 1 + |z| |w| − z̄w − z w̄ − |z| − |w| + z w̄ + wz̄
2
2
= 1 + |z| |w| − |z| − |w| .
In the same way, the right-hand side of the identity is
(1 + |zw|)2 − (|z| + |w|)2 = 1 + 2 |zw| + |z|2 |w|2 − |z|2 − |w|2 − 2 |z| |w|
2
2
2
2
= 1 + |z| |w| − |z| − |w| .
Therefore, using the above relations we reach the desired conclusions.
(ii) Proceed as before:
2
2
|1 + z̄w| − |z + w| = (1 + z̄w)(1 + z w̄) − (z + w)(z̄ + w̄)
= 1 + |z|2 |w|2 + z̄w + z w̄ − |z|2 − |w|2 − z w̄ − wz̄
2
2
2
2
2
2
= 1 + |z| |w| − |z| − |w| = (1 − |z| )(1 − |w| ).
Solution 5.5. It is clear that z = 0 if and only if w = 0, thus we can assume that
both numbers are non-zero. Also suppose that z = w, then we need to show that
z̄w = 1. Simplifying the given identity, we obtain z + zww̄ − w − wz z̄ = 0.
The last equality can be written as z − w = zw(z̄ − w̄). Considering the norm on
both sides, and using the fact that the norm of a complex number is equal to the
norm of its conjugate, it follows that |zw| = 1, that is, zwz̄ w̄ = 1.
2
Multiplying the equality z + zww̄ − w − wz z̄ = 0 by z̄, we obtain |z| + 1 − z̄w −
wz̄ |z|2 = 0. Now, from this last equation we have that (|z|2 + 1)(1 − z̄w) = 0, thus
z̄w = 1.
Solution 5.6. Observe that z12 + z22 + z32 = (z1 + z2 + z3 )2 − 2(z1 z2 + z2 z3 + z3 z1 ).
2
2
2
Then, since
% 0, it follows that z1 + z2 + z3 = −2(z1 z2 + z2 z3 + z3 z1 ) =
$ z1 + z2 + z3 =
−2z1z2 z3 z11 + z12 + z13 = −2z1 z2 z3 (z¯1 + z¯2 + z¯3 ) = 0.
Solution 5.7. Since z1 z̄1 = |z1 |2 = 1, then
1
z1
= z̄1 , and also
1
z2
= z̄2 . Then,
1
+ 1
z̄1 + z̄2
z1 + z2
= z1 1 z21 =
,
1 + z̄1 z̄2
1 + z1 z2
1 + z1 z2
hence
z1 +z2
1+z1 z2
is a real number.
Solution 5.8. If some of a, b, c is zero, the result is clear. Then, suppose that a, b,
c = 0.
It is enough to see that (d − a)(d − b)(d − c) = 0, or equivalently, that
d3 − (a + b + c)d2 + (ab + bc + ca)d − abc = 0 and, by the hypothesis of the exercise,
this is equivalent to showing that (ab + bc + ca)d = abc.
222
Chapter 10. Solutions to Exercises and Problems
Since all numbers have the same norm, define r = |a| = |b| = |c| = |d|. On the
other hand, it is known that d = a+ b + c, then d¯ = a + b + c = ā + b̄ + c̄. Now,
¯
b̄b
dd
āa
c̄c
1
1
1
1
1
1
2 1
d = a + b + c = r
a + b + c , and it follows that d = a + b + c . That is,
d(ab + bc + ca) = abc, as we wanted to prove.
Solution 5.9. We show first that (i) is equivalent to (ii). If a, b, c are collinear,
then arg(b − a) = arg(c − a) or arg(b −
both are on
$ a) =
% arg(c −
$ a) + π, because
%
1
the same line, then it follows that arg c−a
=
arg
(c
−
a)
·
=
arg(c
− a) −
b−a
b−a
c−a
arg(b − a) = 0 or π, which implies that c−a
b−a ∈ R. Reciprocally, if t = b−a ∈ R,
then c − a = t(b − a) or c = (1 − t)a + tb, which means that c is on the straight
line that determines a and b, that is, a, b and c are collinear.
Now, we show that (iii) is equivalent to (iv). In order to do this, note that the
determinant of the matrix in (iv) is equal to bc̄ − cb̄ − a(c̄ − b̄) + ā(c − b). Then
z = cb̄ − cā − ab̄ ∈ R is equivalent to z = z̄, that is, cb̄ − cā − ab̄ = bc̄ − ac̄ − bā,
which is the same as bc̄ − cb̄ − a(c̄ − b̄) + ā(c − b) = 0.
Finally, observe that
c − a b̄ − ā
c−a
cb̄ − cā − ab̄ + |a|2
=
·
.
=
b−a
b − a b̄ − ā
|b − a|2
2
|a|
c−a
Since |a−b|
2 is a real number, b−a ∈ R is equivalent to cb̄ − cā − ab̄ ∈ R, that is,
(ii) is equivalent to (iii).
From the part (ii), it follows that the equation of the line through b and c is
Im( z−c
z−b ) = 0.
Solution 5.10. By Exercise
1
0 = 1
1
5.9, z, i, iz are collinear if and only if
i −i
1−i
1+i
z
z̄ = z z̄ −
z̄ −
z,
2
2
iz −iz̄
2 1+i 2
=
. Then, the complex numbers that satisfy
which is equivalent to z − 1+i
2
2
the condition of the exercise are the points in the circle with center 1+i
2 and radius
1+i √2
=
2
2 .
Solution 5.11. We will construct first the two squares which have as side length
the segment determined by z and w. To do that, consider the points 0 and z − w
as two consecutive vertices of a square. Then one possible third vertex is i(z − w),
which is the rotation with angle 90◦ of the complex number z − w (or it could be
−i(z − w) if we rotate −90◦ ). Finally, the fourth vertex is the sum of the previous
two, that is, (z − w) + i(z − w) (or (z − w) − i(z − w) in the other case). Then, to
calculate the vertices of the squares that are formed with z and w as consecutive
vertices, we should add w to the vertices of the squares found, that is, the vertices
223
10.5 Solutions of Chapter 5
we are looking for are z, w, i(z − w) + w and z + i(z − w) or z, w, −i(z − w) + w
and z − i(z − w).
In the case where z and w are opposite vertices of the square, again translate
one of the vertices to the origin, that is, now the opposite vertices are 0 and
z − w. Now, in a square the diagonals intersect each other in a right angle in their
midpoints, then we need to consider the two complex numbers orthogonal to z − w
and then translate them to the
of z − w. That is, we need to consider
midpoint
z−w
and
−i z−w
Finally,
w to
+
+ z−w
the complex numbers i z−w
2
2
2
2 .
adding
z−w
z−w
all the previous
vertices,
we
get
the
square
with
vertices
w,
i
+
w,
z,
+
2
2
z−w
−i z−w
+
+
w.
2
2
Solution 5.12. Proceed by induction. The basis of induction is n = 2. We know
(cos θ + i sin θ)2 = cos2 θ − sin2 θ + i2 cos θ sin θ = cos 2θ + i sin 2θ,
where in the last equality we used the identity for the sum of angles for sine and
cosine.
Suppose then that the identity is true for some n = k, that is, we have
(cos θ + i sin θ)k = cos kθ + i sin kθ.
Then,
(cos θ + i sin θ)k+1 = (cos θ + i sin θ)(cos θ + i sin θ)k
= (cos θ + i sin θ)(cos kθ + i sin kθ)
= cos θ cos kθ − sin θ sin kθ + i(cos θ sin kθ + cos kθ sin θ)
= cos (k + 1)θ + i sin (k + 1)θ.
Solution 5.13. Equation z +
1
z
= 2 cos θ can be rewritten as
z 2 + 1 = 2z cos θ or z 2 − 2z cos θ + 1 = 0,
√
then z = cos θ ± cos2 θ − 1 = cos θ ± i sin θ.
Using de Moivre’s formula, it follows that z n = cos nθ ± i sin nθ, then
1
1
= cos nθ ∓ i sin nθ.
=
n
z
cos nθ ± i sin nθ
Adding the last two identities leads to z n +
1
zn
= 2 cos nθ.
Solution 5.14. Equation |z 2 + z̄ 2 | = 1 is equivalent to |z 2 + z̄ 2 |2 = (z 2 + z̄ 2 )(z̄ 2 +
z 2 ) = 1, but using |z| = 1, the last equation is equivalent to (z 4 + 1)2 = z 4 , which
can be factored as a difference of squares (z 4 − z 2 + 1)(z 4 + z 2 + 1) = 0. These
two quadratic equations can be solved using directly the general formula to obtain
224
Chapter 10. Solutions to Exercises and Problems
that the solutions are z 2 = 12 ±
we can obtain the values of z.
√
3
2 i
and z 2 = − 21 ±
√
3
2 i,
and from these equalities
Solution 5.15. (i) Let z1 , z2 be the roots of the equation with |z1 | = 1. Since
z1 z2 = ac , we have that |z2 | = ac |z11 | = 1. Now, since z1 + z2 = − ab and |a| = |b|,
it follows that |z1 + z2 |2 = 1. This last equation is equivalent to
(z1 + z2 )(z̄1 + z̄2 ) = 1,
or (z1 + z2 )
1
1
+
z1
z2
= 1.
2
Hence, (z1 + z2 )2 = z1 z2 , that is, − ab = ac , which can be reduced to b2 = ac.
(ii) It follows that b2 = ac and c2 = ab by the previous part. Multiplying both
equalities we have that b2 c2 = a2 bc, and then a2 = bc. Therefore, a2 + b2 + c2 =
ab + bc + ca. This last identity is equivalent to (a − c)2 = (a − b)(b − c). Taking
norms, |c − a|2 = |a − b| · |b − c|. Similarly, we can obtain |b − c|2 = |c − a| · |a − b|
and |a − b|2 = |b − c| · |c − a|. Adding the last equalities, we get |b − c|2 + |c − a|2 +
|a − b|2 = |b − c| · |c − a| + |c − a| · |a − b| + |a − b| · |b − c|, which is equivalent to
(|b − c| − |c − a|)2 + (|c − a| − |a − b|)2 + (|a − b| − |b − c|)2 = 0, and the result
follows.
Solution 5.16. Let z 5 + az 4 + bz 3 + cz 2 + dz + *
e be the polynomial having
* as roots
5
the numbers z1 , z2 , z3 , z4 , z5 . By Vieta, a = i=1 zi = 0 and b = i<j zi zj =
%2
$
*5
1 *5
− 12 i=1 zi2 = 0. Also, we have
z
i
i=1
2
0=
5
#
i=1
*
zi =
5
#
#
1
1
z1 z2 z3 z4 ,
=
z
z1 z2 z3 z4 z5
i=1 i
cyclic
*
then d = 0 and 0 = i<j zi zj = z1 z2 z13 z4 z5 cyclic z1 z2 z3 , hence c = 0. Thus, the
polynomial can be reduced to z 5 + e, which has as roots complex numbers that
are the vertices of a regular pentagon.
Solution 5.17. If a = b, |a − b + c| works; if a = −b, then |a + b + c| works. Suppose
that a is different from b and −b. Consider the numbers a + b, a − b, −a + b and
−a − b, which are the vertices of a rhombus of side 2. Taking as a center each of
these vertices, construct disks of radius 1; these 4 disks cover the circle with center
0 and radius 1. In particular, the point c belongs to one of these disks, then the
distance from the center of such a disk to c is less than or equal to 1.
Second Solution. The numbers a, b, c are the vertices of a triangle, with orthocenter
in a + b + c. If the triangle is acute, then its orthocenter is inside the triangle and
then |a + b + c| ≤ 1. If the triangle is obtuse, without loss of generality, we can
suppose that the obtuse angle is in a, then −a, b, c are the vertices of an acute
triangle with orthocenter −a + b + c, which is inside the triangle, and in this case
|−a + b + c| ≤ 1.
225
10.5 Solutions of Chapter 5
Solution 5.18. If at least one of the numbers a, b, c is 0, the result follows. Consider
b
c
a
, β = |b|
, γ = |c|
, then |α| = |β| = |γ| = 1 and if a |bc| + b |ca| + c |ab| = 0
α = |a|
is divided by |abc|, then α + β + γ = 0, hence α, β, γ are the vertices of an
equilateral triangle. Thus, the angle between two of them is π3 . Using the cosine
2
2
2
2
law, |a − b| = |a| + |b| + |a| |b| ≥ 3 |a| |b|. Similarly, |b − c| ≥ 3 |b| |c| and
2
2
2
2
2
2
2
|c − a| ≥ 3 |c| |a|, then |a − b| |b − c| |c − a| ≥ 33 |a| |b| |c| .
Solution 5.19. Since a, b, c have the same norm and |abc| = 1, it is clear that
= ab + bc + ca.
|a| = |b| = |c| = 1. Then, 1 = a + b + c = a1 + 1b + 1c = ab+bc+ca
abc
The monic polynomial which has zeros a, b, c is,
P (z) = (z − a)(z − b)(z − c) = z 3 − (a + b + c)z 2 + (ab + bc + ca)z − abc
= z 3 − z 2 + z − 1 = (z − 1)(z 2 + 1),
hence, {a, b, c} = {1, i, −i}.
Solution 5.20. Multiplying both sides of the equation x4 + x3 + x2 + x + 1 = 0 by
x−1, we get x5 −1 = 0; then to find the roots of x4 +x3 +x2 +x+1 = 0 is equivalent
to finding the roots of x5 − 1 = 0, which are different from 1. These
rootsarethe
quintic roots of unity, given by w, w2 , w3 , w4 where w = cos 52 π + i sin 52 π .
Second Solution. The equation can be solved dividing by x2 , and then making the
substitution y = x + x1 , and finally using the general formula to solve a quadratic
equation. That is,
x2 +
x2 + 2 +
x+
1
x2
1
x
1
1
+x+ +1=0
2
x
x
1
+ x+
−1=0
x
2
+ x+
1
x
−1=0
y 2 + y − 1 = 0.
The roots of this last equation are y1 =
solving the two equations
x+
1
= y1
x
√
−1+ 5
,
2
and
x+
y2 =
√
−1− 5
.
2
It is left to find x
1
= y2 ,
x
which are equivalent to x2 − y1 x + 1 = 0 and x2 − y2 x + 1 = 0. Solving these two
equations we find the four roots we are looking for:
√
√
√
√
10 + 2 5
10 + 2 5
−1 + 5
−1 + 5
+i
,
x2 =
−i
,
x1 =
4
4
4
4
√
√
√
√
−1 − 5
10 − 2 5
10 − 2 5
−1 − 5
x3 =
+i
,
x4 =
−i
.
4
4
4
4
226
Chapter 10. Solutions to Exercises and Problems
As a result of having two different methods for solving equation x4 +x3 +x2 +x+1 =
0, we can conclude that
√
√
−1 + 5
2
10 + 2 5
2
π + i sin
π =
+i
.
cos
5
5
4
4
Solving for the real and the imaginary part, we get
√
√
−1 + 5
10 + 2 5
◦
◦
cos 72 =
, sin 72 =
.
4
4
Solution 5.21. Note that the polynomial x6 + 2x5 + 2x4 + 2x3 + 2x2 + 2x + 1 can
be factored as
x6 + 2x5 + 2x4 + 2x3 + 2x2 + 2x + 1 = (x + 1)(x5 + x4 + x3 + x2 + x + 1).
In this way we have to find the roots of the equation
(x + 1)(x5 + x4 + x3 + x2 + x + 1) = 0,
where it is clear that x = −1 is one of the roots. The other roots are complex
numbers and can be calculated following the trick used in the previous problem.
Multiply x − 1 by x5 + x4 + x3 + x2 + x + 1 = 0 to get equation x6 − 1 = 0 and
find the roots distinct from 1. These roots are the 6th roots of unity, which can
be calculated using de Moivre’s formula (5.3).
Solution
=
3m+2 for some positive integer m, then the complex number
5.22. If n
2π
+
i
sin
is a solution with norm 1. Conversely, if z is a solution with
cos 2π
3
3
norm 1, then z̄ = z1 is also a solution. Then, z n + z + 1 = 0 = z n + z n−1 + 1, which
implies that z n−2 = 1, z 2 + z + 1 = 0, z 3 = 1 with z = 1, hence n = 3m + 2 for
some positive integer m.
Second Solution. Let P (z) = z n + z + 1 = 0. If P (w) = 0, with |w| = 1, then
w = cos θ+i sin θ, and then, using de Moivre’s formula (5.3), wn = cos nθ+i sin nθ,
it follows that 0 = (cos nθ + cos θ + 1) + i(sin nθ + sin θ). Then sin2 nθ = sin2 θ and
cos2 nθ = cos2 θ + 2 cos θ + 1, and from this cos θ = − 21 . It follows that w3 = 1
and w2 + w + 1 = 0, therefore wn = w2 , and then n ≡ 2 (mod 3).
Conversely, if n ≡ 2 (mod 3), for w = 1, with w a root of unity of order 3,
P (w) = 0. Then P (z) = z n + z + 1 = (z 2 + z + 1)Q(z), for some polynomial Q(z)
with integer coefficients.
Solution 5.23. (i) Let S = 1 + w + w2 + · · · + wn−1 . Multiplying by w both sides
of the equality, we get
Sw = w + w2 + · · · + wn−1 + wn ,
and subtracting from S the last equality, we obtain S − Sw = 1 − wn . Therefore
n
S = 1−w
1−w = 0.
227
10.5 Solutions of Chapter 5
(ii) Let S = 1 + 2w + 3w2 + · · · + nwn−1 . Multiplying by w we get
Sw = w + 2w2 + 3w3 + · · · + nwn .
Then, S − Sw = 1 + w + w2 + w3 + · · · + wn−1 − nwn = n−nwn (by the first part,
2
1 + w + w2 + w3 + · · · + wn−1 = 0). Therefore, S = −nw
1−w , hence 1 + 2w + 3w +
n
· · · + nwn−1 = w−1
.
Solution 5.24. (i) Observe that if w = 1 is a nth root of unity, then z n − 1 =
(z − 1)(z − w) . . . (z − wn−1 ). Hence,
zn − 1
= (z − w)(z − w2 ) . . . (z − wn−1 ) = z n−1 + z n−2 + · · · + z + 1.
z−1
Now, let z = 1 in the previous equality in order to obtain
(1 − w)(1 − w2 ) . . . (1 − wn−1 ) = n.
(ii) Consider the polynomial P (z) = (z − w)(z − w2 ) . . . (z − wn−1 ), and since
n
−1
= z n−1 +z n−2 +· · ·+z+1.
z −1 = (z−1)(z−w) . . . (z−wn−1 ), we get P (z) = zz−1
n
Note now that
P ′ (z)
P (z)
=
1
z−w
+
′
P (1)
P (1) .
1
z−w 2
+···+
1
z−w n−1 ,
then it is enough to calculate
Since P ′ (z) = (n − 1)z n−2 + (n − 2)z n−3 + · · · + 2z + 1, we obtain P ′ (1) =
′
(1)
. Now, from P (1) = n we can conclude that PP (1)
= n−1
1+2+· · ·+(n−1) = (n−1)n
2
2 .
Solution 5.25. (i) Note that
(a + bω + cω 2 )(a + bω 2 + cω)
= a2 + abω 2 + acω + bcω 2 + abω + b2 + acω 2 + abω + c2
= a2 + b2 + c2 + (ab + bc + ca)ω + (ab + bc + ca)ω 2
= a2 + b2 + c2 + (ab + bc + ca)(ω + ω 2 )
= a2 + b2 + c2 + (ab + bc + ca)(−1).
(ii) Substitute the equation obtained in (i).
Solution 5.26. The number of common vertices is given by the number of common
roots of z 1982 −1 = 0 and z 2973 −1 = 0. Then, by Theorem 5.4.1, it follows that the
number we are looking for is the greatest common divisor of (1982, 2973) = 991.
Solution 5.27. The roots of x2 + x + 1 are w = ei
w3 = 1 and 1 + w + w2 = 0, we obtain
2π
3
and w2 . Using the relations
n = 3k ⇒ w6k + w3k + 1 = 1 + 1 + 1 = 3,
n = 3k + 1 ⇒ w6k+2 + w3k+1 + 1 = w2 + w + 1 = 0,
n = 3k + 2 ⇒ w6k+4 + w3k+2 + 1 = w4 + w2 + 1 = w + w2 + 1 = 0.
Therefore the answer is for all n that are not multiples of 3.
228
Chapter 10. Solutions to Exercises and Problems
Solution 5.28. Use that x, y are of the form
a b c
c a b ,
b c a
in order to see that the product of both numbers is of the same form.
Second Solution. Use that
x = a3 + b3 + c3 − 3abc = (a + b + c)(a + bω + cω 2 )(a + bω 2 + cω)
= P (1)P (ω)P (ω 2 ),
for P (z) = cz 2 + bz + a and ω a cubic root of unity.
Then, x belongs to S if and only if x = P (1)P (ω)P (ω 2 ), hence, if
x = P (1)P (ω)P (ω 2 ) and y = Q(1)Q(ω)Q(ω 2 )
for Q(z) another polynomial of degree 2, we have that xy = R(1)R(ω)R(ω 2 ) with
R(z) = P (z)Q(z). Note that R(z) is of degree 4, and after dividing R(z) by z 3 − 1,
we get that R(z) = (z 3 − 1)L(z) + R1 (z) with R1 (z) of degree at most 2 and with
xy = R1 (1)R1 (ω)R1 (ω 2 ), then xy ∈ S .
Solution 5.29. Let w = e2πi/5 , then w5 = 1. Evaluating in the original equation
w, w2 , w3 , w4 , we obtain the following four equations:
P (1) + wQ(1) + w2 R(1) = 0
P (1) + w2 Q(1) + w4 R(1) = 0
P (1) + w3 Q(1) + wR(1) = 0
P (1) + w4 Q(1) + w3 R(1) = 0.
Now, if these equations are multiplied by −w, −w2 , −w3 , −w4 , respectively, we
obtain:
−wP (1) − w2 Q(1) − w3 R(1) = 0
−w2 P (1) − w4 Q(1) − wR(1) = 0
−w3 P (1) − wQ(1) − w4 R(1) = 0
−w4 P (1) − w3 Q(1) − w2 R(1) = 0.
Using 1 + w + w2 + w3 + w4 = 0 and adding the equations, we get 5P (1) = 0, that
is, x − 1 divides P (x).
Solution 5.30. If z is a root of P (z), then z 2 is also a root. Hence, if |z| > 1 there
will be an infinite number of roots, which is impossible since P (z) is a polynomial.
If 0 < |z| < 1, the same will happen, and there will be an infinite number of roots.
Then, all roots are 0 or they belong to the unit circle.
229
10.6 Solutions of Chapter 6
If P (z) is a constant polynomial, then the constant polynomials P (z) = 1 and
P (z) = 0 satisfy the equation.
If P (z) = az + b, with a = 0, substituting in the given equation, we get (az +
b)(−az + b) = az 2 + b, that is, az 2 + b = −a2 z 2 + b2 . Since a = 0, it follows that
a = −1 and b2 = b, then b = 0 or b = 1, and in this case there are two polynomials,
P (z) = −z and P (z) = 1 − z.
If P (z) = az 2 + bz + c, with a = 0, then
P (z)P (−z) = (az 2 + bz + c)(az 2 − bz + c) = a2 z 4 + (2ac − b2 )z 2 + c2 .
Comparing with P (z 2 ) = az 4 + bz 2 + c, we obtain a2 = a, 2ac − b2 = b and c2 = c.
Since a = 0, it follows that a = 1; for c2 = c we have the solutions c = 0 and
c = 1. For each of the values of c, we get two values for b: if c = 0, then b = 0
and b = −1; for c = 1, we get b = 1 and b = −2. Thus, in this case we have 4
polynomials that satisfy the given equation: P (z) = z 2 , P (z) = z 2 −z = −z(1−z),
P (z) = z 2 − 2z + 1 = (1 − z)2 and P (z) = z 2 + z + 1.
10.6 Solutions to exercises of Chapter 6
Solution 6.1. Observe that if f (x) + f
f
1
1−x
+f
Hence, 2f (x) = x +
x−1
x
x−1
x
−
1
1−x
=
=
$
1
1−x
1
1−x
%
= x, then
and f
x−1
x
+ f (x) =
x−1
.
x
−x3 +x−1
x(1−x) .
Solution 6.2. Prove by induction that f (n) = n.
Let m = n = 1 in order to see that f (1) = 1, and since f (2) = 2, by (ii)
and using the induction hypothesis, it follows that f (2k) = 2f (k) = 2k and
f (2k + 2) = f (2)f (k + 1) = 2(k + 1) = 2k + 2.
Finally, by (iii), 2k = f (2k) < f (2k + 1) < f (2k + 2) = 2k + 2, hence f (2k + 1) =
2k + 1. Then f (n) = n, for all n ∈ N.
Solution 6.3. Taking −x instead of x in the original equation, we obtain −xf (−x)−
2xf (x) = −1. Then
xf (x) + 2xf (−x) = −xf (−x) − 2xf (x),
hence, 3xf (x) = −3xf (−x). This we can substitute in the original equation to
obtain xf (x) = 1.
1
Solution 6.4. Taking −x instead of x in the original equation, we obtain −x
f (x) +
$ %
1
−1
1
2
f −x = −x, then f (x)−xf x = x . Now, taking x instead of x in the original
230
Chapter 10. Solutions to Exercises and Problems
+ f (x) = x1 . Adding the last two equalities leads
equation, we have that xf −1
x
1
2
to 2f (x) = x + x . Hence, the only function that satisfies the original equation is
f (x) =
1
2
1
+ x2 .
x
Solution 6.5. Taking y = −x in the original equation, we get xf (x) + xf (−x) =
2xf (0), for all x. Then f (x) + f (−x) = 2f (0) for all x = 0.
Taking x+y and x in the functional equation, we get (x+y)f (x+y)−xf (x) =
yf (2x + y), and taking 2x + y, −x, we obtain (2x + y)f (2x + y) + xf (−x) =
(3x + y)f (x + y).
Then, the last two equations can be rewritten as
x (f (x + y) − f (x)) = y (f (2x + y) − f (x + y))
(2x + y)(f (2x + y) − f (x + y)) = x(f (x + y) − f (−x)).
Multiplying the first equation by 2x + y and the second one by y, and reducing
we get (2x + y)x(f (x + y) − f (x)) = yx(f (x + y) − f (−x)). Canceling out x
on both sides of the equation, simplifying and solving for 2xf (x + y), leads to
2xf (x + y) = (2x + y)f (x) − yf (−x). Substituting the value of f (−x), gives us
2xf (x + y) = (2x + y)f (x) − y(2f (0) − f (x))
= 2(x + y)f (x) − 2yf (0).
That is,
xf (x + y) = (x + y)f (x) − yf (0)
xf (x + y) − xf (0) = (x + y)f (x) − yf (0) − xf (0)
x(f (x + y) − f (0)) = (x + y)(f (x) − f (0)).
Now if x = 1, then f (1+y)−f (0) = (1+y)(f (1)−f (0)). Substituting 1+y = x, we
get f (x) = f (0)+ x(f (1)− f (0)). Then, if we define the constants m = f (1)− f (0),
b = f (0), the functions that satisfy the equation have the form f (x) = mx + b.
Clearly, the functions of the form f (x) = mx + b satisfy the original functional
equation.
Solution 6.6. First, note that if f (x + 1) = f (x) + 1, by induction it follows that
f (x+n) = f (x)+n, for all n ∈ N. Moreover, f (x2 ) = f (x)2 and f (x+n) = f (x)+n
imply that f ((x + n)2 ) = f (x + n)2 = (f (x) + n)2 . Then f (x2 + 2xn + n2 ) =
f (x2 + 2xn) + n2 = f (x)2 + 2f (x)n + n2 , and then f (x2 + 2xn) = f (x)2 + 2f (x)n.
Taking x = 0 and n = 1 in the last equation, we get f (0)2 +f (0) = 0, then
$ f (0) =
% 0.
2
Moreover, taking x = pq and n = q in the last equation, leads to f pq2 + 2p =
$ %2
$ 2
%
$ 2%
2
f pq + 2qf ( pq ) = f ( pq2 ) + 2qf ( pq ) and, since f pq2 + 2p = f pq2 + 2p, then
f ( qp ) = pq . That is, f (x) = x for all x ∈ Q+ ∪ {0}.
231
10.6 Solutions of Chapter 6
Solution 6.7. Suppose that x, y, z are different numbers. The equation can be
rewritten as x(f (y)− f (z))+ yf (z) = f (x)(y − z)+ zf (y); now subtracting on both
sides yf (y), we get x(f (y) − f (z)) + y(f (z) − f (y)) = f (x)(y − z) + f (y)(z − y),
hence
f (y) − f (z)
f (x) − f (y)
=
.
x−y
y−z
Then, the slope between points (x, f (x)) and (y, f (y)) is equal to the slope between
points (y, f (y)) and (z, f (z)), then every 3 points of the graph of f are collinear;
hence the graph of f is a line and therefore f (x) = mx + b, for some real numbers
m and b. In fact, m is the common slope and b = f (0). Clearly, the affine functions
f (x) = mx + b satisfy the equation.
Second Solution. Taking y = −1, z = 1, we get xf (−1) − f (1) + f (x) = −f (x) +
f (−1) + xf (1), and solving for f (x) the result is
f (x) =
f (1) − f (−1)
f (−1) + f (1)
x+
.
2
2
Solution 6.8. The equality g(f (x)) = −x, for any real number x, guarantees that
g(f (g(x))) = −g(x). If to the equality f (g(x)) = −x, we apply g to both sides,
we obtain g(f (g(x))) = g(−x). Then g(−x) = −g(x), hence g is odd. Similarly we
can prove that f (−x) = −f (x).
Solution 6.9. Taking y = 0 in the original equation, we have f (f (x)) = f (x)−f (0).
But since f is surjective, given any real number y there exists x with f (x) = y,
then f (y) = y − f (0). Taking y = 0 in this last equation, we get f (0) = 0. Thus
f (x) = x for any real number x. It is clear that the function f (x) = x satisfies the
equation.
Solution 6.10. Taking x = 0 in the original equation gives us f (0) = 0. Now, if
x = 1, we have f (f (y)) = y, then f is bijective. Taking f (y) as y in the equation
and using f (f (y)) = y, we get f (xy) = xf (y). By symmetry in the variables x,
y, also it is true that f (xy) = yf (x), and then xf (y) = yf (x). Hence for x, y
f (x)
= f (y)
is constant and equal to
different from 0, it follows that f (x)
x
y , then
x
f (1), thus f (x) = f (1)x. Using f (x) = f (1)x in the original equation, it follows
that xy = f (1)xf (y) = f (1)2 xy for x, y ∈ R, then f (1)2 = 1. Hence f (x) = x or
f (x) = −x are the only continuous solutions of the equation.
Solution 6.11. Since m − n + f (n) ≥ 1 holds for all n ∈ N, then f (n) ≥ n. Letting
F (n) = f (n) − n, we can rewrite the functional equation as
F (m + F (n)) = F (m) + n, for all m, n ∈ N.
Taking m = 1 and adding 1 to both sides of the last equation, we have that
F (1 + F (n)) + 1 = F (1) + n + 1, for n ∈ N. If now we apply F on both sides and
232
Chapter 10. Solutions to Exercises and Problems
we use the new equation, we get F (1) + 1 + F (n) = F (n + 1) + 1, then
F (n + 1) = F (n) + F (1), for all n ∈ N.
That is, F (n) = F (1)n for all n ∈ N, then (m+nF (1))F (1) = mF (1)+n, for all n,
m ∈ N. In this last equation, taking n = m = 1, leads to F (1)(F (1)+1) = F (1)+1,
hence F (1) = 1, and then f (n) = 2n for all n ∈ N. Clearly f (n) = 2n satisfies the
functional equation.
Solution 6.12. In Example 6.2.4, we proved that the function is bijective.
With y = 1 and using the injectivity, it follows that
$ % f (1) = 1, and then
f (f (y)) = y1 . Applying f to both sides, we get
1
f (y)
=f
1
y
.
+
For x, y ∈ Q , take z such that f (z) = y, then
f (xy) = f (xf (z)) =
f (x)
= f (x)f (f (z)) = f (x)f (y).
z
In this way a function that satisfies the functional equation must satisfy the two
equations
1
and f (xy) = f (x)f (y).
f (f (x)) =
x
One particular solution can be defined as follows. Let p1 , p2 , . . . be the ordered
prime numbers and we define the function on the prime numbers as follows:
pi+1 , if i is odd
f (pi ) =
1
pi−1 , if i is even
and for a rational number r = pn1 1 · · · pnk k , the function is defined as
f (r) = f (p1 )n1 · · · f (pk )nk ,
where nk ∈ Z.
Solution 6.13. Taking x = y in the original equation we get xf (x) = x(f (x))2 ,
then x(f (x)2 − f (x)) = 0 for any real number x. Hence, for x = 0, we have
f (x)2 = f (x), so that for every real number x it follows that f (x) = 0 or 1.
If for all x = 0 we have f (x) = 0, then by continuity it follows that f (0) = 0,
and then f is identically zero on the real numbers.
If for some x0 = 0 it happens that f (x0 ) = 1, then taking x0 in the original
equation we obtain x0 f (y)+y = (x0 +y)f (y), hence y = yf (y) for all real numbers
y. Then, f (y) = 1 for all y = 0, and by continuity f (0) = 1, which guarantees that
f is identically 1 on the real numbers.
Therefore, the only functions that satisfy the equation are the constant functions 0 and 1.
233
10.6 Solutions of Chapter 6
Solution 6.14. (i) We should expect that the period is related to a, then it is a
good idea to iterate the function. By doing it we get
1
+
2
1
= +
2
f (x + 2a) =
1
= +
2
f (x + a) − f (x + a)2
1
+
2
f (x) − f (x)2 −
1
− f (x)
2
2
1
−
4
1
= + f (x) −
2
f (x) − f (x)2 − (f (x) − f (x)2 )
1
.
2
Since f (x) ≥ 12 for all x, then we have f (x) = f (x + 2a) for all x. Hence, f is
periodic with period 2a.
(ii) To find an example observe that f (x) ≥ 12 for all x and, on the other hand,
2
the original equation guarantees that f (x + 2) − 12 = f (x + 1)(1 − f (x + 1)) ≤
1 2
2 , where the inequality follows from the geometric and the arithmetic mean
inequality. Therefore, a possible example is, for n ∈ Z, the function
1
, if 2n ≤ x < 2n + 1
f (x) = 2
1, if 2n + 1 ≤ x < 2n + 2.
Solution 6.15. Suppose that the equation has at least one real solution x. Then
m(a + b) = |x − a| + |x − b| + |x + a| + |x + b|
≥ |(x − a) − (x + b)| + |(x − b) − (x + a)| = 2(a + b),
and since a + b > 0, it follows that m ≥ 2.
Conversely, suppose m ≥ 2, then the equation has at least one real solution.
In fact, if we define
f (x) = |x − a| + |x − b| + |x + a| + |x + b|,
observe that f (0) = 2(a + b) ≤ m(a + b) and f (ma + mb) = 4m(a + b) > m(a + b).
By the intermediate value theorem27 , there exists x such that f (x) = m(a + b).
Solution 6.16. Without loss of generality we can assume that f (0) = 0, since the
function g(x) = f (x) − f (0) satisfies the equation and g(0) = 0.
Taking y = 0 in the equation, we have f (x2 ) = xf (x). Using this last equation
and taking y = 1, we get xf (x) − f (1) = (x + 1)(f (x) − f (1)), hence f (x) = f (1)x.
This means all functions that satisfy the original equation are of the form f (x) =
f (1)x + f (0). It is easy to check that the affine functions f (x) = mx + b satisfy
the equation.
27 See
[21].
234
Chapter 10. Solutions to Exercises and Problems
Solution 6.17. Taking x = y = 1 in the original equation, it follows that f (1)2 −
f (1) = 2, then f (1) = 2. Now, if we let y = 1, we get f (x)f (1) − f (x) = x + x1 ,
then f (x) = x + x1 . And it is clear that f (x) = x + x1 satisfies the equation.
Solution 6.18. (i) In Example 6.2.2 we proved that these functions are injective. As we have seen in the example, taking x = y, leads to f (xf (x)) = x2 , in
particular f (f (1)) = 1. Taking x = f (1) in the last equation, gives us f (1)2 =
f (f (1)f (f (1))) = f (f (1)) = 1, hence f (1) = 1. Letting y = 1 in the original
equation we obtain f (x) + f (f (x)) = 2x.
If z > 0, taking x = zf (z) and y = 1z in the functional equation, we get
f
1
z
zf (z)f
1
1
f (zf (z)) = 2zf (z) .
z
z
+f
Then, using f (zf (z)) = z 2 , it follows that
f
1
z
zf (z)f
= f (z),
but since f is injective, it follows that f (z)f z1 = 1.
If now we take x = z, y = z1 in the original functional equation, we get
f
but since f
1
z
=
1
f (z) ,
zf
f
1
z
+f
1
f (z) = 2,
z
then
f
Since f (z)f
1
z
z
f (z)
+f
f (z)
z
= 2.
= 1, it also follows that
z
f (z)
·f
f (z)
z
= 1,
then f
z
f (z)
=f
f (z)
z
=1
and again the injectivity of f guarantees that f (z) = z.
(ii) In Example 6.2.3, we proved that these functions are surjective. Then, there
exists a number x0 such that f (x0 ) = 0. Letting x = x0 in the original equation,
we obtain f (y) = 2x0 + f (f (y) − x0 ), therefore if we make z = f (y) − x0 we get
f (z) = z − x0 .
Hence, the functions that satisfy the equation must have the form f (z) =
z + c, for some constant c.
Solution 6.19. First note that 2 must not be in the image in order to consider the
(x)−3
. If for some x, f (x) = 1 then f (x + a) = −2
quotient ff (x)−2
−1 = 2, hence 1 is not in
235
10.6 Solutions of Chapter 6
the image either. Now, observe that
2f (x) − 3
f (x + a) − 3
=
f (x + a) − 2
f (x) − 1
f (x + 2a) − 3
= f (x).
f (x + 3a) = f (x + 2a + a) =
f (x + 2a) − 2
f (x + 2a) =
Then, f (x) is periodic of period 3a.
Solution 6.20. Let T be the period of f . Suppose that T = pq , where p and q
are relatively prime positive integers. Then, qT = p is also a period of f . Let
n = kp + r, where k and r are integers and 0 ≤ r < p − 1.
Then f (n) = f (kp + r) = f (r), and f (n) ∈ {f (1), f (2), . . . , f (p − 1), f (p)}, for
all positive integers n, which is a contradiction with the fact that {f (n) | n ∈ N}
has an infinite number of elements.
Solution 6.21. Letting x = y = 0, we get
f (0) =
2f (0)
,
1 − f (0)2
which makes sense if f (0) = ±1. Then f (0)3 + f (0) = 0, hence f (0) = 0.
Now take g(x) = arctan f (x), then tan g(x) = f (x) which is well defined for
x ∈ (−1, 1). Substituting in the equation (6.7) we obtain
tan g(x + y) =
tan g(x) + tan g(y)
= tan(g(x) + g(y)).
1 − tan g(x) tan g(y)
The last equality follows from the tangent formula for the sum of two angles. Now,
apply the inverse tangent function on both sides of the equation to obtain
g(x + y) = g(x) + g(y) + k(x, y)π,
where k(x, y) is a function that only takes integer values. On the other hand, since
f (0) = 0 we have g(0) = 0 and then k(0, 0) = 0. But since k is a continuous
function, k(x, y) = 0 for all x, y ∈ R, we get the equation
g(x + y) = g(x) + g(y),
which is the Cauchy equation whose continuous solution is g(x) = αx. Hence, the
solution of equation (6.7) is the function f (x) = tan αx.
Solution 6.22. Since
tan u + tan v
= tan(u + v),
1 − tan u tan v
236
Chapter 10. Solutions to Exercises and Problems
we can take x = tan u and y = tan v with xy = 1, which is true if and only if
tan u tan v = 1, that is, u − v = π2 . The equation becomes
f (tan u) + f (tan v) = f (tan(u + v)),
then f ◦ tan is additive and continuous, therefore f (tan u) = cu, which implies
that f (x) = c arctan x.
Solution 6.23. The function f (x) ≡ −1 is a solution of the functional equation. If
now we define g(x) = f (x) + 1, substituting we get
g(x + y) − 1 = g(x) − 1 + g(y) − 1 + [g(x) − 1][g(y) − 1]
= g(x) − 1 + g(y) − 1 + [g(x)g(y) − g(x) − g(y) + 1]
= g(x)g(y) − 1.
Therefore g(x) satisfies the Cauchy equation g(x + y) = g(x)g(y) and then g(x) =
ax , with a ∈ R+ and f (x) = ax − 1.
Solution 6.24. Suppose that there exists a function that satisfies
f (f (n)) = n + 1.
(10.9)
Applying f to both sides of the equation, we obtain f (n + 1) = f (f (f (n))) =
f (n) + 1. Let us see by induction that f (n + 1) = f (1) + n.
The case n = 1 is obvious, since f (n + 1) = f (n) + 1 and after substituting
n = 1, we get f (1 + 1) = f (1) + 1. Suppose the result true for n − 1 and prove it
for n. Since it is true for n − 1, the following holds:
f (n + 1) = f (n) + 1 = f ((n − 1) + 1) + 1 = (f (1) + n − 1) + 1 = f (1) + n,
then f (n + 1) = f (1) + n for all n ∈ N. From this last equation and by equation
(10.9), we get
n + 1 = f (f (n)) = f (n) − 1 + f (1) = n − 1 + f (1) − 1 + f (1) = n − 2 + 2f (1).
Thus f (1) = 32 , which is a contradiction, since the image of f are the natural
numbers. Therefore, no f exists that satisfies equation (10.9).
Solution 6.25. If x ≥ 2, then f (x) = f (x − 2 + 2) = f ((x − 2)f (2))f (2) = 0, this
together with (iii), implies that f (x) = 0 if and only if x ≥ 2. For 0 ≤ y < 2,
we have f (y) = 0 and 0 = f (2) = f (2 − y + y) = f ((2 − y)f (y))f (y), then
f ((2 − y)f (y)) = 0, hence (2 − y)f (y) ≥ 2.
2
, we get f (x + y) = f (xf (y))f (y) = f (2)f (y) = 0, then x + y ≥ 2,
Taking x = f (y)
but this implies 2 ≥ (2 − y)f (y). Since (2 − y)f (y) ≥ 2, then (2 − y)f (y) = 2, that
2
is, f (u) = 2−y
for 0 ≤ y < 2.
237
10.6 Solutions of Chapter 6
Hence,
f (y) =
2
2−y ,
for 0 ≤ y < 2
for y ≥ 2.
0,
It is not difficult to see that f satisfies the conditions of the exercise.
Solution 6.26. It is clear that f (x) = 0 and f (x) = 1 are solutions; let us see that
there are no other solutions.
Taking x = y = 0 in the functional equation, it follows that f (0) = [f (0)]2 ,
then either f (0) = 0 or f (0) = 1. Let us analyze the two cases:
(i) f (0) = 0. Letting y = 0 in the functional equation, we get f (x) = f (x)f (0) =
0, then f (x) = 0 for all x ∈ R.
ii) f (0) = 1. First, we will see that f (x) = 0 for all x ∈ R. Letting x = y in
the original equation, we obtain 1 = f (0) = f (2x)f (x), then f (x) = 0 for all
x ∈ R.
Substituting x and y by 2u and u, respectively, we obtain f (u) = f (3u)f (u), since
f (u) = 0, then f (3u) = 1. Finally, letting 3u = x, we get f (x) = 1 for all x ∈ R.
Therefore the only solutions are f (x) = 0 and f (x) = 1.
Solution 6.27. By Theorem 6.5.4, for a function f we have
∆n f (x) =
n
#
(−1)k
k=0
n
f (x + n − k) when h = 1.
k
Moreover, by Example 6.5.3, if P (x) = a0 +a1 x+· · ·+an xn , then ∆n P (x) = an n!,
when h = 1.
n
. The coefficient
Consider f (x) = P (x) = (n − x)n*
of xn in this polynomial
n
n
n
n
k n
is (−1) , then (−1) n! = ∆ P (x) = k=0 (−1) k (k − x) . Therefore, letting
x = 0 in the last equation, we get
n
#
k=0
(−1)k k n
n
k
= (−1)n n!.
Solution 6.28. We prove that the function f is injective. If f (n) = f (m), then
f (f (f (n))) + f (f (n)) + f (n) = f (f (f (m))) + f (f (m)) + f (m), hence 3n = 3m,
that is, n = m.
Evaluating in n = 0, it follows that f (f (f (0))) + f (f (0)) + f (0) = 0, then
f (f (f (0))) = f (f (0)) = f (0) = 0.
It is evident that f (n) = n satisfies the equation; let us see that it is the
only solution. By induction suppose that f (k) = k for 0 ≤ k < n. Since f is
injective, f (n) cannot take any of the values 0, 1, . . . , n − 1, then f (n) ≥ n and
also f (f (n)) ≥ n and f (f (f (n))) ≥ n. Thus, f (f (f (n))) + f (f (n)) + f (n) ≥ 3n.
238
Chapter 10. Solutions to Exercises and Problems
By hypothesis, the equality must hold, then f (f (f (n))) = f (f (n)) = f (n) = n,
which proves that f (n) = n, for all n ∈ N ∪ {0}.
Solution 6.29. Let us prove that f (x) = x. First, note that f is injective because
if f (x) = f (y), then
x = f n (x) = f n (y) = y,
hence x = y and f is injective. Now, let us see that f is increasing. Suppose that
there exist x1 and x2 such that x1 < x2 and f (x1 ) ≥ f (x2 ); since f is continuous
in [0, x1 ], then by the intermediate value theorem28 there is c ∈ [0, x1 ] such that
f (c) = f (x2 ), which is a contradiction, since f is injective.
Now, assume that x < f (x), then
f (x) < f (f (x)) = f 2 (x) < · · · < f n (x) = x,
which is a contradiction. Similarly, if we suppose that x > f (x) we reach another
contradiction, therefore f (x) = x is the only function that satisfies the conditions.
Solution 6.30. Note that f is injective. If for all x, y ∈ R we have that f (x) =
f (y), then f n (x) = f n (y), hence −x = −y. Also, f is surjective because −x =
f (f n−1 )(x).
Since f (−x) = f (f n (x)) = f n (f (x)) = −f (x), we have that f is odd, therefore f (0) = 0.
But if f is bijective and continuous, then it is monotone. Let us prove that
f cannot be increasing. If x < y implies that always f (x) < f (y), then −x =
f n (x) < f n (y) = −y, and y < x, which is a contradiction. Thus, if f is decreasing,
then x and f (x) should have different signs for x = 0 (note that x > 0, implies
that f (x) < 0, and x < 0, implies that f (x) > 0). Then, for x = 0, xf (x) < 0 and
then x, f (x), f 2 (x), . . . , f n (x) alternate signs, but if f n (x) = −x, then n is odd.
Let x > 0 and assume that f (x) > −x. Since f is decreasing and odd, we
have
f (f (x)) < f (−x) = −f (x) < x,
again, since f (x) < −f (f (x)), being decreasing and odd
f (f (x)) > f (−f (f (x))) = −f (f (f (x))).
Continuing in this way, we get
x > −f (x) > f 2 (x) > −f 3 (x) > · · · > −f n (x) = x,
which is a contradiction, therefore f (x) ≤ −x.
Similarly, we can show that for x > 0 it is not possible that f (x) < −x. Then
f (x) = −x for x > 0. Now, using that f is odd, we conclude that f (x) = −x, for
all x ∈ R. And f (x) = −x satisfies the functional equation.
28 See
[21].
239
10.7 Solutions of Chapter 7
10.7 Solutions to exercises of Chapter 7
Solution 7.1. Proceed by induction. For n = 1, we have a1 = 1 < 47 and for n = 2,
7 2
49
we have a2 = 3 = 48
16 < 16 = 4 , which proves the induction basis.
n−2
For n > 2, suppose the result valid for n − 2 and n − 1, that is, an−2 < 47
n−1
. By direct calculation and using the induction hypothesis, it
and an−1 < 74
follows that
an = an−1 + an−2
7
4
n−1
<
7
4
n−2
<
7
4
+
49
16
n−2
7
4
=
=
7
4
n−2
7
+1
4
=
7
4
n−2
11
4
n
.
Solution 7.2. Adding 2n to both sides of an = 3an−1 + 2n−1 , we get an + 2n =
3an−1 + 3 · (2n−1 ) = 3(an−1 + 2n−1 ), for all n ≥ 2. Setting bn = an + 2n , we
obtain bn = 3bn−1 = · · · = 3n−1 b1 . Since b1 = a1 + 2 = 1 + 2 = 3, it follows that
bn = 3n−1 · 3 = 3n , hence an = 3n − 2n .
Solution 7.3. Since an+1 = 1 + a1 a2 . . . an , it follows that a1 a2 . . . an = an+1 − 1,
then a1 a2 . . . an−1 = an − 1; hence an+1 − 1 = (an − 1)(an ) > 0, therefore
1
an+1 − 1
=
1
1
1
.
=
−
(an − 1)(an )
an − 1 an
Finally, we get
1
1
1
1
+ ··· +
=1+
+ ···+
a1
an
a2
an
1
1
1
1
=1+
−
−
+ ···+
a2 − 1 a3 − 1
an − 1 an+1 − 1
1
1
1
−
= 2−
< 2.
=1+
a2 − 1 an+1 − 1
an+1 − 1
Solution 7.4. By definition an+1 an−1 = a2n + 1. Consider an+2 an = a2n+1 + 1. If
we subtract from this last equation the original identity, we get
an+2 an − an+1 an−1 = a2n+1 − a2n ,
which can be rewritten as an (an+2 + an ) = an+1 (an+1 + an−1 ). Therefore,
an+2 + an
an+1 + an−1
=
.
an+1
an
240
Chapter 10. Solutions to Exercises and Problems
Now, the sequence bn =
an+1 +an−1
an
with n ≥ 2 is constant29 and since
a3 + a1
b2 =
=
a2
a22 +1
a1
+ a1
a2
= 3,
n−1
it follows that an+1a+a
= 3, then an+1 = 3an − an−1 , for every n ≥ 2. Now,
n
the principle of mathematical induction helps us to conclude that every an is an
integer and the original equation implies that each an is positive.
Solution 7.5. We have a2 − a1 ≥ 1 and an+2 − an+1 ≥ (an+1 − an ) + 1, where the
second inequality follows by applying the condition to (n, n + 1, n + 1, n + 2) for all
n. By induction, it is possible to show that an+1 − an ≥ n for all n ≥ 1. Therefore,
an+1 ≥ n + an and a1 ≥ 1, and again by induction we have an ≥ 12 (n2 − n + 2).
Since the sequence an = 12 (n2 − n + 2) satisfies the conditions of the problem (the
condition ai + al > aj + ak in this case becomes i2 + l2 > j 2 + k 2 , where we take
i = d − y, l = d + y, j = d − x, k = d + x, with 0 ≤ x < y), the smallest value of
a2008 is 2 015 029.
Solution 7.6. Note that a1 = 1 − a0 implies that a0 = 1 − a1 .
Now, a2 = 1 − a1 (1 − a1 ) = 1 − a1 a0 , and then by induction we have an =
1 − a0 a1 . . . an−1 . That is,
an+1 = 1 − an (1 − an ) = 1 − an (a0 . . . an−1 ) = 1 − a0 . . . an−1 an .
The proof is finished using induction. For n = 0, 1 the identity follows immediately.
Now suppose the statement holds for n and consider
1
1
+ ···+
a0
an+1
1
1
+ ···+
= (a0 . . . an )
a0
an
(a0 . . . an+1 )
an+1 + (a0 . . . an+1 )
1
an+1
= an+1 + a0 . . . an = 1.
Solution 7.7. For n = 0, we have x20 = y0 + 2. Now, use induction. Suppose that
x2k = yk + 2 and prove that x2k+1 = yk+1 + 2.
Indeed, x2k+1 = (x3k − 3xk )2 = (x2k )3 − 6(x2k )2 + 9(x2k ). Using the induction hypothesis, we have
x2k+1 = (yk + 2)3 − 6(yk + 2)2 + 9(yk + 2) = yk3 − 3yk + 2 = yk+1 + 2.
Solution 7.8. For n = 1, we have 1 + 4a1 a2 = 1 + 4(1)(12) = 49 = 72 . Now, we
use induction to show that for n ≥ 2, we have
1 + 4an an+1 = (an+1 + an − an−1 )2 .
That is, for n = 2 we get 1 + 4a2 a3 = 1 + 960 = 961 = 312 , and (a3 + a2 − a1 )2 =
(20 + 12 − 1)2 = 312 .
29 See
Example 7.1.4.
10.7 Solutions of Chapter 7
241
For the inductive step, suppose that 1 + 4an an+1 = (an+1 + an − an−1 )2 .
Note that
(an+2 + an+1 − an )2 = (2an+1 + 2an − an−1 + an+1 − an )2
= (2an+1 + an+1 + an − an−1 )2
= 4a2n+1 + 4an+1 (an+1 + an − an−1 ) + (an+1 + an − an−1 )2
= 4a2n+1 + 4an+1 (an+1 + an − an−1 ) + (1 + 4an an+1 )
= 4an+1 (2an+1 + 2an − an−1 ) + 1 = 4an+1 an+2 + 1,
as we wanted to prove.
√
Solution 7.9. Note that a1 = 2, a2 < 4, a3 < 4 + 3 · 4 = 4. We show by induction
that a√n < 4 for all
√ n ∈ N. For n = 0, 1, 2 and 3, it is clear. Suppose that an < 4,
then 4 + 3an < 4 + 3 · 4 = 4, and an+1 < 4. Hence, the sequence is bounded by 4.
Solution 7.10. Since an+1 = an + a12 , then a3n+1 = a3n + 3 + a33 + a16 > a3n + 3.
n
n
n
Since √
a32 = 1 + 3 + 3 + 1 > 2 · 3, by induction it follows that a3n √
> 3n. Therefore,
an > 3 3n and the sequence is not bounded. Moreover, a9000 > 3 27000 = 30.
Solution 7.11. Suppose that all terms of the sequence are rational positive numbers, an = pqnn , with (pn , qn ) = 1. Then
p2n+1
pn
pn + qn
= a2n+1 = an + 1 =
+1=
,
2
qn+1
qn
qn
2
2
2
that is, qn+1
(pn + qn ) = qn · p2n+1 . Then, note that qn |qn+1
and qn+1
|qn , therefore
2
qn+1 = qn for all n.
n
Then, qn+1 = (q1 )1/2 is a positive integer for all n. This happens only if
q1 = 1 and then qn = 1 √for all n, meaning that an is an integer for all n. Now,
if an = 1, then an+1 = 2, which is a contradiction. Then, an > 1 for all n. It
follows that a2n+1 − a2n = an + 1 − a2n = 1 + an (1 − an ) < 0 and an+1 < an for all
n, that is, we have an infinite decreasing sequence of positive integers, which is a
contradiction. Therefore, the sequence must contain irrational numbers.
Solution 7.12. It is not difficult to see that the constant sequences
{an )= A}, the
(
linear sequences {an = Bn} and the sequences of the form an = Cn2 , with A,
B, C fixed numbers, solve the recurrence. Then, also the sequences
)
(
an = A + Bn + Cn2
are solutions.
)Given the initial conditions, the solutions are {an = 1}, {an = n}
(
and an = n2 , respectively.
Solution 7.13. Since the sequence is bounded, some terms are repeated infinitely
many times. Let K be the greatest number that is repeated infinitely many times
in the sequence, and let N be a positive integer such that ai ≤ K for i ≥ N .
242
Chapter 10. Solutions to Exercises and Problems
Choose m ≥ N such that am = K. We will prove that m is the period of the
sequence, that is, ai+m = ai for all i ≥ N .
First, we suppose that ai+m = K for some i. Since ai + am is divisible by
ai+m = K, then ai = K = ai+m .
Now, if ai+m < K, choose j ≥ N such that ai+j+m = K, then it follows that
ai+m + aj < 2K. Since ai+m + aj is divisible by ai+j+m = K, then ai+m + aj = K
and therefore aj < K. Since ai+j+m = K, the argument in the previous paragraph
implies that ai+j+m = ai+j = K, and then K divides ai + aj . It follows that
ai + aj = K, since ai ≤ K and aj < K. Therefore, ai+m = K − aj = ai .
Solution 7.14. The sequence an = n! satisfies the given recursion because n(n! +
(n − 1)!) = n(n + 1)(n − 1)! = (n + 1)!.
The number of derangements of n + 1 elements can be found as follows:
Consider the permutations of n+ 1 elements without fixed points; the first element
can be any of the n elements different from the first. Since there are n elements
left, the dn+1 permutations can be divided into n groups according to which one
was in the first place of the n elements different from the first. The groups have the
same number of elements. Take one of the groups, say the one where the second
element was in the first place.
The permutations are divided in two, when 1 goes to 2 and otherwise. In the
first case, there are dn−1 derangements and in the second 1 is moved to any place
different from 2 and the rest will move freely to a different place from the first,
then there are dn such permutations, hence dn+1 = n(dn + dn−1 ).
The sequences are different since the first terms are not equal, that is, d0 = 1,
d1 = 0 and a0 = a1 = 1.
Solution 7.15. (i) Note that
dn − ndn−1 = −(dn−1 − (n − 1)dn−2 ) = (dn−2 − (n − 2)dn−3 ) = · · ·
= (−1)n−2 (d2 − 2d1 ) = (−1)n−2 (1 − 2 · 0) = (−1)n .
(ii) A direct application of the formula in Example 7.2.4, leads to
dn = n(n − 1) · · · 2 · d1 +
= (n!)d1 +
n−2
#
n−2
#
j=1
n(n − 1) · · · (j + 2)(−1)j+1 + (−1)n
n!
n!
(−1)j+1 + (−1)n
(j
+
1)!
n!
j=1
(−1)2
(−1)3
(−1)n
+
+ ···+
2!
3!
n!
2
1
(−1)
(−1)n
(−1)
+
+ ···+
= n! 1 +
1!
2!
n!
= n!
.
243
10.7 Solutions of Chapter 7
Solution 7.16. The
characteristic√equation of the recursion is x2 − x − 1 = 0 which
√
1+ 5
has roots r = 2 and s = 1−2 5 . The solutions of the equation have the form
$ √ %n
$ √ %n
Ln = A 1+2 5 + B 1−2 5 .
$ √ %2
$ √ %2
√
Since 1 = L1 = A+B
+ 25 (A − B) and 3 = L2 = A 1+2 5 + B 1−2 5 ,
2
$ √ %n $ √ %n
then A = B = 1, hence Ln = 1+2 5 + 1−2 5 , for all n.
b0
> 0 and by induction bn > 0. Consider
Solution 7.17. If b0 > 0, then b1 = 1+b
0
1
an = bn ; the recursive equation takes the form an+1 = an + 1, which has as a
b0
solution the sequence an = a0 +n. Then bn = 1+nb
is the solution of the equation.
0
Solution 7.18. Notice that
an+3 =
1
1
1
=
=
1 − an+2
1 − 1−a1n+1
1 − 1− 1 1
= an .
1−an
Solution 7.19. Suppose that an = bn+1
bn , then the recursion takes the form bn+2 =
4bn+1 −4bn which is linear of order 2. Its characteristic polynomial is λ2 −4λ+4 = 0,
which has as unique solution λ = 2. Then, bn = (A + nB)2n for some numbers A
n+1
is the solution of the equation.
= (A+(n+1)B)2
and B. Hence an = (A+(n+1)B)2
(A+nB)2n
(A+nB)
Clearly an converges to 2.
Solution 7.20. By Proposition 7.2.13, it is enough to observe the following inequalities, where we use the fact that ak+1 ≤ 2ak , for k = 1, 2, . . . , n,
an+1 ≤ 2an = an + an
≤ an + 2an−1 = an + an−1 + an−1
..
.
≤ an + an−1 + · · · + a2 + 2a1
= an + an−1 + · · · + a2 + a1 + 1.
Solution 7.21. Denote the sequence by {pn }; prove that pn+1 ≤ 2pn , for any n ≥ 1.
For n = 1 it is immediate, since 2 = p2 = 2p1 = 2. For n ≥ 2, we use Bertrand’s
postulate30 , which says that given an integer m > 1, there exists a prime number
p such that m < p < 2m.
For pk with k ≥ 2, it follows, again by Bertrand’s postulate, that there exists a
prime number p with pk < p < 2pk . But this prime number is greater than or
equal to the prime number after pk , that is pk+1 ≤ p. Then, pk+1 ≤ 2pk and then,
using the previous exercise, we have the result.
30 See
[15].
244
Chapter 10. Solutions to Exercises and Problems
Solution 7.22. Let a1 , a2 , . . . , an be the integer weights of each of the golden
pieces and suppose that a1 ≤ a2 ≤ · · · ≤ an . By hypothesis,
a1 + a2 + · · · + an = 2n and an ≤ a1 + a2 + · · · + an−1 .
If an = a1 + a2 + · · · + an−1 , we are done. Case a1 ≥ 2 is clear since ai ≥ 2 for all
i, and the condition a1 + a2 + · · · + an = 2n, implies that a1 = a2 = · · · = an = 2.
Since n is even, we can perform the required partition.
Now, suppose a1 = 1 and an < a1 + a2 + · · · + an−1 , then
an ≤ 1 + a1 + a2 + · · · + an−1 .
Then, it is enough to show that ak+1 ≤ 1 + a1 + a2 + · · ·+ ak , for k = 1, 2, . . . , n− 2.
Suppose that the previous statement is not true, that is, let ak+1 > 1 + a1 + a2 +
· · · + ak for some k ∈ {1, 2, . . . , n − 2}. Then ak+1 ≥ k + 2, and ai ≥ k + 2 for every
i = k + 1, k + 2, . . . , n. Moreover, we know that ai ≥ 1 for i = 1, 2, . . . , k, then
2n = a1 + a2 + · · · + an ≥ k + (n − k)(k + 2) = −k 2 + (n − 1)k + 2n.
This implies that k 2 − (n − 1)k ≥ 0, therefore k ≤ 0 or k ≥ n − 1, and both
contradict that k ∈ {1, 2, . . . , n − 2}, and so the result follows.
Now, to make the division follow the ideas of the proof of Proposition 7.2.13.
Solution 7.23. It is possible to show by induction that 1 ≤ an ≤ 2 and that
an ≥ an+1 , for all n ≥ 1. Therefore, the limit of the
√ sequence is equal to some
number L (see Theorem 7.4.7) which satisfies L = L, hence L = 1.
Solution 7.24. (i) Observe that
n
#
i=0
1
1
−
ai
ai+1
=
1
1
1
1
1
1
−
+
−
+ ···+
−
a0
a1
a1
a2
an
an+1
a2 − a1
an+1 − an
a1 − a0
+
+ ···+
a0 a1
a1 a2
an+1 an
d
d
d
+
+ ···+
=
a0 a1
a1 a2
an+1 an
n
# 1
.
=d
a a
i=0 i i+1
=
On the other hand, the series
n
#
i=0
1
1
−
ai
ai+1
Therefore,
*n
=
*n
1
i=0 ( ai
−
1
ai+1 )
is telescopic, hence
1
1
1
1
1
1
1
1
−
+
−
+ ···+
−
=
−
.
a0
a1
a1
a2
an
an+1
a0
an+1
1
i=0 ai ai+1
=
1
d
$
1
a0
−
%
an+1 .
1
245
10.7 Solutions of Chapter 7
(ii) Observe that
n
#
1
1
−
ai ai+1
ai+1 ai+2
i=0
1
1
1
1
1
1
−
+
−
+ ···+
−
a0 a1
a1 a2
a1 a2
a2 a3
an an+1
an+1 an+2
a3 − a1
an+2 − an
a2 − a0
+
+ ···+
=
a0 a1 a2
a1 a2 a3
an an+1 an+2
n
#
2d
2d
2d
1
=
.
+
+ ···+
= 2d
a0 a1 a2
a1 a2 a3
an an+1 an+2
aa a
i=0 i i+1 i+2
$
%
*
On the other hand, the series ni=0 ai a1i+1 − ai+11ai+2 is telescopic, then
=
n
#
i=0
Therefore,
1
1
−
ai ai+1
ai+1 ai+2
*n
1
i=0 ai ai+1 ai+2
1
2d
=
$
=
1
a0 a1
1
1
−
.
a0 a1
an+1 an+2
−
1
an+1 an+2
%
.
(iii) Since a1 = a0 + d, a2 = a1 + d = a0 + 2d, . . ., an+1 = a0 + (n + 1)d, it follows
from part (i) that
∞
#
i=0
1
1
= lim
n→∞ d
ai ai+1
1
1
−
a0
a0 + (n + 1)d
=
1
.
da0
(iv) Since a1 = a0 + d, a2 = a1 + d = a0 + 2d, . . ., an+1 = a0 + (n + 1)d, we have
from part (ii) that
∞
#
i=0
1
1
= lim
n→∞ 2d
ai ai+1 ai+2
1
1
−
a0 a1
(a0 + (n + 1)d)(a0 + (n + 2)d)
=
1
.
2da0 a1
Solution 7.25. Use the recurrence formula for the Fibonacci sequence in order to
obtain the following equalities:
(i)
∞
∞
#
#
fn+1 − fn−1
fn
=
=
f
f
fn−1 fn+1
n=2
n=2
n=2 n−1 n+1
∞
#
= lim
N →∞
(ii)
∞
#
1
1
−
fn−1
fn+1
1
1
1
1
+
−
−
f1
f2
fN
fN +1
=
1
1
+
= 2.
f1
f2
∞
∞
#
#
1
fn
fn+1 − fn−1
=
=
f
f
f
f f
f
f f
n=2 n−1 n+1
n=2 n−1 n n+1
n=2 n−1 n n+1
=
∞
#
n=2
1
1
−
fn−1 fn
fn fn+1
= lim
N →∞
1
1
−
f1 f2
fN fN +1
=
1
= 1.
f1 f2
246
Chapter 10. Solutions to Exercises and Problems
Solution 7.26. In step 0, the equilateral triangle has perimeter 3. In step 1, the
curve is formed by 4 · 3 segments of length 13 , then it has perimeter 31 · 4 · 3. In
step 2, the curve is formed by 42 · 3 segments of length 312 . Then its perimeter is
1
4n
2
has perimeter Pn = 31n · 4n · 3 = 3n−1
.
32 · 4 · 3. In general, in step n the curve
4n
= ∞, since Pn is a geometric progression
In this way, limn→∞ Pn = limn→∞ 3n−1
with ratio r = 43 > 1.
√
The equilateral triangle of side 1 has area equal to 43 . In step 1 the curve encloses
an area equal to the sum of the area of the original triangle and the area of the
are%constructed on every side of the
three equilateral triangles of sides 13 , which
$
original triangle. That is, A1 =
√
3
4
√
3
4
+3
·
1
32
.
1
In step 2 another $
4 × 3 equilateral
to the
%
$triangles
% of side 9 are added
%
$ √structure,
√
√
√
√
3
3
3
3
1
1
then A2 = 4 + 3 4 · 32 + 4 · 3 4 · 34 . In general, An = 4 + 3 43 · 312 +
$√
%
%
$√
1
4 · 3 43 · 314 + · · · + 4n−1 · 3 43 · 32n
.
%%
$
√ $
2
n−1
Notice that we can write An = 43 1 + 13 1 + 94 + 942 + · · · + 94n−1 . In this
way, we have that the sum inside the second parenthesis is the
sum of a$ geometric
%%
√ $
progression with ratio r = 94 . Therefore, limn→∞ An = 43 1 + 13 1−1 4
=
9
√
√
3
1 + 53 = 2 5 3 .
4
Solution 7.27. (i) For n ≥ 1, it follows that 2n1+j ≥
Since the inequality is strict for j = 2n , we have
2n
1
2n+1 ,
for all j = 1, 2, . . . , 2n .
1
2n
1
1
1
+ n
+ · · · + n+1 > n+1 = .
+1 2 +2
2
2
2
(ii) The sum can be written as
1
1
1
1
+
+ ··· + 2 = +
n n+1
n
n
+
+
+
Since
1
kn+1
+ ··· +
1
(k+1)n
≥
n
(k+1)n
=
1
k+1 ,
1
1
+ ···+
n+1
2n
1
1
+ ··· +
2n + 1
3n
..
.
1
1
+ ···+ 2
(n − 1)n + 1
n
.
for each k = 2, 3, . . . , n − 1, we have
1
1
1 1
1
1
1
+
+ ···+ 2 ≥ + + + ···+ .
n n+1
n
n 2 3
n
In order to reach the conclusion, observe that for n ≥ 2,
1
1
1
2 + 3 + 6 = 1.
1
n
+
1
2
+
1
3
+ ··· +
1
n
≥
247
10.7 Solutions of Chapter 7
1
1
+ n1 + n+1
>
(iii) Simplifying the inequality n−1
2
2
n − 1 < n , which is always true for n ≥ 1.
(iv) From part (i), we get 1 +
1
1
1
2n +2 + · · · + 2n+1 > 2 then
1+
1
2
> 12 ,
1
3
+
1
4
>
3
n,
we see that it is equivalent to
1 1
2, 5
+ ··· +
1
8
> 21 , . . . ,
1
2n +1
+
1 1
1
n+1
+ + · · · + n+1 >
,
2 3
2
2
hence we can conclude that the harmonic series is divergent.
Solution 7.28. Use Abel’s summation formula to obtain
n
#
k 2 q k−1 =
k=1
n−1
#
(k 2 − (k + 1)2 )(1 + q + · · · + q k−1 ) + n2 (1 + q + · · · + q n−1 )
k=1
n−1
#
=−
(2k + 1)
k=1
n−1
# k
qk − 1
q−1
qn − 1
q−1
+ n2
n−1
#
qk − 1
qn − 1
q −1
−2
(k + 1)
+ n2
q−1
q−1
q−1
k=1
k=1
n−1
n−1
#
#
1
2
k
k
q −1 −
(k + 1)(q − 1)
=
q−1
q−1
=
k=0
+ n2
k=0
qn − 1
q−1
n
2
qn − 1
−
−
=
(q − 1)2
q−1 q−1
+n
2
qn − 1
q−1
n
#
k=1
n(n + 1)
kq k−1 −
2
.
(10.10)
Using Example 7.3.5, we can see that equation (10.10) is equal to
2
qn − 1
−
(q − 1)2 q − 1
=
nq n
qn − 1
−
q − 1 (q − 1)2
+
−n + n(n + 1) + n2 (q n − 1)
q−1
n2 q n
2(q n − 1) (1 − 2n)q n − 1
.
+
+
3
2
(q − 1)
(q − 1)
q−1
Therefore,
n
#
k=1
k 2 q k−1 =
2(q n − 1) (1 − 2n)q n − 1
n2 q n
.
+
+
3
2
(q − 1)
(q − 1)
q−1
248
Chapter 10. Solutions to Exercises and Problems
Solution 7.29. Observe that
∞
#
n=0
n(n − 1)xn =
∞
#
n=0
n2 xn −
∞
#
nxn .
n=0
Now, use Examples 7.3.3 and 7.3.4, to obtain
∞
#
n2 xn =
n=0
∞
#
n=0
n(n − 1)xn +
∞
#
nxn =
n=0
x
2x2
+
.
(1 − x)3
(1 − x)2
Solution 7.30. In order to prove (i), (ii) y (iii) use the fact that the series are
1
geometric series with ratios 43 , −1
3 and 3 , respectively. Then, their sums are equal
1
to 1−r , where r is the ratio.
∞ n
*
(iv) Use Example 7.3.3 to obtain
= 2.
n
n=1 2
(v) Use Exercise 7.29 to prove that
2
∞
1
#
2 12
8 4
n2
2
=
+
1 3
1 2 = + = 6.
n
2
2
2
n=1
2
2
10.8 Solutions to exercises of Chapter 8
Solution 8.1. The product of the coefficients of the polynomials is
4
2
2
7
1
1
8
4
8 4
32 16 56 8
2 7 1
14 2
8 8
48 25 57 8
Therefore, the product of the polynomials is R(x) = 8x5 + 8x4 + 48x3 + 25x2 +
57x + 8.
Evaluating in x = 2, we have P (2) = 4 · 23 + 2 · 22 + 7 · 2 + 1 = 55, Q(2) =
2
2 · 2 + 2 + 8 = 18 and R(2) = 8 · 25 + 8 · 24 + 48 · 23 + 25 · 22 + 57 · 2 + 8 = 990.
Solution 8.2. Each factor of P (x) is a geometric progression, then
P (x) = (1 − x + x2 − · · · + x100 )(1 + x + x2 + · · · + x100 )
(−x)101 − 1
x101 − 1
x101 + 1
=
−x − 1
x−1
x+1
2 101
(x ) − 1
= 1 + x2 + · · · + x200 .
=
x2 − 1
=
x101 − 1
x−1
249
10.8 Solutions of Chapter 8
Solution 8.3. Apply the division algorithm to obtain H(x) and R(x), then
x8 − 5x3 + 1 = (x5 − x4 + x3 − 2x2 + 3x − 9)(x3 + x2 + 1) + 11x2 − 3x + 10.
Hence H(x) = x5 − x4 + x3 − 2x2 + 3x − 9 and R(x) = 11x2 − 3x + 10.
Solution 8.4. Let P (x) = nxn+1 − (n + 1)xn + 1, evaluating in x = 1, P (1) =
n − (n + 1) + 1 = 0, that is, x = 1 is a zero of P (x), then x − 1 divides P (x). In
fact
P (x) = (x − 1)(nxn − xn−1 − · · · − x − 1).
If Q(x) = nxn − xn−1 − xn−2 − · · · − x − 1, then Q(1) = n − n = 0 implies
that x − 1 divides Q(x), hence (x − 1)2 divides P (x).
Solution 8.5. It is clear that P1 (x) = 1 + x has as the only root −1 and P2 (x) has
roots −1 and −2. By induction, we will see that the roots of Pn (x) are −1, −2,
. . . , −n. Suppose that Pn (x) has roots −1, −2, . . . , −n, then
Pn (x) =
(x + 1)(x + 2) · · · (x + n)
.
n!
Hence,
x(x + 1)(x + 2) · · · (x + n)
(n + 1)!
(x + 1)(x + 2) · · · (x + n) x(x + 1)(x + 2) · · · (x + n)
+
=
n!
(n + 1)!
(x + 1)(x + 2) · · · (x + n)(n + 1) + x(x + 1)(x + 2) · · · (x + n)
=
(n + 1)!
(x + 1)(x + 2) · · · (x + n)(x + n + 1)
,
=
(n + 1)!
Pn+1 (x) = Pn (x) +
which shows that the roots of Pn+1 (x) are −1, −2, . . . , −(n + 1).
Solution 8.6. Since P (0) = 0, then P (1) = P 2 (0) + P (0) + 1 = 1. Now, evaluate in
x = 1, the identity P (x2 +x+1) = P 2 (x)+P (x)+1 to obtain P (3) = 3. Evaluating
in x = 3, we get that P (32 + 3 + 1) = P 2 (3) + P (3) + 1 = 9 + 3 + 1 = 13, then
P (13) = 13.
Now, if we define xn+1 = x2n + xn + 1 and P (xn ) = xn , then P (xn+1 ) =
2
P (xn + xn + 1) = P 2 (xn ) + P (xn ) + 1 = x2n + xn + 1 = xn+1 . This process
constructs an infinite number of fixed points of P (x) which are different, since
xn+1 − xn = x2n + 1 > 0. But a polynomial cannot have an infinite number of fixed
points unless P (x) = x.
250
Chapter 10. Solutions to Exercises and Problems
Solution 8.7. Let P (x) = an xn + · · · + a1 x + a0 be a polynomial with a0 = 0. Since
n
n−1
1
1
1
1
n
n
+ an−1
+ · · · + a1
x P
an
= x
+ a0
x
x
x
x
= an + an−1 xn−1 + · · · + a1 xn−1 + a0 xn ,
we have xn P x1 = P (x) if and only if an + an−1 xn−1 + · · · + a1 xn−1 + a0 xn =
an xn + an−1 xn−1 + . . . + a1 x + a0 if and only if the “complementary coefficients”
are equal, that is, ai = an−i , for all i = 0, . . . , n.
Solution 8.8. Use the previous exercise to see that P (−1) = 0, then P (x) =
(x + 1)Q(x), for a polynomial Q(x) of degree n − 1. Since
(x + 1)Q(x) = P (x) = xn P
it follows that Q(x) = xn−1 Q
follows that Q(x) is reciprocal.
1
x
= xn
1
+1 Q
x
1
x
,
1
x . Hence, using again the previous exercise, it
Solution 8.9. A reciprocal
since an = a0 = 0,
polynomial does not have 0 as a root,
then a = 0. Since an P a1 = P (a) = 0, it follows that P a1 = 0, then a1 is also a
zero of P (x).
Solution 8.10. If n is odd, it follows that x2n−2 + x2n−4 + · · · + x4 + x2 + 1 can be
factored as (xn−1 + xn−2 + xn−3 + · · · + 1)(xn−1 − xn−2 + xn−3 − xn−4 + · · · + 1),
which proves that the polynomial in the left is divisible by 1 + x + · · · + xn−1 .
If n is even, −1 is a root of 1 + x + x2 + · · · + xn−1 , but it is not a root
of 1 + x2 + x4 + · · · + x2n−2 . Then 1 + x2 + x4 + · · · + x2n−2 is not divisible by
1 + x + x2 + · · · + xn−1 .
Solution 8.11. Suppose that n ≥ m. Recall that Euclid’s algorithm is used to find
the greatest common divisor of (m, n), in the following way:
n = ms1 + r1
m = r1 s2 + r2
..
..
.
.
rj−1 = rj sj+1 + 0,
with 0 ≤ ri < ri−1 and r0 = m, hence (m, n) = rj .
Notice that
xn − 1 = xms1 +r1 − 1
= xr1 (xms1 − 1) + xr1 − 1
= xr1
xms1 − 1
xm − 1
(xm − 1) + xr1 − 1,
251
10.8 Solutions of Chapter 8
then the division of xn −1 by xm −1, leaves remainder xr1 −1, then (xn −1, xm −1) =
(xm − 1, xr1 − 1). Proceeding in the same way, we obtain (xn − 1, xm − 1) =
(xm − 1, xr1 − 1) = (xr1 − 1, xr2 − 1) = · · · = (xrj−1 − 1, xrj − 1) = xrj − 1.
Therefore, (xn − 1, xm − 1) = x(n,m) − 1.
Solution 8.12. The problem is equivalent to finding all pairs (m, n) such that
(xmn+n − 1)(x − 1)
(xm+1 − 1)(xn − 1)
is a polynomial. Notice that xn − 1 and xm+1 − 1 are divisors of xmn+n − 1. These
factors can only have x − 1 as a common factor, but by the previous exercise,
(xm+1 −1, xn −1) = x(m+1,n) −1, therefore it will be enough to have (m+1, n) = 1.
Solution 8.13. If a is an integer with P (a) = 0, then P (x) = (x − a)Q(x), for
some polynomial Q(x) with integer coefficients, and P (0) = −aQ(0) and P (1) =
(1 − a)Q(1). But if a is an integer, then either a or 1 − a is even, and so one of
P (0) and P (1) is even, which is a contradiction.
Solution 8.14. Consider the polynomial with roots x, y, z, that is, P (u) = (u −
x)(u − y)(u − z) = u3 + au2 + bu + c. Then, by Vieta’s formulas,
xyz
, c = −xyz,
a = −x − y − z = −w, b = xy + yz + zx =
w
therefore, b = − wc = ac . Hence, P (u) = u3 + au2 + bu + ab = (u + a)(u2 + b), and
u = −a = w is a root. Without loss of generality, we can assume that it is x, that
is, x + y + z = x, from where y + z = 0. This last equality implies that y = −z
and, then b = −y 2 , hence the other roots are y and −y. That is, the roots are
(x, y, −y) and therefore the solutions of the system are the triplets (x, y, −y) and
its permutations, with x, y ∈ R.
Solution 8.15. Since the coefficients of the polynomial are integers, it is enough
to see that it is irreducible over Z[x]. The polynomial has no integer roots, since
a root of x4 − x3 − 3x2 + 5x + 1 must divide 1, and then it must be 1 or −1, but
P (1) = 3 and P (−1) = −5. Or, by Exercise 8.13, and because P (0) and P (1) are
odd, P (x) = 0 has no integer solutions.
Hence, if P (x) can be factored, it must be into two monic quadratic polynomials, as follows: x4 − x3 − 3x2 + 5x + 1 = (x2 + bx + c)(x2 + dx + e), with b, c,
d, e integers.
Equating coefficients, it follows that:
b + d = −1
c + e + bd = −3
be + cd = 5
ce = 1.
252
Chapter 10. Solutions to Exercises and Problems
The last identity implies c = e = 1 or c = e = −1. Now, the third identity takes
the form b + d = 5 or b + d = −5. In any case, these equalities are in contradiction
with the first identity.
Second Solution. It is enough to see that the polynomial is irreducible over Z2 [x].
But in Z2 [x], the polynomial can be written as x4 + x3 + x2 + x + 1. It is clear
that 0 and 1 are not roots of this polynomial, then if it is reducible it should be
decomposed as the product of two irreducible quadratic polynomials. But the only
irreducible quadratic polynomial in Z2 [x] is x2 +x+1, and (x2 +x+1)(x2 +x+1) =
x4 + x2 + 1 = x4 + x3 + x2 + x + 1. This completes the proof.
Solution 8.16. Consider the integers x0 = n and xk+1 = P (xk ), for k ≥ 0. If
k = 1, the result is immediate. Suppose now xk = x0 , with k ≥ 2, and define di =
xi+1 − xi . Since di = xi+1 − xi divides P (xi+1 ) − P (xi ) = xi+2 − xi+1 = di+1 , for
all i = 0, 1, . . . , k − 1, and since dk = d0 = 0, it follows that |d0 | = |d1 | = · · · = |dk |.
Suppose that d0 = d1 . In this case d2 = d0 , otherwise x3 − x2 = −(x2 − x1 ),
then x3 = x1 and the sequence of iterates takes the form x0 , x1 , x2 , x1 , x2 , . . . .
Hence it should not exist xk , with k ≥ 2, that coincides with x0 . Similarly, if
dj = d0 , for every j, then xj = x0 + jd0 = x0 , for every j, which is a contradiction.
Hence d0 = −d1 , that is, x1 − x0 = −(x2 − x1 ). Therefore x0 = x2 .
Solution 8.17. Suppose that an = 1. If r1 , . . . , rn are the roots, by Vieta’s formulas
it follows that r12 r22 · · · rn2 = 1 and
#
r12 + r22 + · · · + rn2 = (r1 + · · · + rn )2 − 2
ri rj = a2n−1 − 2an−2 ≤ 3.
1≤i<j≤n
The inequality between the arithmetic and the geometric mean guarantees that
r12 + r22 + · · · + rn2 ≥ n n r12 r22 · · · rn2 = n.
Both inequalities imply that n ≤ 3.
In the case n = 3, from the eight polynomials of the form x3 ± x2 ± x ± 1,
the only ones that have three roots are x3 − x ± (x2 − 1) = (x2 − 1)(x ± 1). In the
case n = 2, only the polynomials x2 ± x − 1 have its two roots real. In the case
n = 1, the only polynomials are x ± 1.
Solution 8.18. If P (a) = b, P (b) = c and P (c) = a, then P (P (P (a))) = P 3 (a) = a,
by Exercise 8.16, and P 2 (a) = a. But on the other hand, P (P (a)) = P (b) = c,
then c = a, which is a contradiction.
Solution 8.19. Suppose that r1 , . . . , rn are all the roots of P (x) and that all are
real; by Vieta’s formulas it follows that
n
#
i=1
ri = −2n and
#
1≤i<j≤n
ri rj = 2n2 .
(10.11)
253
10.8 Solutions of Chapter 8
*
2
On the other hand, the Cauchy–Schwarz inequality guarantees that ( ni=1 ri ) ≤
*
*
*n 2 *n
2
n
n
−1
2
2
i=1 ri ) . Hence
i=1 1 , and then −
i=1 ri ≤ n (
i=1 ri
#
1≤i<j≤n
1
ri rj =
2
≤
n
#
ri
i=1
1
1
−
2 2n
2
n
1# 2
r
2 i=1 i
n 2
#
n−1
=
ri
(−2n)2 = 2n(n − 1),
2n
i=1
−
contradicting equation (10.11).
Solution 8.20. Calculate the derivative of the polynomial P (x), that is, P ′ (x) =
3x2 − 2x − 8, which has roots x = 2 and x = − 43 . Since P (2) = 0, then 2 is a
multiple root.
Solution 8.21. The fact that P (x) is divisible by (x + 1)2 is equivalent to the
fact that −1 must be a root of multiplicity at least 2 of P (x), that is, P (−1) =
P ′ (−1) = 0. This is equivalent to
−1 + a − b + c = 0
3 − 2a + b = 0.
Then, the solution’s triplets are (a, b, c) = (t, 2t − 3, t − 2), with t ∈ R.
Solution 8.22. The polynomial of the Lagrange interpolation formula is
P (x) =
=
n
#
k=0
n
#
(−1)n−k 2k
x(x − 1)(x − 2) · · · (x − k + 1)(x − k − 1) · · · (x − n)
k(k − 1) · · · (k − k + 1)(k − k − 1) · · · (k − n)
(−1)n−k 2k
x(x − 1)(x − 2) · · · (x − k + 1)(x − k − 1) · · · (x − n)
.
k! (n − k)!
k=0
Then,
P (n + 1)
n
#
(n + 1)n · · · (n + 1 − (k − 1))(n + 1 − k − 1) · · · (n + 1 − n)
=
(−1)n−k 2k
k! (n − k)!
k=0
=
=
n
#
k=0
n
#
k=0
n
(−1)n−k 2k
#
(n + 1)!
(n + 1)!
=
(−1)n−k 2k
k! (n − k)!(n + 1 − k)
k! (n + 1 − k)!
k=0
(−1)n−k 2k
n+1
.
k
254
Chapter 10. Solutions to Exercises and Problems
On the other hand,
1 = (2 − 1)n+1 =
n+1
#
k=0
n+1
k
2k (−1)n+1−k
= −P (n + 1) + 2n+1 .
Therefore, P (n + 1) = 2n+1 − 1.
√ √
Solution 8.23. (i) First observe that x ∈ [− 5, 5], since the left-hand side is
non-negative. Squaring both sides of the equation and rearranging, we get
x4 − 10x2 + x + 20 = 0,
which can be factorized as
(x2 + x − 5)(x2 − x − 4) = 0.
Then, x2 + x − 5 =
x2 −x − 4 = 0. The solutions of these
are, for
√
0 or √
equations
1
the first x1,2 = 2 −1 ± 21 and for the second x3,4 = 12 1 ± 17 . Only two
√ √
of them
are in
the
interval
[− √5, 5], and therefore the solutions of the equation
√
are 12 −1 + 21 and 12 1 − 17 .
(ii) As in the previous part, squaring both sides of the equation, it follows that
x4 − 2ax2 + x + a2 − a = 0,
which is a quadratic equation in a,
a2 − (2x2 + 1)a + x4 + x = 0.
The discriminant of the quadratic equation is (2x − 1)2 , so that the roots of the
equation are a1 = x2 + x and a2 = x2 − x + 1. It follows that,
a2 − (2x2 + 1)a + x4 + x = (a − x2 − x)(a − x2 + x − 1) = 0.
2
Now, solving the quadratic equations x2 + x −
√ a = 0 and x − x + 1 − a = 0,
√
1±
1+4(a−1)
give us the roots x = −1± 2 1+4a and x =
. We can show that the
√2
√
√ √
−1− 1+4a
−1+ 1+4a
root
is not. On the other hand,
2 √ is always in [− a, a] while
2
√ √
1+ 1+4(a−1)
3
would be in [− a, a], if 4 ≤ a ≤ 1 and the root x =
the root
2
√
√ √
1− 1+4(a−1)
belongs to [− a, a] only when a ≥ 1.
2
√
Solution 8.24. It is clear that x ≥ 0, a + x ≥ 0 and a − a + x ≥ 0. Squaring both
sides of the equation leads to
√
√
a + x = a − x2 .
x2 = a − a + x which is equivalent to
255
10.8 Solutions of Chapter 8
Then, a − x2 ≥ 0, and after squaring, it follows that
a + x = (a − x2 )2 ,
which is equivalent to the equation P (x) = x4 − 2ax2 − x + a2 − a = 0. In order
to solve this equation, consider a as the variable and x as the parameter, then
a2 − (2x2 + 1)a + x4 − x = 0.
The discriminant of the equation is (2x2 + 1)2 − 4(x4 − x) = (2x + 1)2 , then the
roots of the equation are a1 = x2 − x and a2 = x2 + x + 1. This implies that we
can factorize P (x) as
P (x) = (x2 − x − a)(x2 + x + 1 − a).
The positive roots of the equation x2 − x − a = 0 do not satisfy the condition
√
a − x2 ≥ 0, since a − x2 = −x. The roots of x2 + x + 1 − a = 0 are x1 = −1+ 24a−3
√
and x2 = −1− 24a−3 . Only x1 can be non-negative, and this happens when a ≥ 1.
Solution 8.25. First, observe that x is a positive number. Taking conjugates on
the left-hand side of the equation leads to
√
√
2 x
m x
√
√ =
√ ,
x+ x+ x− x
x+ x
√
then m is also a positive number. Dividing by x and simplifying the equation,
√
√
gives us (2 − m) x +
√ x = m x√− x. Then 2 − m ≥ 0, and hence,
√ after
squaring, (2−m)2 (x+ x) = m2 (x− x) which is equivalent to (2−m)2 ( x+1) =
√
√
√
2
−2m+2
m2 ( x − 1). Solving for x, we get x = m2(m−1)
, hence m > 1 and the solution
to the equation is x =
(m2 −2m+2)2
4(m−1)2 ,
for 1 < m ≤ 2.
Solution 8.26. From the first equation, we get
x=
y + 76 −
y + 11 which is equivalent to x(
y + 76 +
y + 11) = 65.
Then,
65
√
x= √
.
y + 76 + y + 11
65√
65√
and z = √x+76+
. Without loss of
Similarly, we can obtain y = √z+76+
z+11
x+11
generality,
we
can
assume
that
x
≤
y
≤
z.
From
these
inequalities
and since the
√
function x is increasing, it follows that
√
√
√
√
x + 76 + x + 11 ≤ y + 76 + y + 11 ≤ z + 76 + z + 11.
256
Chapter 10. Solutions to Exercises and Problems
Then, taking inverses and multiplying by 65, it follows that y ≤ x ≤ z. Hence,
x = y. From this equality, we obtain
65
65
√
√
= √
= z.
x= √
y + 76 + y + 11
x + 76 + x + 11
Thus x = y = z.
Then, in order to find the triplets, we need to find the solutions to the
equation
√
√
x x + 76 + x + 11 = 65,
(10.12)
√
√
where x is a positive real number. Since the function x x + 76 + x + 11 is
monotone increasing for positive real numbers, there is at most one solution to
the equation (10.12). Since 5 is a solution, the only triplet that solves the system
is (5, 5, 5).
Solution 8.27. Consider a, b and c as the unknowns, then we have a system of linear
equations. Multiplying the first equation by x and then by y, next substituting cx
in the second equation and cy in the last equation, we obtain
1
1
−
xz
y
1
1
a(xy − z) − b(y 2 + 1) = − − .
x yz
a(x2 + 1) − b(xy + z) =
(10.13)
(10.14)
Now, multiply equation (10.13) by y 2 + 1 and equation (10.14) by −(xy + z) and,
add them. It follows that
a(x2 + y 2 + z 2 + 1) =
x2 + y 2 + z 2 + 1
,
xz
1
1
1
hence a = xz
. Similarly, we obtain b = yz
and c = xy
. Since a, b, c are positive, x,
1
1
y, z must have the same sign. Therefore, abc = (xyz)2 , hence xyz = ± √abc
. That
is, the solutions to the system are
√
b
a
c
,√
,√
abc
abc
abc
and
a
c
b
, −√
, −√
−√
abc
abc
abc
.
Solution 8.28. Let Q(x) = xk + a1 xk−1 + a2 xk−2 + · · · + ak−1 x + ak and P (x) =
x2 + px + q. Consider the equality
(xk + a1 xk−1 + · · · + ak )2 + p(xk + a1 xk−1 + · · · + ak ) + q
−(x2 + px + q)k − a1 (x2 + px + q)k−1 − · · · − ak = 0.
Calculating the coefficients of x2k , x2k−1 , . . . , x1 , x0 , we get a system of equations
for a1 , a2 , . . . , ak , p, q. It is quite difficult to write this system, and to solve it
even more. However, some useful observations can be made without solving the
system.
257
10.8 Solutions of Chapter 8
When the polynomials are expanded, notice that the coefficients b1 , b2 , . . . ,
bk of the powers of x2k−1 , x2k−2 , . . . , xk , are
b1 = 2a1 + R1 (p, q) = 0,
b2 = 2a2 + R2 (p, q, a1 ) = 0,
..
..
.
.
bk = 2ak + Rk (p, q, a1 , . . . , ak−1 ) = 0,
where each Ri is some algebraic expression in terms of p, q, a1 , . . . , ak−1 . The first
of these equations implies that a1 can be expressed in terms of p and q; the second
equation implies that a2 can be expressed in terms of p, q and a1 , and therefore,
only in terms of p and q. Similarly, we can conclude that all coefficients of the
polynomial Q(x) that commute with P (x) can be expressed in a unique way in
terms of p and q, which is what we wanted to prove.
Solution 8.29. First, let us see that the polynomial Q(x) = P (P (x)) commutes
with P (x). In fact, Q(P (x)) = P (P (P (x))) = P (Q(x)). This polynomial Q(x)
has degree 4 and, by the previous exercise, it is the only polynomial of degree 4
that commutes with P (x). Similarly, it can be shown that the only polynomial of
degree 8 that commutes with P (x) is R(x) = P (P (P (x))).
Solution 8.30. Let S(x) = Q(R(x)) and T (x) = R(Q(x)). Since P (x) commutes
with both Q(x) and R(x), it follows that P (S(x)) = P (Q(R(x))) = Q(P (R(x))) =
Q(R(P (x))) = S(P (x)). Therefore, P (x) commutes with S(x). Similarly, it can be
shown that P (x) commutes with T (x). Since S(x) and T (x) are monic polynomials
of the same degree (if Q(x) and R(x) have degrees k and l, respectively, then S(x)
and T (x) have degrees kl), by Exercise 8.28, it follows that S(x) = T (x), that is,
Q(R(x)) = R(Q(x)).
Solution 8.31. Let P (x) = ax + b and Q(x) = cx + d. The condition P (Q(x)) =
Q(P (x)) implies that acx + ad + b = acx + bc + d, that is, d(a − 1) = b(c − 1), and
from here we proceed by cases.
First, if a = 1, then b(c − 1) = 0, and b = 0 or c = 1. If b = 0, it follows that
P (x) = x and Q(x) = cx + d commute and they have the common fixed point
d
. Now, if c = 1, then P (x) = x + b and Q(x) = x + d, which clearly commute.
− c−1
Second if a = 1, then d =
b(c−1)
a−1
and we have the polynomials, P (x) = ax + b,
b
and Q(x) = cx + b(c−1)
with fixed point
a−1 with fixed point − a−1 , if c = 1.
If c = 1, then d = 0, hence P (x) = ax + b and Q(x) = x, a case that was already
considered.
b
,
− a−1
It is clear that if P (x) = x + α and Q(x) = x + β, then they commute. Now,
if there is x0 such that P (x0 ) = Q(x0 ) = x0 , then P (x) = a(x − x0 ) + x0 and
Q(x) = b(x − x0 ) + x0 . A direct calculation proves that the last two polynomials
commute, which is the end of the proof.
258
Chapter 10. Solutions to Exercises and Problems
Solution 8.32. Notice that
Pa (Qa (x)) = Pa (Q(x − a) + a) = P (Q(x − a) + a − a) + a = P (Q(x − a)) + a.
Similarly, Qa (Pa (x)) = Q(P (x − a)) + a. Then, Pa (x) and Qa (x) commute if and
only if P (Q(x − a)) = Q(P (x − a)), which is the case.
Solution 8.33. The system can be rewritten as
x5 + y 5 = σ15 − 5σ13 σ2 + 5σ1 σ22 = 33
σ1 = 3.
Substituting the value of σ1 in the last equation, we obtain the equation
15σ22 − 5 · 27σ2 + 9 · 27 = 33.
Simplifying the equation, we get σ22 − 9σ2 + 14 = 0.
The solutions are σ2 = 2 and σ2 = 7. Now, in order to obtain the values that
we are looking for, we need to solve, for σ2 = 2 and σ2 = 7, the system
x + y = 3 and xy = σ2 .
√
√
Solution 8.34. Define y = 4 x and z = 4 97 − x, then the equation can be rewritten
as y + z = 5. We have to solve the system of equations
y+z =5
4
y + z 4 = 97 − x + x = 97.
Now, y 4 + z 4 = σ14 − 4σ12 σ2 + 2σ22 = 97. Substituting the value of σ1 , we have to
solve the quadratic equation 2σ22 − 100σ2 + 54 = 97. The solutions are σ2 = 44
and σ2 = 6.
To obtain the values of y and z, we must find the solutions to the systems of
equations
y+z =5
y+z =5
yz = 44,
yz = 6.
Solution 8.35. We have
x3 + y 3 = σ13 − 3σ1 σ2 = c.
(10.15)
From the second equation, we have σ12 − 2σ2 = b. Solving for σ2 , we get σ2 =
since σ1 = a. Substituting in equation (10.15)
a3 − 3a
a2 − b
2
=c
a3 − 3ab + 2c = 0.
a2 −b
2 ,
259
10.8 Solutions of Chapter 8
Solution 8.36. Substituting the value of z 2 in the second equation, it follows that
x2 + y 2 + xy = b2 .
From the first equation, we get
x + y − a = −z.
(10.16)
Squaring both sides leads to
(x + y)2 − 2a(x + y) + a2 = z 2
(x + y)2 − 2a(x + y) + a2 = xy
2
2
(10.17)
2
x + xy + y − 2a(x + y) = −a .
Since x2 + y 2 + z 2 = x2 + y 2 + xy = b2 , substituting in equation (10.17) and
2
+b2
. Substituting this value in (10.16), we
solving for x + y, results in x + y = a 2a
2
2
2
2 2
2
2
−b )
+b
−b
, that is, xy = (a 4a
. Observe, since z and a
and x + y = a 2a
get z = a 2a
2
2
2
are positive, that a > b . By Vieta’s formulas, x and y are roots of the equation
w2 −
a2 + b 2
2a
w+
(a2 − b2 )2
= 0.
4a2
(10.18)
But the solutions of equation (10.18) are
a2 +b2
2a
w1,2 =
±
a2 +b2 2
2a
−4
$
(a2 −b2 )2
4a2
%
.
2
Since we are looking for real solutions, the discriminant must be positive, that is
∆=
a2 + b 2
2a
2
−
1
4(a2 − b2 )2
= 2 (3a2 − b2 )(3b2 − a2 ) > 0.
4a2
4a
Since 3a2 > b2 , then 3b2 − a2 > 0. Hence, 3b2 > a2 > b2 , which means |b| < a <
√
3|b|.
Solution 8.37. Take σ1 = x + y + z, σ2 = xy + yz + zx and σ3 = xyz. Then
1 2
(a − b2 ), σ3 =
2
where the third equation was obtained from the
tion (4.8).
Solve now,
σ1 = a,
σ2 =
1
a(a2 − b2 ),
2
given factorization in equa-
1
1
u3 − au2 + (a2 − b2 )u − a(a2 − b2 ) = 0
2
2
+
,
1 2
2
2
(u − a) u + (a − b ) = 0.
2
2
2
b2 −a2
Its solutions are u1 = a, u2 = b −a
and
u
=
−
3
2
2 .
260
Chapter 10. Solutions to Exercises and Problems
Solution 8.38. The equation is equivalent to xy + x2 y 2 = 3xy 2 + 3x2 y, which is
equivalent to xy(1 + xy − 3x − 3y) = 0. If (x, y) = (m, 0) or (x, y) = (0, m), then
they are solutions for any integer m.
If xy = 0, 1 + xy − 3x − 3y = 0. Since x = 3 or y = 3 does not satisfy the equation,
then we can divide by x − 3 or y − 3. Solving for y,
y=
8
3x − 1
=3+
,
x−3
x−3
so that y is an integer if x − 3 divides 8. Then, x − 3 = ±1, x − 3 = ±2, x − 3 = ±4
and x − 3 = ±8. Therefore, x ∈ {−5, −1, 1, 2, 4, 5, 7, 11}, that is, the solutions are
(m, 0), (0, m), for any integer m, and
(x, y) ∈ {(−5, 2), (2, −5), (−1, 1), (1, −1), (4, 11), (11, 4), (5, 7), (7, 5)}.
Solution 8.39. Let k be a fixed number and consider the solutions of the equation
(a, b), with a, b ∈ N, a ≥ b, and from this set of solutions choose the one for which
a + b is minimum.
2
and, since a is a
If we can show that a = b, then 1 + a1 = k2 , hence a = k−2
positive integer, k = 3 or 4.
Let us see that b = a in the following way. Suppose a > b and notice that a is a
b+1
solution of the equation x+1
b + x = k, or equivalently, it is a root of the quadratic
2
2
equation x − (kb − 1)x + b + b = 0. If a1 is the other root, by Vieta’s formulas, it
2
follows that a+a1 = kb−1 and aa1 = b2 +b. Hence, a1 +b = b a+b +b. Since a > b, it
2
follows that a ≥ b+1 so that b ≥ b a+b = a1 . Therefore, the pair (b, a1 ) is a solution
of the original equation. Since (a, b) is the solution with minimum sum a + b, it
2
follows that a+b ≤ b+a1 = b+ b a+b so that b2 +b ≥ a2 ≥ (b+1)2 = b2 +2b+1, which
implies b + 1 ≤ 0. But this contradicts the fact that b is positive, therefore a = b.
Solution 8.40. Consider the case k > 3. Suppose that the integers a, b, c satisfy
equation a2 + b2 + c2 = kabc; then at least one of them is positive, and the other
two are both positive or both negative. In the negative case, we can change the
sign to both numbers to obtain a solution where all are positive. Then, without
loss of generality, suppose that all three numbers are positive.
Now, let us prove that the three numbers are distinct. Suppose that the
previous statement is false, for instance a = b. Then 2a2 + c2 = ka2 c, that is,
c2 = a2 (kc − 2). Therefore, kc − 2 is a perfect square and then there is an integer
number d ≥ 1 such that kc = 2 + d2 . Substituting the value of kc in the equation
2a2 + c2 = ka2 c, it follows that c2 = d2 a2 or c = da. Now, d2 = kc − 2 = k(da) − 2,
hence 2 = d(ka − d) and this implies d divides 2, that is, d = 1 or 2. In both cases
ka = 3, contradicting the fact that k > 3.
Then, suppose that a > b > c ≥ 1. The triplet (kbc − a, b, c) is also a
solution of the equation x2 + y 2 + z 2 = kxyz, with kbc − a a positive integer, since
a(kbc − a) = b2 + c2 and a > 0.
10.9 Solutions of Chapter 9
261
Consider now the polynomial P (x) = x2 − (kbc)x + b2 + c2 . The roots of this
polynomial are a and kbc − a, moreover
P (b) = 2b2 + c2 − kb2 c ≤ 2b2 + c2 − kb2 < 3b2 − kb2 = (3 − k)b2 < 0.
The interval where P (x) is negative is the interval with end points the roots
of the polynomial. Then b is between a and kbc − a, and since b < a, it follows
that kbc − a < b. Hence
max (kbc − a, b, c) = b < a = max (a, b, c).
Repeating this construction, we obtain a decreasing sequence of positive integers,
which is something impossible, then there are no solutions of the original equation
for k > 3.
Now, let k = 2. Suppose that a2 + b2 + c2 = 2abc, where a, b, c are integers.
Since a2 + b2 + c2 is even, not all numbers a, b, c are odd. If exactly one of them
is even, reducing modulo 4, we get 2 ≡ 0 (mod 4), a contradiction. Therefore, the
three numbers are even, and they can be written as a = 2a′ , b = 2b′ and c = 2c′ , so
that a′2 +b′2 +c′2 = 4a′ b′ c′ . The last equation is the case k = 4, which has no integer
solutions except (0, 0, 0), and from this we get (a, b, c) = (2a′ , 2b′ , 2c′ ) = (0, 0, 0).
For k = 3, a solution is (1, 1, 1) and, for k = 1, consider multiples of 3 to
reduce it to the case k = 3.
10.9 Solutions to problems of Chapter 9
√
Solution 9.1. Note that b = a2 +2a+1 = (a+1)2 ∈ Q and b > 0, then a = −1± b.
On the other hand,
√
√
a3 − 6a = (−1 ± b)3 − 6(−1 ± b)
√
√
√
= −1 ± 3 b − 3b ± b b + 6 ∓ 6 b
√
= 5 − 3b ± (b − 3) b.
√
Since
a3 − 6a and 5 − 3b are√rational numbers, we have (b − 3) b ∈ Q. If b = 3,
√
b ∈ Q and,
√ then a = −1 ± b ∈ Q is a contradiction. Therefore b = 3 and then
a = −1 ± 3.
√
Moreover, it is clear that if a = −1± 3, the numbers a2 +2a = 2 and a3 −6a = −4
are rational.
Solution 9.2. First, we make the following substitution to simplify the notation,
a = 3 pq 2 , b = 3 qr2 and c = 3 rp2 . We have to prove that a1 + 1b + 1c = ab+bc+ca
abc
262
Chapter 10. Solutions to Exercises and Problems
is a rational number. Since abc = pqr is a rational number, it is enough to prove
that ab + bc + ca is a rational number. Notice that
(a + b + c)3 = a3 + b3 + c3 + 3(ab + bc + ca)(a + b + c) − 3abc.
(10.19)
Since a + b + c is a rational number, so is (a + b + c)3 and clearly a3 = pq 2 ,
b3 = qr2 and c3 = rp2 are rational numbers. It is now straightforward, from
equation (10.19) and since a + b + c = 0, that ab + bc + ca is a rational number.
Solution 9.3. Since a = a(2 − a) − a(1 − a), we have that the numbers a(2 − a)
and a(1 − a) cannot be both rational numbers, then one of them is an irrational
number; this one will define the number b, that is, b is 2 − a or 1 − a. Notice that
a + (2 − a) = 2 and a + (1 − a) = 1 are rational numbers, and a(2 − a) or a(1 − a)
is an irrational number (if −a(1 − a) is an irrational number, then also a(1 − a) is
an irrational number).
Similarly, since a1 = a + a2 − a + a1 , the numbers a + a2 and a + a1 cannot
be both rational numbers; then one of them is an irrational number and b′ is
one of the numbers a2 or a1 . Notice that, ab′ = 1 or 2 is a rational number and
a + b′ = a + a2 or a + a1 is an irrational number.
Solution 9.4. For each i, there exist among 1, 2, . . . , m, ⌊m/xi ⌋ multiples of xi .
None of them is a multiple of xj for j = i, since the least common multiple of
xi and xj is greater than m. Then, there exist ⌊m/x1 ⌋ + ⌊m/x2 ⌋ + · · · + ⌊m/xn ⌋
different numbers in {1, 2, . . . , m}, and these numbers are divisible by some of the
numbers x1 , x2 , . . . , xn . None of these last numbers can be 1 (unless n = 1, and
in this case, the result is immediate). Therefore,
m
m
m
+
+ ··· +
≤ m − 1.
x1
x2
xn
Since
m
xi
< ⌊ xmi ⌋ + 1 for each i, we have
m
1
1
1
+
+ ···+
x1
x2
xn
< m + n − 1.
If we prove that n ≤ (m + 1)/2, then
1
3
1
1
n−1
< .
+
+ ···+
<1+
x1
x2
xn
m
2
To prove that n ≤ (m + 1)/2, observe that the largest odd divisors of x1 ,
x2 , . . . , xn are all different, because otherwise, if two numbers have the same
greatest odd divisor, one of them will be a multiple of the other, which will be a
contradiction to the hypothesis. Therefore, n is less than or equal to the quantity
of odd numbers among 1, 2, . . . , m, and the inequality follows.
263
10.9 Solutions of Chapter 9
√
Solution 9.5. Let m = ⌊ n⌋ and a = n − m2 . We have m ≥ 1, since n ≥ 1. Now,
from n2 + 1 = (m2 + a)2 + 1 ≡ (a − 2)2 + 1 mod (m2 + 2), it follows that the
condition of the problem is equivalent to the fact that (a − 2)2 + 1 is divisible by
m2 + 2. Since
0 < (a − 2)2 + 1 ≤ max{22 , (2m − 2)2 } + 1 ≤ 4m2 + 1 < 4(m2 + 2),
then (a − 2)2 + 1 = k(m2 + 2), for k = 1, 2 or 3. Now, we prove that none of these
cases occur.
Case 1. When k = 1, we have (a − 2)2 − m2 = 1, and this implies that a − 2 = ±1
and m = 0, but this contradicts the fact that m ≥ 1.
Case 2. When k = 2, we have (a − 2)2 + 1 = 2(m2 + 2), but any perfect square
is congruent to 0, 1, 4 modulo 8, and therefore (a − 2)2 + 1 ≡ 1, 2, 5 mod 8,
meanwhile 2(m2 + 2) ≡ 4, 6 mod 8, then this case does not occur.
Case 3. When k = 3, we have (a − 2)2 + 1 = 3(m2 + 2). Since any perfect square
is congruent to 0, 1 modulo 3, we have (a − 2)2 + 1 ≡ 1, 2 mod 3, meanwhile
3(m2 + 2) ≡ 0 mod 3, then this case is also impossible.
Solution 9.6. It is easy to prove that when a = 0 or b = 0 or a = b or a and b are
both integers, the identity follows.
Suppose now that a, b do not accomplish any of the above conditions. We
a
have, for n = 1, that ab = ⌊a⌋
⌊b⌋ , then b is a rational number different from zero.
p
a
Suppose that b = q , with (p, q) = 1.
If p is different from 1 and −1, then p divides ⌊na⌋, for all n, in particular it
divides ⌊a⌋, therefore a = kp + ǫ, for some k ∈ N and 0 ≤ ǫ < 1. Notice that ǫ = 0,
otherwise a = kp and b = kq = ⌊b⌋ are integers.
Then, since there exists n ∈ N, with 1 ≤ nǫ < 2, we have that ⌊na⌋ =
⌊knp + nǫ⌋ = knp + 1 is not divisible by p, which is a contradiction.
Similarly, it cannot happen for q to be different from 1 and −1. Therefore p,
q ∈ {±1}, but since a = b, we have that b = −a, then ⌊−a⌋ = − ⌊a⌋, which is only
possible if a is an integer number. Therefore, there are no more pairs of numbers
(a, b) that satisfy the conditions the problem.
Solution 9.7. As we proved in Example 1.3.3, we have
n 2n 3n
(m − 1)n
(m − 1)(n − 1)
+
,
+
+ ··· +
=
m
m
m
m
2
and we want to find the value of the sum
4
n 2n 4 3n 4
(m − 1)n
+
+
+ ···+
= X.
m
m
m
m
Adding both equations, term by term, we get
2n 3n
(m − 1)n
(m − 1)(n − 1)
n
+
+
+ ···+
=X+
,
m
m
m
m
2
(10.20)
264
Chapter 10. Solutions to Exercises and Problems
n
on the left-hand side of the equation (10.20) and using Gauss addifactorizing m
tion formula, we have that the sum on the left-hand side is
(m − 1)m
2
n
n
(1 + 2 + · · · + (m − 1)) =
m
m
=
n(m − 1)
.
2
Replacing this value in equation (10.20) and solving for X, we have
X=
n(m − 1) (m − 1)(n − 1)
m−1
(m − 1)
−
=
(n − n + 1) =
.
2
2
2
2
Solution 9.8. Since (a + b + c)3 = a3 + b3 + c3 + 3(a + b)(b + c)(c + a), it follows
that
1
1
√
+1
+ √
3
3
a
b
3
=
1
1
√
+√
3
3
a
b
1 1
+ +1+3
a b
1
√
+1
3
b
1
1+ √
3
a
= 10 + 3(18) = 64,
therefore
1
√
3
a
+
1
√
3
b
= 3, then the solutions are (a, b) =
1
8, 1
1
, 1, 8 .
Solution 9.9. From the first identity, we get a − b = 1c − 1b = b−c
bc . Similarly,
a−b
a−b
c−a
b − c = ca and c − a = ab . Therefore a − b = (abc)2 and, since a and b are
different, we have (abc)2 = 1, so that abc = ±1. It is clear that the numbers
(a, b, c) = (1, − 12 , −2) and (a, b, c) = (−1, 12 , 2) satisfy the identities and, with
these triplets, we obtain the two possible values of abc.
1
Solution 9.10. Since (a+ b)(a+ c) = a(a+ b + c)+ bc = bc
+ bc ≥ 2 and the equality
holds when bc = 1,√
it follows that the minimum value is 2, and it is reached when
b = c = 1 and a = 2 − 1.
Solution 9.11. The equation is equivalent to (bc+a)(ca+b)(ab+c) = (ab+bc+ca−
abc)2 . We expand the equation and cancel out terms to obtain abc(a2 +b2 +c2 +1) =
2abc(a + b + c) − 2abc(ab + bc + ca), which is equivalent to (a + b + c − 1)2 = 0.
Therefore, a + b + c = 1.
Solution 9.12. First prove that
b−c c−a a−b
+
+
a
b
c
=−
(b − c)(c − a)(a − b)
abc
and do the same for
a
b
c
+
+
b−c c−a a−b
= −9
Therefore, the value we are looking for is 9.
abc
.
(b − c)(c − a)(a − b)
265
10.9 Solutions of Chapter 9
Solution 9.13. Notice that
a
1
1
b
c
1
+
+
+
+
b−c c−a a−b
b−c c−a a−b
a
b
c
=
+
+
(b − c)2
(c − a)2
(a − b)2
a+b
b+c
c+a
+
+
+
(b − c)(c − a) (c − a)(a − b) (a − b)(b − c)
0=
and that
a+b
b+c
c+a
+
+
= 0.
(b − c)(c − a) (c − a)(a − b) (a − b)(b − c)
Hence, the value we are looking for is 0.
Solution 9.14. Notice that a2 + 1 = a2 + ab + bc + ca = (a + b)(a + c). Similarly,
b2 + 1 = (b + c)(b + a) and c2 + 1 = (c + a)(c + b). Then (a2 + 1)(b2 + 1)(c2 + 1) =
((a + b)(b + c)(c + a))2 .
Solution 9.15. The condition a1 + 1b + 1c = 0, implies that abc = 0 and ab+bc+ca = 0.
Since (a+b+c)2 = a2 +b2 +c2 +2(ab+bc+ca), it is clear that a2 +b2 +c2 = (a+b+c)2 .
Solution 9.16. Notice that the identity of the hypothesis implies that ab+bc+ca =
a2
a
a2
= a2 −2(ab+ca)
= 3a−r
, where r = 2(a + b + c). Then,
0, so a2 +2bc
a2
a2
b2
c2
a
b
c
+ 2
+ 2
=
+
+
+ 2bc b + 2ca c + 2ab
3a − r 3b − r 3c − r
3
=
since (3a − r)(3b − r)(3c − r) = 27abc +
From the equality
a2
a2 +2bc
a2
27abc + r2
= 1,
(3a − r)(3b − r)(3c − r)
r3
2 .
bc
+ 2 a2 +2bc
= 1, we can also conclude that
bc
ca
ab
+ 2
+ 2
= 1.
+ 2bc b + 2ca c + 2ab
Solution 9.17. Notice that
a+1
b+1
c+1
+
+
ab + a + 1 bc + b + 1 ca + c + 1
b+1
c+1
1
c+1
a(1 + bc)
+
+
=1+
+
=
a(b + 1 + bc) bc + b + 1 ca + c + 1
bc + b + 1 ca + c + 1
1
c+1
1 + b(c + 1)
=1+
+
=1+
= 2.
b(c + 1 + ca) ca + c + 1
b(c + 1 + ca)
266
Chapter 10. Solutions to Exercises and Problems
Solution 9.18. The equation is equivalent to a2 − 2ab + b2 − ac + bc = 0. Defining
y = a − b and factorizing, we obtain y 2 − cy = 0. The roots of this quadratic
equation are 0 and c, but 0 is not possible, since a and b are different. Hence,
a − b = c.
Solution 9.19. If the numbers x1 , . . . , xn solve the system, then
1
0 = x1 + x22 + · · · + xnn − n − x1 + 2x2 + · · · + nxn − n(n + 1)
2
= (x22 − 2x2 + 2 − 1) + (x33 − 3x3 + 3 − 1) + · · · + (xnn − nxn + n − 1).
But, using the inequality between the geometric and the arithmetic mean, for each
k ≥ 2 and x ≥ 0, we have
√
k
xk + k − 1 = xk + 1 + · · · + 1 ≥ k xk = kx,
with equality if and only if x = 1.
Then, since each term of the sum xkk − kxk + k − 1 ≥ 0 and the total sum
is zero, we have that each term of the sum is zero, and this happens if each xk = 1.
Then, x2 = · · · = xn = 1 and, recalling the first equation, we also have that x1 = 1.
Solution 9.20. Observe that
1+
1
x
1+
1
y
≥9
⇔ (x + 1)(y + 1) ≥ 9xy
⇔ 2 ≥ 8xy
⇔ (x + y)2 ≥ 4xy
⇔ (x − y)2 ≥ 0.
Solution 9.21. The inequality is equivalent to
x3 + y 3 + z 3 − 3xyz ≥
9
|(x − y)(y − z)(z − x)|.
4
But since
x3 + y 3 + z 3 − 3xyz =
it is enough to prove that
"
!
1
(x + y + z) (x − y)2 + (y − z)2 + (z − x)2 ,
2
" 9
!
1
(x + y + z) (x − y)2 + (y − z)2 + (z − x)2 ≥ |(x − y)(y − z)(z − x)|.
2
4
Let p = |(x − y)(y − z)(z − x)|, using the inequality between the geometric and
the arithmetic mean, we have that
(x − y)2 + (y − z)2 + (z − x)2 ≥ 3 3 p2 .
(10.21)
267
10.9 Solutions of Chapter 9
Now, since |x − y| ≤ x + y, |y − z| ≤ y + z, |z − x| ≤ z + x, we get
2(x + y + z) ≥ |x − y| + |y − z| + |z − x|.
Applying again the inequality between the geometric and the arithmetic mean, we
obtain
√
(10.22)
2(x + y + z) ≥ 3 3 p.
Hence, the result follows from inequalities (10.21) and (10.22).
Solution 9.22. If a ≥ 1, since b + c > a ≥ 1, we have that
ab + bc + ca = a(b + c) + bc > 1 + bc > 1,
which is a contradiction to the fact that ab+bc+ca = 1, therefore a < 1. Similarly,
we can prove that b < 1 and c < 1.
Thus, (1 − a)(1 − b)(1 − c) > 0, then
1 + ab + bc + ca > a + b + c + abc.
Adding 2 on both sides of the last inequality, and using the fact that ab+bc+ca = 1,
we obtain 3 + ab + bc + ca > 2 + a + b + c + abc, that is,
4 > 1 + a + b + c + ab + bc + ca + abc = (a + 1)(b + 1)(c + 1).
Solution 9.23. Notice that
a+b+c−
abc
(a + b)(b + c)(c + a)
=
ab + bc + ca
ab + bc + ca
2
a + b2 + c2 + 3(ab + bc + ca)
,
=
ab + bc + ca
1
1
1
+ b+c
+ a+c
= 1 is equivalent
the last equality is valid, since the condition a+b
to (a + b)(b + c)(c + a) = (a + b)(b + c) + (b + c)(c + a) + (c + a)(a + b) =
a2 + b2 + c2 + 3(ab + bc + ca).
Then, the result to be proved is equivalent to a2 + b2 + c2 ≥ ab + bc + ca, which is
valid using the inequality between the geometric mean and the arithmetic mean.
Solution 9.24. Since the expression is a symmetric function in a, b and c, we can
assume, without loss of generality, that c ≤ b ≤ a. In such a case, a (b + c − a) ≤
b (a + c − b) ≤ c (a + b − c).
For example, the first inequality can be justified as follows:
a (b + c − a) ≤ b (a + c − b) ⇔ ab + ac − a2 ≤ ab + bc − b2
⇔ (a − b) c ≤ (a + b) (a − b)
⇔ (a − b) (a + b − c) ≥ 0.
268
Chapter 10. Solutions to Exercises and Problems
By the rearrangement inequality, see Example 7.3.6, we have
a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c)
2
≤ ba(b + c − a) + cb(c + a − b) + ac(a + b − c)
a (b + c − a) + b2 (c + a − b) + c2 (a + b − c)
≤ ca(b + c − a) + ab(c + a − b) + bc(a + b − c).
!
"
Therefore, 2 a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c) ≤ 6abc.
Solution 9.25. Squaring and rearranging the given inequality, it is equivalent to
n
#
i=1
But, since
*n
x2i − 2
i=1
yi2 =
n
#
xi yi +
i=1
*n
2
i=1 zi ,
n
#
i=1
yi2 ≤
n
#
i=1
x2i − 2
n
#
xi zi +
i=1
n
#
zi2 .
i=1
the inequality we have to prove is equivalent to
n
#
i=1
xi zi ≤
n
#
xi yi ,
i=1
which is the rearrangement inequality (see Example 7.3.6).
Solution 9.26. Let (a1 , a2 , . . . , an ) be a permutation
of (x1 , x2 , . .%. , xn ) with a1 ≤
$
1
1
1
, that is, bi =
a2 ≤ · · · ≤ an , and let (b1 , b2 , . . . , bn ) = n2 , (n−1)
2 , · · · , 12
1
(n+1−i)2 ,
for the indices i = 1, . . . , n.
Consider the permutation (a′1 , a′2 , . . . , a′n ) of (a1 , a2 , . . . , an ), defined by a′i =
xn+1−i , for i = 1, . . . , n.
Using the rearrangement inequality (see Example 7.3.6), we have
x2
xn
x1
+ 2 + · · · + 2 = a′1 b1 + a′2 b2 + · · · + a′n bn
2
1
2
n
≥ an b1 + an−1 b2 + · · · + a1 bn
= a1 bn + a2 bn−1 + · · · + an b1
a2
an
a1
= 2 + 2 + ···+ 2.
1
2
n
Since 1 ≤ a1 , 2 ≤ a2 , . . . , n ≤ an , we get
1
x1 x2
xn
a1 a2
an
1
2
n
1 1
+
+···+ 2 ≥ 2 + 2 +···+ 2 ≥ 2 + 2 +···+ 2 = + +···+ .
12 22
n
1
2
n
1
2
n
1 2
n
Solution 9.27. First, define x2i = 0, x2i−1 = 12 , for all i = 1, . . . , 50. Then, we
2
25
have S = 50 · 21 = 25
2 . Now we prove that always S ≤ 2 .
269
10.9 Solutions of Chapter 9
Let 1 ≤ i ≤ 50, under the conditions of the problem, we have x2i−1 ≤
1 − x2i − x2i+1 and x2i+2 ≤ 1 − x2i − x2i+1 . Using the inequality between the
geometric mean and the arithmetic mean, we get
x2i−1 x2i+1 + x2i x2i+2 ≤ (1 − x2i − x2i+1 )x2i+1 + x2i (1 − x2i − x2i+1 )
(x2i + x2i+1 ) + (1 − x2i − x2i+1 )
2
= (x2i + x2i+1 )(1 − x2i − x2i+1 ) ≤
2
=
1
.
4
Adding these inequalities, for i = 1, 2, . . . , 50, we obtain
S=
50
#
1
25
.
(x2i−1 x2i+1 + x2i x2i+2 ) ≤ 50 · =
4
2
i=1
Solution 9.28. (i) From bx + by ≤ ax + by ≤ ab, it follows that x + y ≤ a.
(ii) We have
√
x+
√
ax
+
a
y=
by
≤
b
ax + by
≤
ab
1 1
+
a b
1 1
+
a b
=
√
a + b.
The first inequality is given by the Cauchy–Schwarz inequality (see Example 4.2.3),
and the second one follows from the hypothesis ax + by ≤ ab.
Solution 9.29. Let x =
a
a−b ,
y=
b
b−c ,
z=
(x − 1)(y − 1)(z − 1) =
=
c
c−a ,
b
a−b
a
a−b
= xyz.
then
c
b−c
b
b−c
a
c−a
c
c−a
When we expand and cancel out terms, we obtain x + y + z = xy + yz + zx + 1.
Then
2a − b
a−b
2
+
2b − c
b−c
2
+
2c − a
c−a
2
= (x + 1)2 + (y + 1)2 + (z + 1)2
= 3 + x2 + y 2 + z 2 + 2(x + y + z)
= 3 + x2 + y 2 + z 2 + 2(xy + yz + zx + 1)
= 5 + (x + y + z)2 ≥ 5.
270
Chapter 10. Solutions to Exercises and Problems
Solution 9.30. Suppose a ≤ b ≤ c, then a2 ≤ bc, therefore
a2 + 1 ≤ bc + 1 ≤ 2a.
(10.23)
On the other hand, since (a−1)2 ≥ 0, we have a2 ≥ 2a−1. The last two inequalities
imply that a2 = 2a − 1, that is, a = 1.
The original second inequality can be rewritten, using that a = 1, as bc ≤ 1. But
since 1 ≤ b ≤ c, we also have that bc ≥ 1, therefore bc = 1.
The first and third inequalities becomes b + 1 ≤ 2c and c + 1 ≤ 2b, therefore
(b + 1)(c + 1) ≤ 4bc = 4. If we expand and cancel out some terms, we get b + c ≤ 2.
The inequality between the geometric mean and the arithmetic mean, and the
2
≤ 1, then the equality holds,
previous inequality, guarantee that 1 = bc ≤ b+c
2
which is true only if b = c. Since bc = 1, then b = c = 1. Therefore, a = b = c = 1
is the only solution.
Solution√9.31. Without
√ loss of generality, we can assume that a = max{a, b, c}.
Then c( b − 1) ≤ a( b − 1) = c, hence b ≤ 4.
√
√
√
We also have b( c − 1) = a ≥ b, then c ≥ 4. Now, 4 ≤ c ≤ c( c− 1) ≤ c( a− 1) =
b ≤ 4, then b = c = 4 and also a = 4. Therefore, there is a unique triplet that
satisfies the equations, that is, (4, 4, 4).
Solution 9.32. If {x1 , . . . , xn } is a real solution of the system, it is clear that also
{x2 , x3 , . . . , xn , x1 } is a solution. But, by hypothesis, we only have one solution,
then x1 = x2 = · · · = xn . The system reduces to only one equation ax2 + (b −
1)x + c = 0, which has a unique solution if (b − 1)2 − 4ac = 0.
Reciprocally, if (b − 1)2 − 4ac = 0, the polynomial P (x) = ax2 + (b − 1)x + c =
2
a x + b−1
has only one solution.
2a
*n
Adding the equations of the system, we have
i=1 P (xi ) = 0, but since
either all the numbers
have
the
same
sign
or
are
zero,
then P (xi ) = 0 for every
, and the system has as a unique solution x1 = x2 = · · · =
xi . Hence xi = − b−1
2a
xn = − b−1
2a .
Solution 9.33. If one of the variables x, y or z is equal to 1 or −1, then we obtain
the solutions (1, 1, 1) or (−1, −1, −1), respectively. Now we will see that these are
the only solutions of the system.
Let f (t) = t2 + t − 1. If one of the variables x, y or z is greater than 1, for
example x > 1, then we have x < f (x) = y < f (y) = z < f (z) = x, which is not
possible. Therefore x, y, z ≤ 1.
If one of the variables x, y or z is less than
x < −1. Since
! −1, choose
"
f (t) = t + 12 − 45 ≥ − 45 , we then get x = f (z) ∈ − 45 , −1 . But,
f
+
5
− , −1
4
=
−11
−1,
16
+
5
⊂ (−1, 0) and f ((−1, 0)) = − , −1 ,
4
271
10.9 Solutions of Chapter 9
!
then it follows that y = f (x) ∈ (−1, 0), z = f (y) ∈ − 45 , −1 and x = f (z) ∈
(−1, 0), which is a contradiction. Then, −1 ≤ x, y, z ≤ 1.
If −1 < x, y, z < 1, then x > f (x) = y > f (y) = z > f (z) = x, which is not
possible. Therefore, there are no other solutions.
Solution 9.34. Add the n equations to obtain
n
#
i=1
x2i +
n
#
i=1
xi − n =
n
#
xi ,
i=1
*n
from where i=1 x2i = n.
On the other hand, rewrite the equations as follows:
x21 + x1 = x2 + 1
x22 + x2 = x3 + 1
..
..
.
.
x2n−1 + xn−1 = xn + 1
x2n + xn = x1 + 1
and multiply the equations to obtain
n
6
i=1
xi (xi + 1) =
n
6
(xi + 1),
i=1
3n
so that i=1 xi = 1 if xi = −1, for all 1 ≤ i ≤ n.
*n
3n
From the two equations, i=1 x2i = n and i=1 xi = 1, we obtain, using the
inequality between the geometric mean and the arithmetic mean, that
7
8 n
*n
2
86
x
n
i=1 i
n
1= =
x2i = 1,
≥ 9
n
n
i=1
and it follows that x21 = x22 = · · · = x2n = 1. Then, a possible solution is
(1, 1, . . . , 1).
If some xi = −1, by the symmetry of the equations, we can assume that
x1 = −1, and then it is easy to see that all xi = −1. Hence the only other solution
is (−1, −1, . . . , −1).
Solution 9.35. The only solution, with all the xi ’s equal, is clearly (2, 2, . . . , 2). If
there is another solution, let m and M be the minimum and maximum value of
the xi , respectively. Then m < M , and for some indexes j, k (taken modulo n),
we would have m2 = xj + xj+1 ≥ 2m and M 2 = xk + xk+1 ≤ 2M , from where
2 ≤ m < M ≤ 2, which is absurd. Therefore (2, 2, . . . , 2) is the only solution.
272
Chapter 10. Solutions to Exercises and Problems
Solution 9.36. The condition x + y + z = 0 implies that x3 + y 3 + z 3 = 3xyz. The
condition x−1 + y −1 + z −1 = 0 guarantees that x−3 + y −3 + z −3 = 3x−1 y −1 z −1 .
Now, since x6 + y 6 + z 6 = (x3 + y 3 + z 3 )2 − 2x3 y 3 z 3 (x−3 + y −3 + z −3 ), we have
x6 + y 6 + z 6 = 9x2 y 2 z 2 − 6x2 y 2 z 2 = 3x2 y 2 z 2 . Thus the result follows if we use
again x3 + y 3 + z 3 = 3xyz.
Solution 9.37. Since the sum of the numbers is zero, d = −a − b − c, therefore
bc − ad = bc + a(a + b + c) = a(a + b) + c(a + b) = (a + c)(a + b)
ac − bd = ac + b(a + b + c) = b(a + b) + c(a + b) = (b + c)(a + b)
ab − cd = ab + c(a + b + c) = c(c + a) + b(c + a) = (b + c)(c + a).
Then, (bc − ad)(ac − bd)(ab − cd) = (a + b)2 (b + c)2 (c + a)2 .
Solution 9.38. We have
# a2 + b 2
# a4
# a2 + b 2
# a4 + ab(a2 + b2 )
# a3
−
=
+
=
bc
a+b
abc
c
abc
cyclic
cyclic
cyclic
cyclic
cyclic
1 4
=
(a + b4 + c4 + ab(a2 + b2 ) + bc(b2 + c2 ) + ca(c2 + a2 ))
abc
1
(a + b + c)(a3 + b3 + c3 ) = 0.
=
abc
Solution 9.39. Since
a3 + b3 + c3 − 3abc =
and since a + b + c > 2, then
!
"
1
(a + b + c) (a − b)2 + (b − c)2 + (c − a)2
2
a + b + c = p and
!
"
(a − b)2 + (b − c)2 + (c − a)2 = 2.
Suppose that !a ≥ b ≥ c. If a > b > c, then a" − b ≥ 1, b − c ≥ 1 and a − c ≥ 2, and
this leads to (a − b)2 + (b − c)2 + (c − a)2 ≥ 6 > 2, which is absurd, therefore
a = b = c + 1 or a − 1 = b = c. Thus, we have that the prime number is of the form
p = 3c + 2, in the first case or of the form p = 3c + 1, in the second case. Then the
p+1 p−2
p+2 p−1 p−1
triplet is ( p+1
3 , 3 , 3 ) in the first case, and ( 3 , 3 , 3 ) in the second case.
Solution 9.40. The equation is equivalent to
(3x)3 + (−3y)3 + (−1)3 − 3(3x)(−3y)(−1) = 1646,
which can be factorized as (3x − 3y − 1)(9x2 + 9y 2 + 1 + 9xy + 3x − 3y) = 2 · 823.
Now, the first factor on the left-hand side is smaller than the second factor and,
since 823 is a prime number and 3x − 3y − 1 ≡ 2 mod 3, we get 3x − 3y − 1 = 2
and 9x2 + 9y 2 + 1 + 9xy + 3x − 3y = 823. Solving the system for positive real
numbers leads to x = 6 and y = 5.
273
10.9 Solutions of Chapter 9
Solution 9.41. Using the identity (4.8), the condition x3 + y 3 + z 3 − 3xyz = 1 is
equivalent to
(10.24)
(x + y + z)(x2 + y 2 + z 2 − xy − yz − zx) = 1.
Let A = x2 + y 2 + z 2 and B = x + y + z. Observe that B 2 − A = 2(xy + yz + zx).
By identity (4.9) we have that B > 0. The equation (10.24) becomes
B A−
B2 − A
2
= 1,
then 3A = B 2 + B2 . Since B > 0, we apply the inequality between the geometric
mean and the arithmetic mean to obtain 3A = B 2 + B2 = B 2 + B1 + B1 ≥ 3, that
is, A ≥ 1. The minimum A = 1 is reached, for example, with (x, y, z) = (1, 0, 0).
Solution 9.42. Equations x +
y
z
zx + y = 2z,
= 2, y +
z
x
= 2, z +
xy + z = 2x,
x
y
= 2, imply that
yz + x = 2y
and that
xyz + y 2 = 2yz,
xyz + z 2 = 2zx,
xyz + x2 = 2xy.
Therefore,
xy + yz + zx = x + y + z,
2
2
2
3xyz + (x + y + z ) = 2(xy + yz + zx).
We also have that
yzx
= (2 − x)(2 − y)(2 − z)
1=
zxy
= 8 − 4(x + y + z) + 2(xy + yz + zx) − xyz.
(10.25)
(10.26)
(10.27)
If we define a = x + y + z, we have by equation (10.25), that xy + yz + zx = a
and we also get that
x2 + y 2 + z 2 = (x + y + z)2 − 2(xy + yz + zx) = a2 − 2a.
Now, from equation (10.26), it follows that 3xyz = −a2 + 4a. Finally, by equation
(10.27), we can conclude that
1 = 8 − 4a + 2a −
−a2 + 4a
.
3
Hence, we get the equation a2 − 10a + 21 = 0, which has roots a = 3 and a = 7.
Therefore, x + y + z is equal to 3 or 7.
But, if x + y + z = 7, since x + yz + y + xz + z + xy = 6, we have that yz + xz + xy = −1,
which is not possible for x, y, z positive numbers.
The sum x + y + z = 3, can be achieved with x = y = z = 1, which are also
solutions of the equations, hence the only possible value for x + y + z is 3.
274
Chapter 10. Solutions to Exercises and Problems
Solution 9.43. Let d be the common difference of the progression {an }. Note that,
for j = 0, 1, . . . , n − 1, we have that
√
√
√
√
aj−1 − aj
aj − aj−1
1
1
.
√ =√
√ ·√
√ =
√
aj−1 + aj
aj−1 + aj
aj−1 − aj
aj − aj−1
Use the fact that aj − aj−1 = d, for all j, in order to get
√
√
√
n √
n
#
#
aj − aj−1
an − a0
1
=
.
√ =
√
aj−1 + aj
d
d
j=1
j=1
Finally, observe that
√
√
an − a0
an − a0
a0 + nd − a0
n
= √
= √
=√
√
√
√ .
d
d( a0 + an )
d( a0 + an )
a0 + an
Solution 9.44. Let d > 0 be the common difference of the progression. Suppose
that a = b − d, b and c = b + d are the lengths of the sides of the triangle. Since
c2 = a2 + b2 , we have (b + d)2 = (b − d)2 + b2 , therefore b = 4d. On the other
2
(b−d)4d
2
hand, the area of the triangle is a·b
= 12d
2 =
2
2 = 6d , and it is also equal to
3d+4d+5d
31
= 6d. From this,
the inradius r multiplied by the semiperimeter s =
2
r · 6d = 6d2 , and then r = d.
Solution 9.45. Suppose that {1, 2, . . . , 9} has been divided into two subsets A
and B such that neither of them contains an arithmetic progression. Suppose that
5 ∈ A. It is clear that 1 and 9 cannot both be in A. Then, we have the following
cases:
(i) If 1 ∈ A and 9 ∈ B. Since {1, 5} ⊂ A, we have that 3 ∈ B; 3, 9 ∈ B imply
that 6 ∈ A; 5, 6 ∈ A imply that 4, 7 ∈ B; 3, 4 ∈ B imply that 2 ∈ A; 7, 9 ∈ B
imply that 8 ∈ A. But, {2, 5, 8} ⊂ A is an arithmetic progression, which is
absurd.
(ii) If 9 ∈ A and 1 ∈ B. This case is analogous to (i).
(iii) If 1, 9 ∈ B. Then we have two subcases:
(1) If 7 ∈ A. In this case, 5, 7 ∈ A imply 6 ∈ B and 3 ∈ B. Therefore,
{3, 6, 9} ⊂ B, which is absurd.
(2) If 7 ∈ B. Since 7, 9 ∈ B, we have that 8 ∈ A; 1, 7 ∈ B imply 4 ∈ A;
4, 5 ∈ A imply 3 ∈ B; 1, 3 ∈ B imply that 2 ∈ A. Therefore, again we
have an arithmetic progression, that is, the progression {2, 5, 8} is in A,
which is a contradiction.
Solution 9.46. Observe that
2
(a + b + c)3 − a2 (b + c) − b2 (c + a) − c2 (a + b)
9
1
= (a + b − 2c) (2a − b − c) (a − 2b + c) .
9
31 See
[5].
10.9 Solutions of Chapter 9
275
Solution 9.47. Suppose the number of prime numbers is not infinite. Let p be the
greatest prime number of the progression. Consider the number n = 4p! − 1, which
belongs to the progression. Since n > p, the number is a composite number and
it does not have prime divisors of the form 4k − 1 (the factors of this form belong
to p!), then its prime divisors are of the form 4k + 1. But the product of factors
of the form 4k + 1 is also a number of the form 4k + 1, and then n must be of this
form as well, which is a contradiction.
Solution 9.48. Divide the set of natural numbers N = {1, 2, 3, . . . } in the following
way
1
4 5 6
11 . . .
2 3
7 8 9 10
...
The sets we are looking for are A = {1, 4, 5, 6, 11, 12, 13, 14, 15, 22, . . .} and B =
{2, 3, 7, 8, 9, 10, 16, 17, 18, 19, 20, 21, 29, . . .}. In fact, each one of them has “gaps”
between numbers as large as we want. Therefore, it is not possible to have in some
of them an arithmetic progression, since the elements of the progression have a
constant difference d, which will be overtaken by a proper gap.
Solution 9.49. The answer is no. If there is an arithmetic progression with difference d, we have that the d consecutive integers
(d + 1)! + 2, (d + 1)! + 3, . . . , (d + 1)! + (d + 1)
are composite numbers. But among them there has to be an element of the progression, because this progression has difference d, which is a contradiction.
Second Solution. Let m > 1 be a number in the progression. Then m + md =
m(d + 1) is also an element of the progression and is not a prime number.
Solution 9.50. Suppose that the arithmetic progression with difference d contains
a perfect square, say a2 . Then the numbers a2 , a2 + d, a2 + 2d, . . . are in the
progression, and then the numbers a2 + (2a + d)d = (a + d)2 are also in the
progression. Now, it is clear that the square numbers of the form (a + kd)2 , for
every k ∈ N, are also in the progression.
Solution 9.51. Suppose that there are 1999 prime numbers, smaller than 12345,
in arithmetic progression. Let p be the first prime number in the progression and
let r be the difference of the progression. Then the progression is p, p + r, p + 2r,
. . . , p + 1998r.
The prime number p cannot be one of the prime numbers 2, 3, . . . , 1997,
because if it is one of them, then p + pr, which is in the progression, is not a prime
number. Therefore, p ≥ 1999.
Since p is an odd number and p + r is prime, then r is even. All the even numbers
are of the form 6n, 6n + 2 or 6n − 2. Let us see now that r cannot be of the
276
Chapter 10. Solutions to Exercises and Problems
form 6n + 2 nor can it take the form 6n − 2. In fact, since p is prime, it is of
the form 6k + 1 or 6k − 1. In any of those four cases, there is in the progression
a multiple of 3, p + r = (6k + 1) + (6n + 2), p + 2r = (6k + 1) + 2(6n − 2),
p + 2r = (6k − 1) + 2(6n + 2), p + r = (6k − 1) + (6n − 2).
Therefore r is of the form 6n and then the progression is
p, p + 6n, . . . , p + 1998(6n).
But p ≥ 1999 and n ≥ 1 imply that p+1998(6n) ≥ 1999+11988 = 13987 > 12345.
Hence, the numbers p + jr cannot be all smaller than 12345.
Solution 9.52. For n < 3, there does not exist a rearrangement with an arithmetic
triplet. For n = 3, the list 2, 1, 3 achieves the task. We will construct an example
for n, using the examples of the previous values of n. On one side of the list, we
put the even numbers between 1 and n, and on the other side the odd numbers. If
the even numbers are j, we rearrange them using the example for the j numbers
and then we have them multiplied by 2. If there are k odd numbers, to order them
we use the example for the k multiplying those numbers by 2 and subtracting 1.
In this way we obtain a rearrangement of the numbers from 1 to n. If on the even
side, 2a, 2b and 2c form an arithmetic triplet, then a, b and c is also an arithmetic
triplet for the case j, which is absurd. If in the odd side, 2a − 1, 2b − a and 2c − 1
is an arithmetic triplet, then a, b and c is also an arithmetic triplet in the example
for the case of k, which is also absurd. Finally, one term on the even side and one
term on the odd side satisfy that their sum is odd, then there is not a third term
in between them such that its double would be this sum. Then, the constructed
rearrangement does not have arithmetic triplets.
Solution 9.53. Since there are 4 solutions for the first equation, a = 0. Let x0 be
the common solution to both equations.
Taking the difference of the equations, it follows that ax40 − ax0 = 0, which can
also be written as ax0 (x30 − 1) = 0.
Then, the common solution is x0 = 0 or x0 = 1.
If x0 = 0 in the first equation, we obtain a = 1, therefore x0 is a solution with
multiplicity at least 2, but this is not possible because we know that there are four
different roots.
Then the common solution is x0 = 1. When we substitute in the first equation, we obtain 2a + b = 1, and then this equation can be rewritten as
ax4 + (1 − 2a)x2 + a − 1 = 0,
which has 1 and −1 as solutions. Therefore
(x − 1)(x + 1)(ax2 − a + 1) = 0.
The quadratic equation ax2 − a + 1 = 0 must have 2 different real solutions, say
r and −r, with r > 0.
277
10.9 Solutions of Chapter 9
There are two cases, a > 1 and a < 0.
If a > 1, then 0 < r < 1 and the roots −1, −r, r, 1 are in arithmetic progression
1
5
9
only when r = 13 . In such case a = 1−r
2 = 8 and b = 1 − 2a = − 4 .
If a < 0, then r > 1, and the numbers −r, −1, 1, r are in arithmetic progression
5
1
1
only if r = 3. In such case a = 1−r
2 = − 8 and b = 4 .
*k
*k
Solution 9.54. Write A =
i=1 ai and B =
i=1 bi . Now we add over i the
corresponding terms in the inequalities
ai n + bi − 1 < ⌊ai n + bi ⌋ ≤ ai n + bi ,
to obtain An + B − k < Xn ≤ An + B. Now, suppose that {Xn } is an arithmetic
progression with common difference d, then nd = Xn+1 −X1 and A+B−k < X1 ≤
A + B. Combine the above inequalities to obtain A(n + 1) + B − k < nd + X1 ≤
A(n + 1) + B or
An − k ≤ An + (A + B − X1 ) − k < nd < An + (A + B − X1 ) < An + k,
from which we conclude that |A − d| < nk , for any integer number n; then A = d.
Since {Xn } is a sequence of integers, d has to be also an integer number, therefore
we conclude that A is an integer number.
Solution 9.55. Consider a partition of {1, . . . , 256} into two subsets, A and B.
Divide {1, . . . , 9} in two subsets A1 and B1 , in the following way: k ∈ A1 (resp.
B1 ) if and only if 2k−1 ∈ A (resp. B). Clearly, A1 ∩ B1 = ∅.
If A1 = ∅ (or B1 = ∅), then B1 (or A1 ) has three numbers in arithmetic progression
a, b and c. Then 2a−1 , 2b−1 and 2c−1 is a geometric progression in B (or A).
If A1 = ∅ and B1 = ∅, then A1 ∪ B1 is a partition of {1, . . . , 9}. By Problem 9.45,
one of the sets, say A1 , contains an arithmetic progression of three terms a, b and
c. Then 2a−1 , 2b−1 and 2c−1 is a geometric progression in A.
Solution 9.56. Remember that the nth term in a geometric progression is an =
a0 · q n , where a0 is the first term
of the progression and q is the ratio.
Since ak+1 = ak 1 + n12 , the given collection of numbers coincide with the
first terms of a geometric progression whose ratio is 1 + n12 and whose first term
is given by 12 .
The nth term of the collection is given by
an =
1
2
1+
1
n2
n
.
n
We have to prove that 1 + n12 < 2, for n > 1. By Newton’s binomial theorem
(Theorem 3.2.3), we have that
1+
1
n2
n
=
n
#
n
i
i=0
1
n2
i
=
n
#
i=0
Ti ,
where Ti =
n
i
1
n2
i
.
278
Chapter 10. Solutions to Exercises and Problems
Now, notice that
Ti
=
Ti+1
n!
i!(n−i)!
n!
(i+1)!(n−i−1)!
·
1
n2i
1
n2i+2
=
(i + 1) 2
· n > 1,
n−i
hence the sequence Ti is decreasing, then Ti < T1 =
1+
1
n2
1
n.
Therefore
n
= 1 + T1 + T2 + T3 + · · · + Tn < 1 + nT1 = 2.
Solution 9.57. Suppose that a1 = a.
By induction we will see that an = a + n − 1. Suppose that an = a + n − 1 and
prove that an+i = a + n + i − 1, for 1 ≤ i ≤ n. Since a2n = an + n, the hypothesis
leads us to an = a + n − 1, then a2n = a + 2n − 1. Now, since the sequence is
increasing, it follows that
a + n − 1 = an < an+1 < · · · < an+i < · · · < a2n = a + 2n − 1.
Hence
an+i = a + n + i − 1
for
1 ≤ i ≤ n.
(10.28)
If a1 = 1, then we get an = n, for all n ≥ 1.
What remains to be proved is the induction basis, that is, a1 = 1.
Suppose that a1 > 1. Let p be the least prime number greater than (a1 + 1)! +
a1 + 1. Of course, this prime number p is an element of the sequence, that is,
p = an = a1 + (n − 1), for some n. By equation (10.28), the an are consecutive
numbers. Moreover, by property (ii), n = p − a1 + 1 is also a prime number.
Since a1 > 1, p − a1 + 1 < p, and by the way we have chosen p, we have that
p−a1 +1 ≤ (a1 +1)!+a1 +1 < p. Then (a1 +1)!+2 ≤ p−a1 +1 ≤ (a1 +1)!+a1 +1.
But this is a contradiction, since among the numbers
(a1 + 1)! + 2, (a1 + 1)! + 3, . . . , (a1 + 1)! + a1 + 1
there are no prime numbers, that is, all the numbers are composite numbers, since
j divides ((a1 + 1)! + j). The contradiction proves that a1 = 1.
Solution 9.58. The proof is by induction over n + m. The statement is clear if
m + n = 2. Suppose that the statement is true for m + n < k, and consider
m + n = k arbitrary numbers a1 , a2 , . . . , am , b1 , b2 , . . . , bn . Define the sets
A = {a1 , a2 , . . . , am }, B = {b1 , b2 , . . . , bn },
C = {c1 , c2 , . . . , cm }, D = {d1 , d2 , . . . , dn }.
We have two cases:
1. If A ∩ C = ∅ or B ∩ D = ∅. For example, suppose that A ∩ C = ∅.
This implies that ai = cj , for some indices i, j ∈ {1, 2, . . . , m}. Without loss of
279
10.9 Solutions of Chapter 9
generality, let i ≤ j. If i = j, one of the terms in the equality we want to prove
is zero and, then we can apply directly the induction hypothesis. If i < j, then
ci < · · · < cj = ai < ai+1 < · · · < aj , then
|ai − ci | + |ai+1 − ci+1 | + · · · + |aj − cj | = (ai − ci ) + (ai+1 − ci+1 ) + · · · + (aj − cj ).
If we change the order of the terms, the value of the sum does not change, and
this value is equal to
(ai+1 − ci ) + (ai+2 − ci+1 ) + · · · + (aj − cj−1 ) + (ai − cj )
= |ai+1 − ci | + |ai+2 − ci+1 | + · · · + |aj−1 − cj−2 | + |aj − cj−1 |,
since ai − cj = 0. Then the result follows from the induction hypothesis, for the
m + n − 1 = k − 1 numbers
a1 < a2 < · · · < ai−1 < ai+1 < · · · < am , b1 < b2 < · · · < bn ,
c1 < c2 < · · · < cj−1 < cj+1 < · · · < cm , d1 < d2 < · · · < dn .
2. If A ∩ C = B ∩ D = ∅. In this case, we have that a1 is in D and b1 is
in C. Without loss of generality, suppose that a1 < b1 . Then a1 has to be equal
to d1 . We will prove that b1 has to be equal to c1 . If b1 = ci , for some i > 1,
then b1 > c1 and, since c1 = bj , for some j > 1, we have that b1 > bj , which is
a contradiction. Therefore, b1 = c1 , and taking into account that ai = d1 , this
implies |a1 − c1 | = |b1 − d1 |. Now, we use the induction hypothesis for the numbers
a2 < a3 < · · · < am , b 2 < b 3 < · · · < b n ,
c2 < c3 < · · · < cm , d2 < d3 < · · · < dn .
Solution 9.59. First, observe that for n ≥ 2, we have
n−1
#
j=1
n−1
#
j
=
(j + 1)!
j=1
1
1
−
j! (j + 1)!
=1−
1
.
n!
(10.29)
Multiplying by n! the equation (10.29), we obtain the desired representation
1+
n−1
#
j=1
j · n!
= n!.
(j + 1)!
Solution 9.60. Remember that every integer number relatively prime to p has a
multiplicative inverse modulo p. Denote the inverse of x modulo p by x−1 . Observe
that
p−1
# p 2 p−1
# p p−1 2
2p
−2=
.
=
p
k
k k−1
k=1
k=1
280
Chapter 10. Solutions to Exercises and Problems
p−1
is an integer number, since it is equal to p1 kp and p divides
Observe that k1 k−1
p
*p−1 $ −1 p−1%2
3
2
. We
k=1 k
k . Then the last sum is congruent, modulo p , to p
k−1
prove that the sum is divisible by p. Observe that, modulo p, we have
p−1
#
k −1
k=1
p−1
k−1
2
≡
p−1
#
k=1
(k −1 )2 {(p − 1)(p − 2) · · · (p − k + 1)
× [(k − 1)(k − 2) · · · (p − k + 1)]−1 }2
≡
≡
p−1
#
k=1
p−1
#
!
"2
(k −1 )2 (−1)(1)−1 (−2)(2)−1 · · · (−(k − 1))(k − 1)−1
(k −1 )2 (−1)2k−2 .
k=1
But the inverse numbers of 1, 2, . . ., p − 1, modulo*p, are the same numbers
p−1
in some other order. Then the sum is congruent to k=1 k 2 , which
is equal to
(p − 1)p(2p − 1)/6. This is divisible by p, since p = 2, 3. Then 2p
−
2
is divisible
p
3
by p . Since
2p − 1
1
2p
−1=
−2 ,
p−1
2
p
3
it follows that 2p−1
p−1 − 1 is divisible by p , as we wanted.
a
b
c
d−c
Solution 9.61. If
∈ S, then by (ii),
1
a
b +1
=
b
a+b
∈ S and
a
b
a
b +1
=
a
a+b ∈
c
d−c+c
c
∈ S, then d−c+c
= dc ∈ S and if d−c
In particular, if
c ∈ S, then
a0
Consider a rational number q0 = b0 , with (a0 , b0 ) = 1 and 0 < q0 <
0
0
enough to see that either b0a−a
or b0a−a
belongs to S.
0
0
1
If q0 = 2 there is nothing to do. If q0 < 21 , then a0 < b0 − a0 . If
b 0 − a0 < a0 .
Let
q1 =
a0
b0 −a0 ,
b0 −a0
a0 ,
if q0 <
S.
= dc ∈ S.
1. Then it is
q0 > 12 , then
1
2
if q0 > 12 ,
then 0 < q1 < 1 and if q1 ∈ S, then q0 ∈ S.
Now, if q1 = ab11 , with (a1 , b1 ) = 1, and considering
a1
1
b1 −a1 , if q1 < 2 ,
q2 =
b1 −a1
1
a1 , if q1 > 2 ,
it follows that 0 < q2 < 1 and if q2 ∈ S, then q1 ∈ S.
This process of going from qk to qk+1 is possible if no qk is equal to 12 ; otherwise,
the proof is complete (qk ∈ 21 ∈ S implies that qk−1 , . . . , q0 ∈ S).
281
10.9 Solutions of Chapter 9
o
, and b1 | b0 − a0 ,
The process cannot be infinite. If q0 < 21 , then ab11 = q1 = b0a−a
0
1
hence b1 ≤ b0 − a0 < b0 . If q0 > 2 , then b1 | a0 and b1 ≤ a0 < b0 . Hence, in any
case b1 < b0 . Similarly, it follows that bk+1 < bk , for all k ≥ 0. Thus, {bk } is a
decreasing infinite sequence of positive integers, which is impossible.
2
2
+b
Solution 9.62. Suppose aab+1
= k and k is not a perfect square. Then a2 − kab +
2
b = k. Suppose (a0 , b0 ) is a solution of the equation.
By symmetry, we can assume that a0 ≥ b0 > 0. We know that a0 is a root of the
quadratic equation a2 − kab0 + b20 − k = 0. Let c1 be the other root of the previous
equation; the two roots satisfy a0 + c1 = kb0 and a0 c1 = b20 − k, then c1 = kb0 − a0
is an integer. Now, since k is not a square, a0 c1 = 0, hence c1 = 0.
If c1 < 0, then c21 − kc1 b0 + b20 ≥ c21 + k + b20 > k, which is a contradiction, therefore
b2 −1
a2 −1
b2 −k
c1 > 0. Moreover, c1 = 0a0 ≤ 0a0 ≤ 0a0 < a0 . Thus (c1 , b0 ) is a positive
solution of a2 − kab + b2 = k, with c1 < a0 . Proceed in the same way to construct
a decreasing sequence of positive integers a0 > c1 > c2 > · · · > 0, but this cannot
happen.
Solution 9.63. Since a, b, c are the roots of P (x), by Vieta’s formulas it follows
that P (x) = x3 − (a + b + c)x2 + (ab + bc + ca)x − abc = x3 − 2007x + 2002, and
then a + b + c = 0, ab + bc + ca = −2007, abc = −2002.
Hence,
a−1
a+1
b−1
b+1
c−1
c+1
abc − (ab + bc + ca) + a + b + c − 1
abc + ab + bc + ca + a + b + c + 1
−2002 − (−2007) − 1
4
1
=
=
=−
.
−2002 + (−2007) + 1
−4008
1002
=
Solution 9.64. If x = a + b + c, y = abc, z = a1 + 1b + 1c are integers, also
ab + bc + ca = abc( a1 + 1b + 1c ) = yz is an integer. Moreover, a, b, c are the roots of
the polynomial w3 − xw2 + yzw − y = 0. Since the coefficients of the polynomial
are integers and the coefficient of w3 is 1, by Gauss’ lemma or by the rational root
theorem, a, b, c are integers. Suppose that 1 ≤ a ≤ b ≤ c, since 1 ≤ a1 + 1b + 1c ≤ a3 ,
it follows that a = 1, 2 or 3. If a = 1, then 1b + 1c ≤ 2, and the only possibilities are
(b, c) = (1, 1) or (2, 2). If a = 2, then 21 ≤ 1b + 1c ≤ 1, hence (b, c) = (3, 6) or (4, 4).
If a = 3, 1b + 1c = 23 , and the only solution is (b, c) = (3, 3). Thus, the solutions are
(a, b, c) = (1, 1, 1), (1, 2, 2), (2, 3, 6), (2, 4, 4), (3, 3, 3).
Solution 9.65. If one of a, b or c is zero, then one of the equations is linear and this
one has a real solution. The discriminant of the equations are 4b2 − 4ca, 4c2 − 4ab,
4a2 − 4bc. Then, given that
"
!
(4b2 − 4ca) + (4c2 − 4ab) + (4a2 − 4bc) = 2 (a − b)2 + (b − c)2 + (c − a)2 ≥ 0,
one of the discriminants is non-negative and therefore the equation with that
discriminant has a real solution.
282
Chapter 10. Solutions to Exercises and Problems
Solution 9.66. Let u, v be the roots of the quadratic polynomial and let α, β, γ
be the roots of the cubic polynomial, with all of these roots being non-negative.
By Vieta’s formulas,
2
uv = − a,
3
α + β + γ = −a, αβ + βγ + γα = b, αβγ = 8.
u + v = 4,
Now, using the inequality between the geometric mean and the arithmetic mean,
4=
4
2
2
=
u+v
2
2
≥ uv
2
= − a,
3
2
2
− a = (α + β + γ) ≥ 2 3 αβγ = 4.
3
3
Hence, on both inequalities the equality holds, and then u = v, α = β = γ and
− 32 a = 4. Hence a = −6, α = β = γ = 2, therefore b = 12. Thus, the only pair is
(a, b) = (−6, 12).
Solution 9.67. We can choose appropriate signs in ±P (±x) in order to assume
that a, b ≥ 0.
(i) There are two cases:
(1) c ≥ 0, |a| + |b| + |c| = a + b + c = P (1) ≤ 1.
(2) c < 0, |a| + |b| + |c| = a + b − c = P (1) − 2P (0) ≤ 3.
Then, 3 is the maximum that is attained with the polynomial P (x) = 2x2 − 1.
(ii) There are four cases:
(1) c ≥ 0, d ≥ 0, |a| + |b| + |c| + |d| = a + b + c + d = P (1) ≤ 1.
(2) c ≥ 0, d < 0, |a| + |b| + |c| + |d| = a + b + c − d = P (1) − 2P (0) ≤ 3.
(3) c < 0, d ≥ 0, |a| + |b| + |c| + |d| = a + b − c + d = 34 P (1) − 31 P (−1) +
8
1
8
1
3 P ( 2 ) + 3 P (− 2 ) ≤ 7.
(4) c, d < 0, |a|+|b|+|c|+|d| = a+b−c−d = 53 P (1)−4P ( 12 )+ 34 P (− 21 ) ≤ 7.
With the polynomial P (x) = 4x3 − 3x, the maximum 7 in this case, is attained.
Solution 9.68. Notice that dx3 +cx2 +bx+a = x3 (d+c x1 +b x12 +a x13 ); then its roots
are the inverse of the roots of ax3 + bx2 + cx + d. If {α, β, γ} are the roots of the
polynomial ax3 + bx2 + cx + d, it follows, by Vieta’s formulas, that α + β + γ = − ab
and α1 + β1 + γ1 = − dc . Using the inequality between the geometric mean and the
bc
= (α + β + γ)( α1 + β1 + γ1 ) ≥ 9.
arithmetic mean, we can conclude that ad
283
10.9 Solutions of Chapter 9
√
Solution 9.69. Given that 2 x2 − 1 = x −
get that
x2 − p, and squaring both sides, we
4(x2 − 1) = x2 − 2x x2 − p + x2 − p
4x2 − 4 = 2x2 − 2x x2 − p − p
2x2 + (p − 4) = −2x x2 − p,
then 2x2 + (p − 4) < 0, since x > 0. Squaring the last equation and simplifying, it
follows that
4x4 + 4x2 (p − 4) + (p − 4)2 = 4x2 (x2 − p)
8x2 (p − 2) + (p − 4)2 = 0
x=±
4−p
8(2 − p)
Since x is a positive real number, then p < 2 and x = √ 4−p
.
8(2−p)
.
Solution 9.70. Let α, β, γ be the roots of P (x) that are positive. From Vieta’s
d
formulas, it follows that α + β + γ = − ab , αβ + βγ + γα = ac , αβγ = − . Since
a
d = P (0) < 0 and αβγ > 0, it follows that a > 0. Dividing by a3 , the inequality
to be proved, it is enough to see that
2
b
a
3
+9
d
a
−7
bc
a2
≤ 0,
that in terms of α, β, γ is
−2 (α + β + γ)3 − 9αβγ + 7 (α + β + γ) (αβ + βγ + γα) ≤ 0.
After simplifying, the left-hand side of the previous inequality is
α2 β + αβ 2 + β 2 γ + βγ 2 + γ 2 α + γα2 ≤ 2 α3 + β 3 + γ 3 .
This inequality follows from the rearrangement inequality applied in the following
way,
α2 β + β 2 γ + γ 2 α ≤ α3 + β 3 + γ 3 , β 2 α + γ 2 β + α2 γ ≤ α3 + β 3 + γ 3 .
Solution 9.71. By Vieta’s formulas, it follows that −a = α+β+γ, b = αβ+βγ+γα
and −c = αβγ. Since α2 = β +γ, then α2 = −α−a. Observe that α = 0, otherwise
α = 0, and then c = 0, contradicting the fact that c is odd. Hence βγ = − αc and
b = α(β + γ) + βγ = α3 − αc .
284
Chapter 10. Solutions to Exercises and Problems
Therefore,
0 = α4 − bα − c = (α + a)2 − bα − c = α2 + 2αa + a2 − bα − c
= −α − a + 2αa + a2 − bα − c = α(2a − b − 1) − (c − a2 + a).
Observe that 2a − b − 1 = 0, otherwise 2a = b + 1, and since b is even, 2a
would be odd, a contradiction. Hence, from the previous equation, it follows that
α = c−a(a−1)
2a−b−1 is rational, and then it is an integer since the polynomial is monic
and it has integer coefficients.
On the other hand, −a = α(α + 1) is even. If β = γ, then 2β = β + γ = −a − α.
If β is rational, it must be an integer; then from equation 2β = −a − α, it follows
that α is even, but then c = −αβγ is even, a contradiction since c is odd.
Solution 9.72. Let x0 be a real solution of equation x2 + px + q = 0. Then
√
√
1+ 1+4·1
1+ 5
−p ± p2 − 4q
≤
=
= s.
x0 =
2
2
2
Notice that x0 could be equal to s if p = q = −1.
If y is a real root of an equation of the form t2 + pt + q = 0, with p, q ∈ [−1, 1],
then any number z with absolute value less than or equal to the absolute value of
y, would be a root too. To see this, let y 2 + py + q = 0, for p, q ∈ [−1, 1], and let
z = αy, where |α| ≤ 1. The equation t2 + αpt + α2 q = 0 has coefficients αp and
α2 q in the interval [−1, 1], since |α| ≤ 1. Moreover, z is a root of this equation,
since
z 2 + αpz + α2 q = (αy)2 + αp(αy) + α2 q = α2 (y 2 + py + q) = 0.
Therefore,
all solutions of the quadratic equations belong to the interval
&
√
√ '
− 1+2 5 , 1+2 5 .
Solution 9.73. Define f : R → N by
1
n−1
f (x) = ⌊x⌋ + x +
+ ···+ x +
− ⌊nx⌋.
n
n
We have to show that f (x) = 0. Observe that
1
n−1
1
1
1
1
1
f x+
= x+
+ x+ +
+ ···+ x + +
− n x+
n
n
n n
n
n
n
1
n−1
= x+
+ ··· + x +
+ ⌊x + 1⌋ − ⌊nx + 1⌋,
n
n
and since ⌊x + k⌋ = ⌊x⌋ + k, for every integer k, it follows that f x + n1 = f (x),
for every real number x. Hence, f is a periodic function with period n1 . In this
285
10.9 Solutions of Chapter 9
way it is enough to study f (x), for 0 ≤ x <
then f (x) = 0 for every real number x.
1
n.
But f (x) = 0 for all these values,
Solution 9.74. The sought for identity can be rewritten as
1
1
n
n
n 1
+
+ 2+
+ · · · + k+1 +
+ · · · = n.
2
2
2
2
2
2
We now use a special case
the Hermite identity (see Problem 9.73 or Example
of
1
1.3.2), then for n = 2, x + 2 = ⌊2x⌋ − ⌊x⌋. This implies that
n
n
n
n
n
+
− 2 + · · · + k − k+1 + · · · = n.
2
2
2
2
2
n
= 0, for k large enough.
This last sum is telescopic and moreover 2k+1
⌊n⌋ −
Solution 9.75. This inequality can be proved using Hermite’s identity, but here we
present an idea for a shorter proof. Let Sn be the right-hand side of the inequality.
Then, if we set S0 = 0, we get
Sn − Sn−1 =
⌊nx⌋
, for all n = 1, 2, . . . .
n
Then
k(Sk − Sk−1 ) = ⌊kx⌋,
for k = 1, 2, . . . , n + 1.
Adding these n + 1 equations, it follows that
−S1 − S2 − · · · − Sn + (n + 1)Sn+1 = ⌊(n + 1)x⌋ + ⌊nx⌋ + · · · + ⌊x⌋.
Now proceed by induction. The basis n = 1 is clear. Suppose that Sk ≤ ⌊kx⌋, for
1 ≤ k ≤ n, then use the last identity for (n + 1)Sn , hence
(n + 1)Sn+1 ≤ ⌊(n + 1)x⌋ + (⌊nx⌋ + ⌊x⌋) + · · · + (⌊x⌋ + ⌊nx⌋) .
Using n times the fact that ⌊u⌋ + ⌊v⌋ ≤ ⌊u + v⌋ for any real numbers u and v, it
follows that (n + 1)Sn+1 ≤ (n + 1)⌊(n + 1)x⌋, which ends the proof.
Solution 9.76. First observe that since f (10) = 0 and f (10) = f (5) + f (2), then
f (5) + f (2) = 0, but f (n) is non-negative, then f (5) = 0 and f (2) = 0. On the
other hand, f (9) = f (3) + f (3) = 0. Then, given that 1985 = 5 · 397, it follows
that f (1985) = 0, since
f (1985) = f (5) + f (397)
= 0 + f (397)
= f (9) + f (397)
= f (9 · 397)
= f (3573) = 0.
286
Chapter 10. Solutions to Exercises and Problems
Solution 9.77. The function must satisfy that f (y) > 0, for y > 0. Then
f (xf (y)) = f (x) −
1
.
xyf (y)
(10.30)
Let a = f (1) > 0. Taking x = 1 and then y = 1 in (10.30), it follows respectively
that
1
1
=a−
,
for y ∈ R+
f (f (y)) = f (1) −
yf (y)
yf (y)
(10.31)
1
+
f (xa) = f (x) −
,
for x ∈ R .
ax
Taking x = 1 in the last equality, f (a) = f (1) − a1 = a − a1 .
Taking x = a in the equation (10.30), it follows that
f (af (y)) = f (a) −
1
1
1
=a− −
.
ayf (y)
a ayf (y)
(10.32)
On the other hand, using equation (10.31),
f (af (y)) = f (f (y)) −
1
1
1
=a−
−
.
af (y)
yf (y) af (y)
(10.33)
Combining equations (10.32) and (10.33), it follows that
1
1
1
1
+
=
+
.
a ayf (y)
yf (y) af (y)
+
Hence f (y) = 1 + a−1
y , for y ∈ R .
This is the only possible solution of the equation. Now substituting in the last
equation, it follows that (a − 1)2 = 1, but since a > 0, the only choice is a = 2,
and then f (x) = 1 + x1 is the only solution.
Solution 9.78. For x ∈ [0, 1], |f (x)| = |f (x) − f (0)| < |x − 0| = x and |f (x)| =
|f (x) − f (1)| < |x − 1| = 1 − x. Then |f (x)| < min {x, 1 − x}, for x ∈ [0, 1].
If |x − y| ≤ 12 , then |f (x) − f (y)| < |x − y| ≤ 12 .
If |x − y| > 21 , without loss of generality, we can assume that 21 ≤ x ≤ 1 and that
y < 12 . Since |f (x)| < min {x, 1 − x} = 1 − x and also |f (y)| < min {y, 1 − y} = y,
it follows that |f (x) − f (y)| ≤ |f (x)| + |f (y)| < 1 − x + y = 1 − (x − y) < 12 .
Solution 9.79. Let F = {f (n)} and G = {g(n)}, for n = 1, 2, . . . . Since g(1) =
f (f (1)) + 1 > 1, then f (1) = 1 and g(1) = 2. Now we prove that if f (n) = k, then
f (k) = k + n − 1
g(n) = k + n
f (k + 1) = k + n + 1.
(10.34)
(10.35)
(10.36)
If we assume for the moment that these statements are true, they can be applied
to f (1) = 1 to obtain g(1) = 2 and f (2) = 3. If we apply equation (10.34) to
287
10.9 Solutions of Chapter 9
f (2) = 3 and to the next numbers, we obtain a chain of results:
f (3) = 4,
f (4) = 6,
f (6) = 9,
f (9) = 14,
f (14) = 22,
f (22) = 35,
f (35) = 56, f (56) = 90,
f (90) = 145, f (145) = 234, f (234) = 378,
....
But f (240) is not in this chain. Observe that equation (10.36) generates larger
numbers; for instance, if we apply it to f (145) = 234, we get f (235) = 380.
Looking at the previous values of the chain, we can see that applying equation
(10.36), f (56) = 90 and then f (91) = 147, f (148) = 239. Finally, f (240) = 388.
It only remains to prove equations (10.34), (10.35) and (10.36). Assuming that
f (n) = k, it follows that the elements in the two disjoint subsets
{f (1), f (2), . . . , f (k)}
and {g(1), g(2), . . . , g(n)}
cover all the natural numbers from 1 to g(n), since g(n) = f (f (n)) + 1 = f (k) + 1.
Counting the elements in the sets, it follows that g(n) = k + n or g(n) = f (n) + n,
which is equation (10.35). Equation (10.34) follows from k + n = g(n) = f (k) + 1.
From equation g(n) − 1 = f (f (n)), notice that g(n) − 1 is an element of F , that
is, two consecutive integers cannot be elements of G. Since k + n is an element
of G, it follows that both k + n − 1 and k + n + 1 are elements of F , moreover,
they are two consecutive elements of F . Therefore, equation (10.34) implies that
k + n + 1 = f (k + 1).
Solution 9.80. Take x = 0 and y = 1 in equations (9.2) and (9.1) to get f (1, 0) =
f (0, 1) = 2. Moreover, if we take x = 0 and y = 0, by equations (9.3) and (9.1),
we get f (1, 1) = f (0, f (1, 0)) = f (1, 0) + 1 = 3. Now, if we take x = 0 and y = 1
in (9.3) and using the previous results, it follows that f (1, 2) = f (0, f (1, 1)) =
f (1, 1) + 1 = 4. We claim that
f (1, y) = y + 2.
(10.37)
We already have verified this equation for y = 0, 1, 2. The inductive step follows
from (9.3):
f (1, k) = f (0, f (1, k − 1)) = f (1, k − 1) + 1 = (k − 1) + 2 + 1 = k + 2.
Now, using induction, show that f (2, y) = 2y + 3. Observe that f (2, 0) = f (1, 1) =
3; now, from f (2, y + 1) = f (1, f (2, y)) = f (2, y) + 2, we obtain the inductive
step. Also, it follows that f (3, 0) = f (2, 1) = 5 and f (3, y + 1) = f (2, f (3, y)) =
2f (3, y) + 3. Then, we get
f (3, y) = 2f (3, y − 1) + 3 = 2(2f (3, y − 2) + 3) + 3
= · · · = 2y f (3, 0) + (2y−1 + · · · + 2 + 1)3
2y − 1
= 2y+3 − 3.
= 2y (23 − 3) + 3
2−1
288
Chapter 10. Solutions to Exercises and Problems
·2
··
Finally, once again by induction, we show that f (4, y) = 22 − 3, where the tower
of numbers 2 has y + 3 floors.
For this, notice that f (4, 0) = f (3, 1) = 24 − 3 = 13 and f (4, y + 1) =
f (3, f (4, y)) = 2f (4,y)+3 − 3, and the inductive step is completed.
Solution 9.81. Since f (n + m) − f (m) − f (n) = 0 or 1, it follows that f (m + n) ≥
f (m) + f (n).
If m = n = 1, then f (2) ≥ 2f (1), but f (2) = 0, that is, 0 ≥ 2f (1), and since
f (n) ≥ 0, then f (1) = 0.
If m = 2 and n = 1, then f (3) = f (2) + f (1) + {0 or 1} = 0 or 1. Since f (3) > 0,
then f (3) = 1.
If m = n = 3, then f (3 + 3) = f (2 · 3) = f (3) + f (3) + {0 or 1}, that is,
f (3 + 3) ≥ 2 · f (3) = 2. Notice that
f (3 + 6) = f (3 + 2 · 3) ≥ f (3) + f (2 · 3) ≥ f (3) + 2 · f (3) = 3f (3).
Hence, f (3n) = f (3 + (n − 1)3) ≥ f (3) + (n − 1)f (3) = n · f (3) = n.
Therefore f (3n) ≥ n, for all n. If for some n0 the inequality is strict, then
for all n ≥ n0 the inequality is also strict.
Since f (9999) = f (3 · 3333) = 3333, it follows that f (3n) = n, for 3 ≤ n ≤
3333, in particular f (3 · 1982) = 1982. Now,
1982 = f (3 · 1982) = f (2 · 1982 + 1982) ≥ f (2 · 1982) + f (1982) ≥ 3 · f (1982),
then 1982 ≥ 3 · f (1982). Thus, f (1982) ≤ 1982
3 < 661.
On the other hand, f (1982) = f (1980 + 2) ≥ f (1980) + f (2) = f (3 · 660) = 660.
Therefore, 660 ≤ f (1982) < 661, that is, f (1982) = 660.
Solution 9.82. First, we show that 1 is in the image of f . For x0 > 0, let
y0 =
1
.
f (x0 )
Then condition (i) states that f (x0 f (y0 )) = y0 f (x0 ) = 1, that is, 1 is in the image
of f . Then there is a value of y such that f (y) = 1. This, together with x = 1 in
(i), imply that f (1 · 1) = f (1) = yf (1). Since f (1) > 0 by hypothesis, then y = 1
and hence f (1) = 1.
Taking y = x in (i), we obtain that f (xf (x)) = xf (x), for all x > 0. Then
xf (x) is a fixed point of f . Now, if a and b are fixed points of f , then using (i)
with x = a, y = b, we guarantee that f (ab) = ba, hence ab is also a fixed point of
f . Thus, the set of fixed points of f is closed under multiplication. In particular,
if a is a fixed point, all non-negative integer powers of a are fixed points, that is,
289
10.9 Solutions of Chapter 9
an is a fixed point for all non-negative integers n. Then, by (ii), there are no fixed
points greater than 1. Since xf (x) is a fixed point, it follows that
xf (x) ≤ 1 or f (x) ≤
Let a = xf (x), then f (a) = a. Now, use x =
1
f (a)
a
f
then
f
1
a
=
1
a
1
a
1
, for all x.
x
and y = a in (i), to obtain
= f (1) = 1 = af
or f
(10.38)
1
xf (x)
=
1
a
,
1
.
xf (x)
This shows that xf1(x) is also a fixed point of f for all x > 0. Then f (x) ≥ x1 . This
and (10.38) imply that f (x) = x1 . This function clearly satisfies the conditions of
the problem.
Solution 9.83. Suppose that ⌊f (y)⌋ = 0 for some y, then the substitution x = 1
implies that f (y) = f (1)⌊f (y)⌋ = 0. Then if ⌊f (y)⌋ = 0 for all y, it follows that
f (y) = 0 for all y. This function obviously satisfies the conditions of the problem.
Now, we have to consider the case when ⌊f (a)⌋ =
0, for some a. Then it follows
from f (⌊x⌋a) = f (x)⌊f (a)⌋, that
f (x) =
f (⌊x⌋a)
.
⌊f (a)⌋
(10.39)
This means that f (x1 ) = f (x2 ) if ⌊x1 ⌋ = ⌊x2 ⌋, then f (x) = f (⌊x⌋). Hence we can
assume that a is an integer.
Now, we have
1
1
f (a) = f 2a ·
= f (2a) f
= f (2a)⌊f (0)⌋,
2
2
this implies that ⌊f (0)⌋ =
0, then we can assume that a = 0. Therefore, equation
(10.39) implies that f (x) = ⌊ff (0)
(0)⌋ = C = 0, for every x. Now, the condition of the
problem is equivalent to equation C = C⌊C⌋, which is true exactly when ⌊C⌋ = 1.
Then, the only functions that satisfy the conditions of the problem are f (x) = 0
and f (x) = C, with ⌊C⌋ = 1.
Solution 9.84. For x = y = 1, we have f (f (1)) = f (1). Now, if we take x = 1 and
y = f (1), and since f (f (1)) = f (1), it follows that [f (1)]2 = f (1). Then there are
two possibilities, either f (1) = 1 or f (1) = 0.
If f (1) = 1, substituting y = 1 in the functional equation, we get f (xf (1)) +
f (x) = xf (1) + f (x), then f (x) = x, for all x ∈ R. But this function is not
bounded, hence f (1) = 1 is not true.
290
Chapter 10. Solutions to Exercises and Problems
If f (1) = 0, taking x = 1, we get f (f (y)) + yf (1) = f (y) + f (y), then
f (f (y)) = 2f (y). If f (y) ∈ Img f , then 2f (y) ∈ Img f , and by induction, f n (y) =
2n f (y) ∈ Img f . We conclude that f (y) ≤ 0, because if f (y) > 0, it will follow
that 2n f (y) ∈ Img f , for all n, which is impossible since the function f is upper
bounded.
that f (f (y)) = 2f (y),
Substituting x by x2 and
x y by f (y) and,
x noticing
we obtain
f
(xf
(y))
+
f
(y)f
f
(y)
.
Then
xf (y) − f (xf (y)) =
=
xf
(y)
+
f
2
2
f (y)f x2 − f x2 f (y) ≥ 0, since f (x) ≤ 0, for all x. All these results together and
equation (9.4) imply that yf (x) ≥ f (xy).
Considering that yf (x) ≥ f (xy), and taking x > 0, y = x1 , we obtain f (x) ≥
0. Since f (x) ≤ 0, then f (x) = 0. Clearly f (x) = 0, for all x ∈ R, satisfies the
functional equation and it is bounded.
Suppose that f is not identically zero, then there exists x0 < 0 such that
f (x0 ) < 0. Let y0 = f (x0 ), then f (y0 ) = f (f (x0 )) = 2f (x0 ) = 2y0 . For any x < 0,
it follows that y0 x > 0, then f (y0 x) = f (2y0 x) = 0. Hence, after substituting y
by y0 in (9.4), we get f (2y0 x) + y0 f (x) = 2y0 x + f (xy0 ), thus f (x) = 2x, for all
x < 0. Hence, the other solution is f (x) = 0 for x ≥ 0, and f (x) = 2x, for x < 0.
Therefore, the only solutions are
0,
if x ≥ 0,
f (x) = 0 and f (x) =
2x,
if x < 0.
It is easy to verify that these functions satisfy the conditions of the problem.
Solution 9.85. First notice that a2n > 1 and a2n+1 < 1, for all n ≥ 1.
The proof of the following statement will be by induction for k ≥ 2.
Pk : For every pair of positive integers a, b with (a, b) = 1 and a + b ≤ k,
there exists an integer n such that an = ab .
If k = 2, the only positive integers a, b that satisfy (a, b) = 1 and a + b ≤ 2, are
a = b = 1, and in such a case a1 satisfies ab = a1 = 1.
Now, suppose that the statement is true for k ≥ 2; we will prove that it is valid
for k + 1. Let a, b be positive integers that satisfy (a, b) = 1 and a + b ≤ k + 1.
If a > b, then a − b and b satisfy that (a − b, b) = 1 and (a − b) + b = a ≤ k, then
by the induction hypothesis, there is an integer n with an = a−b
b . Then
a2n = an + 1 =
a−b
a
+1= .
b
b
If a < b, then b − a and a satisfy that (b − a, a) = 1 and (b − a) + a = b ≤ k; hence
by the induction hypothesis, there is an integer n with an = b−a
a . Then
a2n+1 =
1
1
=
=
a2n
an + 1
1
a
= .
b
+1
b−a
a
291
10.9 Solutions of Chapter 9
Let us now see the uniqueness of the representation.
If an = am > 1, then m and n are even. However, if an = am < 1, then m and n
are odd. In the first case, n = 2n′ and m = 2m′ , and then an = am implies that
an′ = am′ . In the second case, n = 2n′ + 1 and m = 2m′ + 1, then an = am implies
that a 1 ′ = a 1 ′ , and then an′ = am′ . Thus, in any case the equality an = am leads
2n
2m
to another equality an′ = am′ , where the subindices are smaller. This process can
only be applied a finite number of steps to conclude that n′ = m′ = 1 or n′ = 1 or
m′ = 1. In the first case, it follows that n = m, and in the other cases we conclude
that an′ = a1 = 1 or am′ = a1 = 1, but this is not possible, since at the beginning
we pointed out that the an ’s are not equal to 1, when n > 1.
Solution 9.86. Since an+1 =
1
an + n. Then
an
1+nan ,
for n ≥ 0, it follows that
1
an+1
=
1+nan
an
=
1
1
=
+ 999
a1000
a999
1
+ 999 + 998
=
a998
..
=
.
1
+ 999 + 998 + · · · + 1
=
a1
999 · 1000
= 1+
= 499501.
2
Therefore a1000 =
1
499501 .
Solution 9.87. From the definition of xi it follows that for every integer k,
x4k−3 = x2k−1 = −x4k−2 and x4k−1 = x4k = −x2k = xk .
*
Then, if we set Sn = ni=1 xi , it follows that
S4k =
k
#
((x4k−3 + x4k−2 ) + (x4k−1 + x4k )) =
k
#
(0 + 2xk ) = 2Sk ,
(10.40)
(10.41)
i=1
i=1
S4k+2 = S4k + (x4k+1 + x4k+2 ) = S4k .
(10.42)
*n
*n
Also, observe that Sn = i=1 xi ≡ i=1 1 = n mod 2. We will show, by induction on k, that Si ≥ 0, for all i ≤ 4k. The basis is true since x1 = x3 = x4 = 1,
x2 = −1. For the inductive step, suppose that Si ≥ 0, for all i ≤ 4k. Using relations
(10.40), (10.41) and (10.42), we obtain that
S4k+4 = 2Sk+1 ≥ 0,
S4k+3
S4k+2 = S4k ≥ 0,
S4k+2 + S4k+4
= S4k+2 + x4k+3 =
≥ 0.
2
292
Chapter 10. Solutions to Exercises and Problems
Then, it will be enough to show that S4k+1 ≥ 0. If k is odd, then S4k = 2Sk ≥ 0;
since k is odd, Sk is also odd, and then S4k ≥ 2. Therefore S4k+1 = S4k +x4k+1 ≥ 1.
Reciprocally, if k is even, then x4k+1 = x2k+1 = xk+1 . Thus S4k+1 = S4k +x4k+1 =
2Sk + xk+1 = Sk + Sk+1 ≥ 0 and the induction is complete.
Solution 9.88. First, from the problem conditions, it follows that each an (n > s)
can be expressed as an = aj1 + aj2 with j1 , j2 < n, j1 + j2 = n. If, say, j1 > s, then
we can proceed in the same way with aj1 , and so on. Finally, we represent an as
an = ai1 + · · · + aik ,
1 ≤ ij ≤ s,
(10.43)
i1 + · · · + ik = n.
(10.44)
Moreover, if ai1 and ai2 are the numbers in (10.43), obtained on the last step, then
i1 + i2 > s. Hence we can adjust (10.44) as
1 ≤ ij ≤ s,
i1 + · · · + ik = n,
i1 + i2 > s.
(10.45)
On the other hand, suppose that the indices i1 , . . . , ik satisfy conditions (10.45).
Then, writing sj = i1 + · · · + ij , from the problem’s condition we have
an = ask ≥ ask−1 + aik ≥ ask−2 + aik−1 + aik ≥ · · · ≥ ai1 + · · · + aik .
Summarizing these observations, we get the following lemma.
Lemma 10.9.1. For every n > r, we have an = max {ai1 + · · · + aik }, where the
collection (i1 , . . . , ik ) satisfies (10.45).
)
(
Now we write r = max aii | 1 ≤ i ≤ s , and fix some index l ≤ s, such that
s = all .
Consider some integer n ≥ s2 l + 2s and choose an expansion of an in the form
(10.43), (10.45). Then we have n = i1 + · · · + ik ≤ sk, so k ≥ n/s ≥ sl + 2. Suppose
that none of the numbers i3 , . . . , ik equals l. Then by the pigeonhole principle there
is an index 1 ≤ j ≤ s which appears among i3 , . . . , ik at least l times, and surely
j = l. Let us delete these l occurrences of j from (i1 , . . . , ik ), and add j occurrences
of l instead, obtaining a sequence (i1 , i2 , i′3 , . . . , i′k′ ) also satisfying (10.45). Using
the lemma, we have
ai1 + · · · + aik = an ≥ ai1 + ai2 + ai′3 + · · · + ai′k′ ,
or, after removing the common terms, laj ≥ jal , then
leads to laj = jal , hence
al
l
≤
aj
j .
The definition of l
an = ai1 + ai2 + ai′3 + · · · + ai′k′ .
Thus, for every n ≥ s2 l + 2s, we have found a representation of the form (10.43),
(10.45) with ij = l, for some j ≥ 3. Rearranging the indices we may assume that
ik = l.
293
10.9 Solutions of Chapter 9
Finally observe that in this representation the indices (i1 , . . . , ik−1 ) satisfy the
conditions (10.45), with n replaced by n − l. Thus, from the lemma, we get an−l +
al ≥ (ai1 + · · · + aik−1 ) + al = an , which, by the problem’s condition, implies
an = an−l + al for each n ≥ s2 l + 2s, as desired.
Solution 9.89. For each i = 1, 2, . . . , k − 1, let Pi be the set of all prime numbers
congruent to i modulo k. Each prime number (except possibly k) is contained in
exactly one of the sets P1 , P2 , . . . , Pk−1 . Since there are an infinite number of
prime numbers, at least one of these sets is infinite, say Pi . Let p = x1 < x2 < · · · <
−p
,
xn < · · · be its elements arranged in increasing order, and define an = xn+1
k
for n = 1, 2, . . . .
Then the sequence p + kan contains all members of Pi , starting with x2 . The
numbers an are positive integers, p is prime and the sequence {an } is increasing,
hence it satisfies the conditions of the problem.
Solution 9.90. The hypothesis implies a + b + c = 2, ab + bc + ca = −1 and abc = 0.
2
Then a,
x3 − 2x√
− x = 0, which are
√
√ b, c are the roots of the cubic polynomial
0, 1 ± 2. Then we can assume that a = 1 + 2, b = 1 − 2 and c = 0. Hence,
a2 + b2 = 6 and ab = −1.
Now, we have that sn−1 sn+1 = (an−1 + bn−1 )(an+1 + bn+1 ) = a2n + b2n +
n−1 n−1 2
a
b
(a + b2 ) = s2n − 2an bn + (ab)n−1 (a2 + b2 ) = s2n − 2(ab)n + 6(ab)n−1 =
2
sn − 8(−1)n , then s2n − sn−1 sn+1 = |8(−1)n | = 8.
Solution 9.91. (i) We show that the series
x20
x2
x2
+ 1 + 2 + ··· ,
x1
x2
x3
with 1 = x0 ≥ x1 ≥ · · · > 0,
(10.46)
has sum greater than or equal to 4. This clearly implies that some partial sum of
the series is greater than or equal to 3.999.
Let L be the infimum (the greatest lower bound) of the sum of all series of the
form (10.46). Clearly L ≥ 1, since the first term x11 ≥ 1. For all ǫ > 0 we can find
a sequence {xn } such that
L+ǫ>
x20
x2
x2
+ 1 + 2 + ··· .
x1
x2
x3
(10.47)
Setting yn = xn+1
x1 , with n ≥ 0, it follows that 1 = y0 ≥ y1 ≥ y2 ≥ · · · > 0. The
series on the right-hand side of the inequality (10.47) can be written as
1
+ x1
x1
y02
y2
y2
+ 1 + 2 + ···
y1
y2
y3
.
By definition of L, the series inside the parenthesis has a sum greater than or
equal to ≥ L. Hence, by (10.47), we have that L + ǫ > x11 + x1 L.
294
Chapter 10. Solutions to Exercises and Problems
Applying the inequality between the arithmetic mean and √the geometric
mean on the right-hand side of the inequality
we get L + ǫ > 2 L. Since this
√
is true for all ǫ > 0, it follows that L ≥ 2 L. Thus, L2 ≥ 4L, and since L > 0,
this implies L ≥ 4.
(ii) Let xn =
1
2n ,
then
∞
∞
#
#
1
x2n
=
= 4,
n−1
x
2
n=0
n=0 n+1
and the partial sums of this series are less than 4.
Solution 9.92. By (i) and (ii), it follows that all the elements in the sequence are
rational numbers. Suppose that ak = pq , with p and q integers such that (p, q) = 1,
2
2
)
then ak+1 = (2p q−q
. Since (p, q) = 1, it follows that (2p2 − q 2 , q 2 ) = (2p2 , q 2 ) =
2
(2, q 2 ), which is equal to either 1 or 2.
If q > 2, then the denominator of ak+1 is greater than the denominator of ak .
Hence, the sequence of denominators will be increasing and therefore it cannot be
an equality among the terms of the sequence.
Moreover, if |ak | > 1, then writing |ak | = 1+e, it follows that ak+1 = 1+4e+2e2 >
|ak |, which gives an increasing sequence.
Therefore, in order that the terms of the sequence repeat themselves, the first term
must have denominator at most 2 and must be between −1 and 1, this gives us
only 5 possible values:
•
•
•
•
•
a0
a0
a0
a0
a0
= −1, which gives the sequence −1, 1, 1, 1, . . . .
= − 12 , which gives the sequence − 21 , − 21 , − 21 , . . . .
= 0, which gives the sequence 0, −1, 1, 1, 1, . . . .
= 12 , which gives the sequence 12 , − 21 , − 21 , − 21 , . . . .
= 1, which gives the sequence 1, 1, 1, 1, . . . .
Solution 9.93. The first terms of the sequence are a0 = 0, a1 = 1, a2 = 2,
a3 = 5, a4 = 12, a5 = 29, a6 = 70, a7 = 169. Observe that an+1 = 2an + an−1 =
a2 an + a1 an−1 and an+2 = 2an+1 + an = 2(2an + an−1 ) + an = 5an + 2an−1 =
a3 an + a2 an−1 . Then we can conjecture that
an+m = am+1 an + am an−1 .
(10.48)
To show it we use induction on m. For m = 1, 2 it was already proved. Now,
suppose that the equality is true for m = k − 1 and for m = k, and show that the
relation holds for k + 1. Observe that
an+k+1 = 2an+k + an+k−1 = 2(ak+1 an + ak an−1 ) + ak an + ak−1 an−1
= (2ak+1 + ak )an + (2ak + ak−1 )an−1 = ak+2 an + ak+1 an−1 ,
which finishes the induction.
295
10.9 Solutions of Chapter 9
If we let n = m in (10.48), we obtain the equality a2n = an (an+1 + an−1 ).
Also note that if n is even, then an is even, which is easy to deduce using
the formula of the sequence an = 2an−1 + an−2 . Now, if n is odd, then an ≡ 1
mod 4. In order to see this, use induction again. For n = 1, the result follows since
a1 = 1, then assume that a2n−1 ≡ 1 mod 4. Then, since a2n+1 = 2a2n + a2n−1 ,
we only need to observe that a2n is even, which finishes the induction.
Summarizing, if n is odd, that is, if there is no power of 2 dividing n, then
an ≡ 1 mod 4, hence there is no power of 2 dividing an .
If n is even, then n − 1 and n + 1 are odd, hence an+1 + an−1 ≡ 2 mod 4,
that is, in the equality a2n = an (an+1 + an−1 ), only one extra factor of 2 appears
inside the parenthesis, and this ends the proof.
Solution 9.94. Since
|am+1 + an − am+n+1 | ≤
1
,
m+n+1
|am + an+1 − am+n+1 | ≤
1
,
m+n+1
from the two inequalities it follows that
|(am+1 − am ) − (an+1 − an )| ≤
2
2
< .
m+n+1
n
Now, using twice the previous inequality, we get
|(am+1 − am ) − (an+1 − an )|
≤ |(am+1 − am ) − (ak+1 − ak ) + (ak+1 − ak ) − (an+1 − an )|
2
2
4
+ = .
k k
k
Since k is arbitrary, |(am+1 − am ) − (an+1 − an )| is equal to 0, then an+1 − an is
constant.
≤ |(am+1 − am ) − (ak+1 − ak )| + |(ak+1 − ak ) − (an+1 − an )| ≤
Solution 9.95. The function f (x) = nx + nx is decreasing in (0, n] (it can be seen
that f ′ (x) < 0, or that the graph is decreasing with the identity f (x) − f (y) =
√
(x−y)(xy−n2 )
n
, for n ≥ 3.
). First we see, using induction in n, that n ≤ an ≤ √n−1
xyn
√
3
√
For n = 3 is clear, since 3 ≤ a3 = 2 ≤ 2 . Suppose the result true for n,
%
$
√
√ . On the other hand, an+1 = f (an ) ≥ f √ n
=
an+1 = f (an ) ≤ f ( n) = n+1
n
n−1
√
n
√
> n + 1, then the result follows.
n−1
√
n
Now, let us see that an < n + 1. If an+1 = f (an ) ≥ √n−1
is true for n ≥ 3,
n−1
then an ≥ √n−2 for n ≥ 4. Since f is decreasing, then
an+1 = f (an ) ≤ f
n−1
√
n−2
=
(n − 1)2 + n2 (n − 2) √
√
< n + 2,
(n − 1)n n − 2
for n ≥ 4. Hence ⌊a2n ⌋ = n, for n ≥ 5. The case n = 4 follows easily after noticing
that a4 = 13
6 .
296
Chapter 10. Solutions to Exercises and Problems
Solution 9.96. Denote by l(n) the last digit of a positive integer n, that is, the
units digit. The sequence {l(n)} is periodic with period 10. Now, for a fixed positive
integer a, the sequence {l(an )} is periodic with period equal to 1 if a ends in 0, 1,
5, 6; the period is equal to 2 if a ends in 4 or 9; and the period is 4 if a ends in 2,
3, 7, 8.
Since the least common multiple of 10 and 4 is 20, and if
m = (n + 1)n+1 + (n + 2)n+2 + · · · + (n + 20)n+20 ,
then l(m) does not depend on n. We now calculate the last digit of 11 + 22 + 33 +
· · · + 2020 . By the periodicity of the sequence of the form {l(an )}, the last digit of
this number is the same as the one in
1 + 22 + 33 + 44 + 5 + 6 + 73 + 84 + 9 + 1 + 24 + 3 + 42 + 5 + 6 + 7 + 82 + 9.
The units digit of this last number is 4. Therefore, the last digit of a sum of the
form (n + 1)n+1 + (n + 2)n+2 + · · · + (n + 100)n+100 is equal to l(4 · 5) = l(20) = 0.
Thus, bn is periodic with period 100.
√
√
Solution 9.97. Notice that (1 + 3)2n+1 + (1 − 3)2n+1 is an even integer. In
order to see this, simplify the binomials to obtain
√
√
(1 + 3)2n+1 + (1 − 3)2n+1
2n+1
# 2n + 1 $ √ %j
# 2n + 1 $√ %j 2n+1
3 +
=
− 3
j
j
j=0
j=0
=2
n
n
#
#
2n + 1 $√ %2j
2n + 1 j
=2
3
3 .
2j
2j
j=0
j=0
√
√
since −1< 1− 3 < 0, it follows that −1 < (1− 3)2n+1 < 0, therefore
Now,
√ 2n+1
√ 2n+1
√
√ 2n+1
√
1+ 3
+ 1− 3
= 1+ 3
= (1+ 3)2n+1 +(1− 3)2n+1
is an even integer.
√ 2n+1
. To see this, observe that
It can be shown that 2n+1 divides 1 + 3
√ %2n+1 $
√
√
√
√ %2n+1
√
3
+ 1− 3
= (1 + 3)(4 + 2 3)n + (1 − 3)(4 − 2 3)n
√
√
√
√
= 2n (1 + 3)(2 + 3)n + 2n (1 − 3)(2 − 3)n .
√
√
√
√
It is only left to see that 2 divides (1 + 3)(2 + 3)n + (1 − 3)(2 − 3)n ,
and for that use again Newton’s binomial theorem.
$
1+
Solution 9.98. The characteristic polynomial of the recursion an+2 = 3an+1 − 2an
is λ2 −3λ+2, which has as roots λ = 2, 1. Then an = Aλn +B for some real numbers
A and B which are determined by 3 = a1 = A · 2 + B and 5 = a2 = A · 22 + B.
The previous system has solutions A = B = 1, then an = 2n + 1, for n ≥ 1.
297
10.9 Solutions of Chapter 9
Solution 9.99. The relation an+6 = an follows from the relation an+2 = an+1 −
an , since an+6 = an+5 − an+4 = an+4 − an+3 − an+4 = −(an+2 − an+1 ) =
−(an+1 − an − an+1 ) = an .
Solution 9.100. Since a2 = 22 , a3 = 52 , a4 = 132 , we can conjecture that an =
2
. Use strong induction in order to show the result. A relation between 3
F2n−1
Fibonacci numbers that helps is F2n+1 = 3F2n−1 − F2n−3 .
Solution 9.101. Calculate the first terms,
u1 =
1
5
=2+
2
2
u2 = u1 (u20 − 2) − u1 =
u4 =
=
u5 =
2+
1
2
1
23
1
23 + 3
2
1
25 + 5
2
23 +
(22 − 2) − 2 +
$
1
2
=2+
1
2
1 %2
1
−2 − 2+
2
2
1
1
1
1
22 + 2 − 2 +
= 2+
22 − 1 + 2
2
2
2
2
$
1 %2
1
2+
−2 − 2+
2
2
1
1
1
22 + 2 − 2 +
= 25 + 5
2
2
2
$
1
1
1 %2
23 + 3 − 2 − 2 +
= 211 + 11 .
2
2
2
u3 = u2 (u21 − 2) − u1 =
=
1
2
1
2+
2
2+
2+
= 23 +
1
23
This allows us to conjecture that un = 2rn + 2r1n for some numbers rn which must
be determined. If the numbers rn are integers, then ⌊un ⌋ = 2rn .
If un = 2rn + 2r1n , then from the original equation it follows that
2
1
1
1
rn−1
rn
un+1 = 2 + rn
+ rn−1
2
−2 − 2+
2
2
2
1
= 2rn + 2−rn 22rn−1 + 2−2rn−1 − 2 +
2
= 2rn +2rn−1 + 2−rn −2rn−1 + 2rn −2rn−1 + 2−rn+2rn−1 − 2 +
1
2
.
Therefore, if we can find a sequence {rn } that satisfies rn+1 = rn + 2rn−1 and
rn − 2rn−1 = (−1)n , the proof is complete.
The characteristic equation of the recursion rn+1 = rn + 2rn−1 is λ2 − λ− 2 =
0, and its roots are λ = 2, −1, and since r0 = 0 and r1 = 1, it follows that
n
n
is the solution, and it is also solution of the other recursion.
rn = 2 −(−1)
3
Hence, ⌊un ⌋ = 2
2n −(−1)n
3
, for all n ∈ N.
298
Chapter 10. Solutions to Exercises and Problems
√
Solution 9.102. First √
notice that the integers n that satisfy ⌊ 2n⌋ = m are the
ones that satisfy m ≤ 2n < m + 1. Then m2 ≤ 2n < (m + 1)2 , but even numbers
between m2 (inclusive) and (m + 1)2 are m, if m is odd, and m + 1 if m is even.
Hence the sequence an is
1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, . . .,
where the integer k appears k times in the sequence if k is odd or k + 1 times if k
is even.
When we write all the numbers 63, we arrive to the term an with n =
1 + 3 + 3 + 5 + 5 + · · · + 63 + 63 = 2(1 + 3 + · · · + 63) − 1 = 2(32)2 − 1 = 2047.
This last term and the 62 previous ones are equal to 63, in particular a2020 = 63.
Solution 9.103. Since P (0) = 0, it follows that P (x) = xQ(x), for some polynomial
1
Q(x) of degree n − 1 that satisfies Q(j) = j+1
, for j = 1, 2, . . . , n. If R(x) =
(x + 1)Q(x) − 1, then the degree of R(x) is n and R(j) = 0, for j = 1, 2, . . . , n.
Hence, R(x) = a0 (x − 1)(x − 2) . . . (x − n). If m > n,
Q(m) =
a0 (m − 1)(m − 2) . . . (m − n) + 1
.
m+1
Evaluating R(x) in x = −1, it follows that −1 = a0 (−2)(−3) . . . (−n − 1), then
n+1
a0 = (−1)
(n+1)! . Hence,
P (m) = mQ(m) =
(−1)n+1 m(m − 1)(m − 2) . . . (m − n)
m
+
.
(n + 1)!
m+1
m+1
Solution 9.104. Consider Q(x) = xP (x) − 1 = c(x − 1)(x − 2) . . . (x − 2n ). For
x = 1, 2, . . . , 2n , it follows that
1
1
1
Q′ (x)
=
+
+ ··· +
.
Q(x)
x−1 x−2
x − 2n
Since Q(0) = −1 and Q′ (x) = P (x) + xP ′ (x), then
1 1
1
P (0) = Q′ (0) = − − − − · · · − n
1 2
2
=
1
1 − 2n+1
1
= 2 − n.
1
2
1− 2
Solution 9.105.
Lemma. If P (x) is a polynomial of degree less than or equal to n, then
n+1
#
(−1)i
i=0
n+1
P (i) = 0.
i
Proof of the lemma. Proceed by induction on n. For n = 0,
P0 − P0 = 0.
1
1
0 P (0) − 1 P (1) =
299
10.9 Solutions of Chapter 9
Suppose valid the result for all n ≤ k and consider the polynomial P (x) of degree
k + 1. The polynomial P (x) − P (x + 1) has degree less than or equal to k, then
0=
k+1
#
(−1)i
i=0
k+1
#
k+1
(P (i) − P (i + 1))
i
k+2
#
k+1
k+1
(−1)i
P (i)
P (i) +
i−1
i
i=1
i=0
,
+
k+1
#
k+1
k+1
= P (0) +
(−1)i
+
P (i) + (−1)k+2 P (k + 2)
i
i
−
1
i=1
=
=
k+2
#
(−1)i
(−1)i
i=0
k+2
P (i).
i
Hence, for the polynomial P (x) the result holds and therefore the proof of
the lemma is complete.
Now, apply the lemma to the polynomial P (x) of the problem,
0=
n+1
#
(−1)i
i=0
=
n
#
n
#
n+1
n+1
(−1)i
P (i) =
i
i
i=0
n+1
i
−1
+ (−1)n+1 P (n + 1)
(−1)i + (−1)n+1 P (n + 1),
i=0
hence P (n + 1) is 1 if n is even and it is 0 if n is odd.
Solution 9.106. Suppose that there are polynomials Q(x) and R(x) with integer
coefficients such that P (x) = Q(x)R(x). Since P (0) = 3, we can assume, without
loss of generality, that |Q(0)| = 3. If Q(x) = xk + ak−1 xk−1 + · · · + a0 , with
a0 = ±3, R(x) = xl + bl−1 xl−1 + · · · + b0 and P (x) = xn + cn−1 xn−1 + · · · + c0 , it
follows that cj = aj b0 + aj−1 b1 + · · · .
Let j be the smallest index such that 3 does not divide aj , then 3 neither
divides cj , since 3 ∤ b0 . Hence, j ≥ n − 1, and then k ≥ n − 1 and l ≤ 1. Thus
the polynomial R(x) has the form ±x ± 1, but neither 1 or −1 are roots of P (x).
Therefore, P (x) is irreducible over Z[x].
Solution 9.107. Suppose that the degree of Q(x) is n ≤ p − 2. Using the lemma
in Problem 9.105,
0=
p−1
p−1
#
#
p−1
Q(i) mod p,
Q(i) ≡
(−1)i
i
i=0
i=0
since p−1
≡ (−1)i mod p. But this is not possible if Q(0) = 0, Q(1) = 1 and
i
Q(i) ≡ 0, 1 mod p.
300
Chapter 10. Solutions to Exercises and Problems
Solution 9.108. Since P (x) > 0, for x ≥ 0, the polynomial can be decomposed in
the following way
P (x) = a0 (x + a1 ) . . . (x + an )(x2 − b1 x + c1 ) . . . (x2 − bm x + cm ),
with ai > 0, for 0 ≤ i ≤ n, and each quadratic polynomial x2 − bj x + cj has no
real roots.
Since the product of polynomials with positive coefficients is a polynomial with
positive coefficients, and since the factors (x+ai ) already have positive coefficients,
it will be enough to analyze the quadratic factors.
Let x2 − bx + c be a quadratic polynomial, with b2 − 4c < 0. Then
n
#
n i 2
x (x − bx + c)
i
i=0
,
n+2
#+ n
n
n
=
−b
+c
xi
i
−
2
i
−
1
i
i=0
(1 + x)n (x2 − bx + c) =
=
n+2
#
Ci xi ,
i=0
where
+
,
n
n
n
Ci =
−b
+c
i−2
i−1
i
!
"
n! (b + c + 1)i2 − ((b + 2c)n + (2b + 3c + 1))i + c(n2 + 3n + 2)
.
=
i!(n − i + 2)!
Now, Ci will be negative if its discriminant is negative (depends of i). The discriminant is
D = ((b + 2c)n + (2b + 3c + 1))2 − 4(b + c + 1)(c(n2 + 3n + 2))
= (b2 − 4c)n2 − 2U n + V,
where U = 2b2 + bc + b − 4c and V = (2b + c + 1)2 − 4c. But since b2 − 4c < 0,
it will be sufficient to take n large enough, and then we will do this with every
quadratic factor.
Solution 9.109. Notice that
xP (x) = yP (y)
⇔
⇔
a(x4 − y 4 ) + b(x3 − y 3 ) + c(x2 − y 2 ) + d(x − y) = 0
a(x3 + x2 y + xy 2 + y 3 ) + b(x2 + xy + y 2 ) + c(x + y) + d = 0.
301
10.9 Solutions of Chapter 9
2
If u = x + y and v = x2 + y 2 , then x2 + xy + y 2 = u 2+v and the previous equation
becomes auv+ 2b (u2 +v)+cu+d = 0, or equivalently, (2au+b)v = −(bu2 +2cu+2d).
2
Since v ≥ u2 , it is clear that u2 |2au + b| ≤ 2 bu2 + 2cu + 2d, which is
true only for a finite number of values of u. Since there are an infinite number
of pairs of integers (x, y) with xP (x) = yP (y), there is an integer u such that
xP (x) = (u − x)P (u − x) for an infinite number of integers x, but since P (x) is a
polynomial, the latter is true for every real number x.
If u = 0, then u is a root of P (x).
If u = 0, then P (x) = −P (−x); this implies b = d = 0. Then P (x) = ax3 + cx =
x(ax2 + c), hence u = 0 is a root of P (x).
√
Solution 9.110. It is clear that (xyz)2 = abc, then xyz = ± abc, hence x =
y=
√
± abc
,
c
z=
√
± abc
a
√
± abc
,
b
solve the system.
+ b + c), xy = 21 (a + b −
, y =
c), yz = 12 (b + c − a), zx = 12 (c + a − b). Hence x = ± (c+a−b)(a+b−c)
2(b+c−a)
(b+c−a)(a+b−c)
(b+c−a)(c+a−b)
±
,z=±
solve the system.
2(c+a−b)
2(a+b−c)
Solution 9.111. It is clear that xy + yz + zx =
1
2 (a
Solution 9.112. The system is equivalent to the following system:
1
1
1
+
= ,
xy
xz
a
1
1
1
+
= ,
yz yx
b
1
1
1
+
= .
zx zy
c
By the previous problem,
1
=±
x
( 1c +
1
a
− 1b )( 1a + 1b − 1c )
=±
2( 1b + 1c − a1 )
(ab + bc − ca)(bc + ca − ab)
2abc(ca + ab − bc)
and similarly for the other variables.
Solution 9.113. If A, B, C are the expressions on the left-hand sides of the equations, it follows that −A + B + C = (−a + b + c)3 , A − B + C = (a − b + c)3 ,
A + B − C = (a + b − c)3 and −A + B + C = A − B + C = A + B − C = 1. The
system is equivalent to −a + b + c = 1, a − b + c = 1, a + b − c = 1, which has the
unique solution (a, b, c) = (1, 1, 1).
Solution 9.114. Proceed by induction on n. The case n = 1 is trivial. Suppose
that it is true for n > 1. The polynomial Q(x) = P (x + 1) − P (x) has degree n − 1
and takes integer values in the integers, then by the induction hypothesis, there
are integers a0 , . . . , an−1 such that
Q(x) = an−1
x
x
.
+ · · · + a0
0
n−1
302
Chapter 10. Solutions to Exercises and Problems
For any integer x >0, it follows
that P (x) = P(0) +
Q(0) + Q(1) + · · · + Q(x − 1).
x
= k+1
, for any integer k, we obtain
Using the identity k0 + k1 + · · · + x−1
k
the desired representation of P (x),
P (x) = an−1
x
x
+ P (0).
+ · · · + a0
1
n
1
Solution 9.115. Let r be a zero of P (x). Then |r|p −|r| ≤ |rp −r| = p. If |r| ≥ p p−1 ,
then
1
|r|p − |r| = |r|(|r|p−1 − 1) ≥ p p−1 (p − 1) > p,
1
which is a contradiction. Here we have used p p−1 >
p−1
p−1
p
p−1 ,
which follows from
1
p−1
p
.
= ((p − 1) + 1) . Therefore, |r| < p
Suppose that P (x) is the product of two non-constant polynomials Q(x) and R(x)
with integer coefficients. One of these polynomials, say Q(x), has constant term
equal to ±p. On the other hand, the zeros r1 , r2 , . . . , rk of Q(x) satisfy |r1 |, . . . ,
1
|rk | < p p−1 , and moreover r1 · · · rk = ±p, hence k ≥ p, which is impossible.
Solution 9.116. Let P (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 . For every x,
the triplet (a, b, c) = (6x, 3x, −2x) satisfies the condition ab + bc + ca = 0. The
condition in P (x) implies that P (3x) + P (5x) + P (−8x) = 2P (7x), for all x. Now
comparing coefficients on both sides of the equality, it follows that the number
K(i) = (3i + 5i + (−8)i − 2 · 7i ) = 0, if ai = 0. Since K(i) is negative for i odd
and positive for i = 0 or for i ≥ 6 even, then ai = 0, for i = 2, 4. Therefore,
P (x) = a2 x2 + a4 x4 , for any real numbers a2 and a4 . It is easy to see that all
polynomials of the previous form satisfy the conditions set for the problem.
Solution 9.117. We have shown that if for some integer t, Q(t) = t, then P (P (t)) =
t (see Exercise 8.16). If such t also satisfies P (t) = t, the number of solutions is
clearly at most the degree of P (x), which is equal to n.
Let P (t1 ) = t2 , P (t2 ) = t1 , P (t3 ) = t4 and P (t4 ) = t3 , where t1 = ti , for
i = 2, 3, 4. Then we have that t3 − t1 divides t4 − t2 and vice versa; therefore
t3 − t1 = ±(t4 − t2 ). Similarly, we have that t3 − t2 = ±(t4 − t1 ).
Suppose that we have positive signs in both equalities: t3 − t2 = t4 − t1
and t3 − t1 = t4 − t2 . Substracting these equalities, we find t1 − t2 = t2 − t1 ,
which is a contradiction. Then, at least one of the equalities has negative sign.
For each one of those cases, this means that t3 + t4 = t1 + t2 , or equivalently,
t1 + t2 − t3 − P (t3 ) = 0. Let C = t1 + t2 , then it has been proved that each integer
number that is a fixed point of Q(x), different from t1 and t2 , is a root of the
polynomial F (x) = C − x − P (x). This is also valid for t1 and t2 , and since P (x)
has degree n > 1, the polynomial F (x) has the same degree, therefore it has no
more than n roots. Hence, we have reached the result.
10.9 Solutions of Chapter 9
303
Solution 9.118. For the first positive integers, it follows that
P (1) = P (02 + 1) = P (0)2 + 1 = 1
P (2) = P (12 + 1) = P (1)2 + 1 = 2
P (5) = P (22 + 1) = P (2)2 + 1 = 5
P (26) = P (52 + 1) = P (5)2 + 1 = 26.
Then, for x0 = 0 and for n ≥ 1, it follows that xn = x2n−1 + 1. Then P (xn ) =
P (x2n−1 + 1) = P (x2n−1 ) + 1 = x2n−1 + 1 = xn . Hence, P (x) has an infinite number
of fixed points, therefore P (x) = x.
Solution 9.119. Let P (1) = a, then it follows that a2 − 2a − 2 = 0. Since P (x) =
(x − 1)P1 (x) + a, for some polynomial P1 (x), substituting in the original equation
and simplifying leads to (x − 1)P1 (x)2 + 2aP1 (x) = 4(x + 1)P1 (2x2 − 1). For
x = 1, it follows that 2aP1 (1) = 8P1 (1), and together with a = 4, implies that
P1 (1) = 0. Hence, P1 (x) = (x − 1)P2 (x), for some polynomial P2 (x). Then P (x) =
(x − 1)2 P2 (x) + a.
Suppose that P (x) = (x − 1)n Q(x) + a, where Q(x) is a polynomial with Q(1) = 0.
Again, substituting in the original equation and simplifying, we get (x−1)n Q(x)2 +
2aQ(x) = 2(2x + 2)n Q(2x2 − 1), which implies that Q(1) = 0, which is a contradiction. Therefore P (x) = a.
Solution 9.120. It is clear that P (x) and Q(x) have the same degree, say n. The
cases n = 0 and n = 1 are clear. Suppose that R(x) = P (x) − Q(x) = 0 and that
0 < k ≤ n − 1 is the degree of R(x), then
P (P (x)) − Q(Q(x)) = [Q(P (x)) − Q(Q(x))] + R(P (x)).
Writing Q(x) = xn + · · · + a1 x + a0 , we obtain
Q(P (x)) − Q(Q(x)) = [P (x)n − Q(x)n ] + · · · + a1 [P (x) − Q(x)].
The main coefficient of the polynomial Q(P (x)) − Q(Q(x)) is n and it is equal
2
to the coefficient of the term xn −n+k . On the other hand, the degree of the
polynomial R(P (x)) is equal to kn < n2 − n + k. Therefore the main coefficient of
P (P (x))−Q(Q(x)) is n, which is a contradiction with the fact that the polynomial
is zero.
It is left to prove the case R(x) equal to some constant c. Then the condition
P (P (x)) = Q(Q(x)) implies that Q(Q(x) + c) = Q(Q(x)) − c, hence the equality
Q(y + c) = Q(y) − c follows for an infinite number of values of y. Thus Q(y + c) ≡
Q(y) − c, which is only possible for c = 0, which can be proved comparing the
coefficients and using that Q(x) is monic.
Notation
The following notation is standard:
N
Z
Q
Q+
R
R+
I
C
Zp
⇔
⇒
a∈A
A⊂B
|x|
|z|
{x}
⌊x⌋
[a, b]
(a, b)
P (x)
deg(P )
f : [a, b] → R
f ′ (x)
f ′′ (x)
f (n) (x)
f (x)n
the positive integers or the natural numbers
the integers
the rational numbers
the positive rational numbers
the real numbers
the positive real numbers
the irrational numbers
the complex numbers
is {0, 1, . . . , p − 1} with the sum and product modulo p.
if and only if
imply
the element a belongs to the set A
A is subset of B
the absolute value of the number x
the module of the complex number z
the fractional part of the number x
the integer part of the number x
the set of real numbers x such that a ≤ x ≤ b
the set of real numbers x such that a < x < b
the polynomial P in the variable x
the degree of the polynomial P (x)
the function f defined in [a, b] with values in R
the derivative of the function f (x)
the second derivative of the function f (x)
the nth derivative of the function f (x)
the nth power of a function f (x)
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5
305
306
Notation
f n (x)
∆f (x)
the nth iteration of a function f (x)
the diference operator of f (x)
det A
*n
the determinant of a matrix A
the sum a1 + a2 + · · · + an
ai
3ni=1
ai
3i=1
a
i=j i
the product a1 · a2 · . . . · an
the product for all a1 , a2 , . . . , an except aj
max{a, b, . . . }
min{a, b, . . . }
√
x
√
n
x
the maximum value among a, b, . . .
the minimum value among a, b, . . .
exp x = ex
#
f (a, b, . . . )
the exponential function
cyclic
the square root of x
the nth root of the real number x
represents the sum of the function f evaluated in all the
cyclic permutations of the variables a, b, . . .
We use the following notation for the source of the problems:
AMC
APMO
IMO
MEMO
OMCC
OIM
OMM
(country, year)
American Mathematical Competition
Asian Pacific Mathematical Olympiad
International Mathematical Olympiad
Middle European Mathematical Olympiad
Mathematical Olympiad of Central America
and the Caribean
Iberoamerican Mathematical Olympiad
Mexican Mathematical Olympiad
problem corresponding to the mathematical olympiad
celebrated in that country, in that year, in some stage
Bibliography
[1] Andreescu T., Andrica D., Complex numbers from A to . . . Z, Birkhäuser,
2005.
[2] Andreescu T., Gelca R., Mathematical Olympiad Challenges, Birkhäuser,
2000.
[3] Andreescu T., Enescu B., Mathematical Olympiad Treasures, Birkhäuser,
2006.
[4] Barbeau E.J., Polynomials, Springer-Verlag, 1989.
[5] Bulajich Manfrino R., Gómez Ortega J.A., Geometrı́a, Cuadernos de Olimpiadas, Instituto de Matemáticas de la Universidad Nacional Autónoma de
México, Sociedad Matemática Mexicana, 2012.
[6] Bulajich Manfrino R., Gómez Ortega J.A., Valdez Delgado R., Desigualdades,
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Index
Absolute value, 10
properties of, 11
Algorithm
division, 66, 140
Euclid, 142
Arithmetic progression, 33
difference of the, 33
of order 2, 34
Bertrand’s postulate, 243
Binomial
Newton, 54
square, 15
Binomial coefficient, 53
Cartesian Plane, 10
Complex number, 75
argument, 75
conjugate, 75
imaginary part, 75
module, 75
real part, 75
Decimal system, 8
Dense set, 101
Derangement, 128
Descartes’ rule of signs, 155
Determinant
2 × 2 matrix, 18
3 × 3 matrix, 18
properties of, 19
Difference operator, 111
Endpoints, 10
Equation
characteristic, 122
difference, 111
© Springer Internationl Publishing Switzerland 2015
R.B. Manfrino et al, Topics in Algebra and Analysis,
DOI 10.1007/978-3-319-11946-5
Factorial, 53
Factorization, 25
Formula
Abel’s summation, 133
de Moivre, 79
interpolation, 151
Pascal, 54
Function, 89
non-increasing, 98
additive, 103
bijective, 94
bounded, 99
bounded above, 99
bounded below, 99
codomain, 89
constant, 89
continuous, 100
correspondence rule, 89
decreasing, 98
domain, 89
even, 96
graph, 89
identity, 89
image, 89
increasing, 98
injective, 94
iteration, 111
limit, 99
non-decreasing, 98
odd, 96
periodic, 97
range, 89
surjective, 94
Functional equations, 111
Cauchy, 102
Functions
composition, 93
309
310
difference, 91
equality, 89
product, 91
quotient, 91
sum, 91
Geometric progession
ratio of the, 36
Geometric progression, 36
Greater than, 6
Hanoi Towers, 44
Harmonic progression, 34
Identity
Sophie Germain, 27
Imaginary axis, 75
Induction
bases, 43
step, 43
Induction principle
Cauchy’s, 49
simple, 43
strong, 48
Inequalities, 21
Inequality
Cauchy–Schwarz, 69
helpful, 26
Nesbitt, 24
rearrangement, 24
Infinite descent, 57
Integers, 1
Interval
close, 10
open, 10
Irrational number, 4
Koch’s snowflake, 130
Lemma
Gauss, 147
growth, 86
Index
Matrix
2 × 2, 17
3 × 3, 18
Mean
arithmetic, 22, 49
geometric, 22, 49
harmonic, 22
quadratic, 22
Monomial, 139
Natural numbers, 1
Notable product
three variables, 16
two variables, 15
Number line, 3
Numbers
even, 32
fractional part, 14
greater than, 6
integer part, 12
Lucas, 128
odd, 32
square, 32
triangular, 31
Parameters, 153
Period, 97
Polynomial, 139
characteristic, 122
coefficients, 63, 139
commute, 157
conjugate, 154
constant, 63
cubic, 63
cyclotomic, 148
degree, 67, 139
derivative, 150
discriminant, 67
equality, 139
greatest common divisor, 64, 142
homogeneous, 161
integer coefficients, 145
irreducible, 146
linear, 63
main term, 139
monic, 63, 139
311
Index
over the integers, 63
over the rationals, 63
primitive, 147
quadratic, 63
quotient, 64
reciprocal, 143, 144
remainder, 64
root, 63, 139, 141, 145
several variables, 161
solution, 63, 139
symmetric, 161
Tchebyshev, 158
zero, 63
Polynomials
division, 64, 141
equality of, 63
product, 64
product by a constant, 64
subtraction of, 64
sum of, 64
Quadratic polynomial
complex coefficients, 79
Rational numbers, 2
Real axis, 75
Real numbers, 4
Root
primitive, 83
Roots
multiple, 150
multiplicity, 67, 143
second-order equation, 67
unity, 82
Second-order equation, 67
discriminant, 67
Sequence, 115
bounded, 118
complete, 125
convergent, 126, 135
decreasing, 125
divergent, 126, 135
Fibonacci, 51
finite differences, 113
increasing, 124
limit, 126, 135
monotone decreasing, 125
monotone increasing, 124
periodic, 119
properties, 118
recursive, 118, 120
totally complete, 125
Series, 129
convergent, 129
derivative, 132
divergent, 129
geometric, 131
harmonic, 131
power, 131
formal, 131
Smaller than, 6
Smaller than or equal to, 7
Straight line
oriented, 3
Subsequence, 127
Sum
of Gauss, 32
of the cubes, 39
of the squares, 38
partial, 129
telescopic, 40
Theorem
binomial, 54
Eisenstein, 147
factor, 66, 82
fundamental of Algebra, 81, 87
proof, 85
rational root, 146
Vieta
formulas, 65, 145
jumping, 163
Red Wolf