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ElementaryNumberTheory Notes
c
David A. Santos
January 15, 2004
ii
Contents
Preface v
1 Preliminaries 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Well-Ordering . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Mathematical Induction . . . . . . . . . . . . . . . . . . . 4
1.4 Binomial Coefficients . . . . . . . . . . . . . . . . . . . . . 16
1.5 Vi
`
ete’s Formulæ . . . . . . . . . . . . . . . . . . . . . . . 16
1.6 Fibonacci Numbers . . . . . . . . . . . . . . . . . . . . . 16
1.7 Pigeonhole Principle . . . . . . . . . . . . . . . . . . . . . 23
2 Divisibility 31
2.1 Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.2 Division Algorithm . . . . . . . . . . . . . . . . . . . . . . . 34
2.3 Some Algebraic Identities . . . . . . . . . . . . . . . . . . 38
3 Congruences. Z
n
47
3.1 Congruences . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.2 Divisibility Tests . . . . . . . . . . . . . . . . . . . . . . . . . 57
3.3 Complete Residues . . . . . . . . . . . . . . . . . . . . . . 60
4 Unique Factorisation 63
4.1 GCD and LCM . . . . . . . . . . . . . . . . . . . . . . . . 63
4.2 Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.3 Fundamental Theorem of Arithmetic . . . . . . . . . . . 76
iii
iv CONTENTS
5 Linear Diophantine Equations 89
5.1 Euclidean Algorithm . . . . . . . . . . . . . . . . . . . . . 89
5.2 Linear Congruences . . . . . . . . . . . . . . . . . . . . . 94
5.3 A theorem of Frobenius . . . . . . . . . . . . . . . . . . . 96
5.4 Chinese Remainder Theorem . . . . . . . . . . . . . . . . 100
6 Number-Theoretic Functions 105
6.1 Greatest Integer Function . . . . . . . . . . . . . . . . . . 105
6.2 De Polignac’s Formula . . . . . . . . . . . . . . . . . . . . 116
6.3 Complementary Sequences . . . . . . . . . . . . . . . . 119
6.4 Arithmetic Functions . . . . . . . . . . . . . . . . . . . . . 121
6.5 Euler’s Function. Reduced Residues . . . . . . . . . . . . 128
6.6 Multiplication in Z
n
. . . . . . . . . . . . . . . . . . . . . . 134
6.7 M
¨
obius Function . . . . . . . . . . . . . . . . . . . . . . . 138
7 More on Congruences 141
7.1 Theorems of Fermat and Wilson . . . . . . . . . . . . . . 141
7.2 Euler’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 147
8 Scales of Notation 151
8.1 The Decimal Scale . . . . . . . . . . . . . . . . . . . . . . 151
8.2 Non-decimal Scales . . . . . . . . . . . . . . . . . . . . . 157
8.3 A theorem of Kummer . . . . . . . . . . . . . . . . . . . . 161
9 Diophantine Equations 165
9.1 Miscellaneous Diophantine equations . . . . . . . . . . 165
10 Miscellaneous Examples and Problems 169
10.1 Miscellaneous Examples . . . . . . . . . . . . . . . . . . . 170
11 Polynomial Congruences 173
12 Quadratic Reciprocity 175
13 Continued Fractions 177
Preface
These notes started in the summer of 1993 when I was teaching
Number Theory at the Center for Talented Youth Summer Program
at the Johns Hopkins University. The pupils were between 13 and 16
years of age.
The purpose of the course was to familiarise the pupils with contest-
type problem solving. Thus the majority of the problems are taken
from well-known competitions:
AHSME American High School Mathematics Examination
AIME American Invitational Mathematics Examination
USAMO United States Mathematical Olympiad
IMO International Mathematical Olympiad
ITT International Tournament of Towns
MMPC Michigan Mathematics Prize Competition
(UM)
2
University of Michigan Mathematics Competition
STANFORD Stanford Mathematics Competition
MANDELBROT Mandelbrot Competition
Firstly, I would like to thank the pioneers in that course: Samuel
Chong, Nikhil Garg, Matthew Harris, Ryan Hoegg, Masha Sapper,
Andrew Trister, Nathaniel Wise and Andrew Wong. I would also like
to thank the victims of the summer 1994: Karen Acquista, Howard
Bernstein, Geoffrey Cook, Hobart Lee, Nathan Lutchansky, David
Ripley, Eduardo Rozo, and Victor Yang.
I would like to thank Eric Friedman for helping me with the typing,
and Carlos Murillo for proofreading the notes.
Due to time constraints, these notes are rather sketchy. Most of
v
vi CONTENTS
the motivation was done in the classroom, in the notes I presented a
rather terse account of the solutions. I hope some day to be able to
give more coherence to these notes. No theme requires the knowl-
edge of Calculus here, but some of the solutions given use it here
and there. The reader not knowing Calculus can skip these prob-
lems. Since the material is geared to High School students (talented
ones, though) I assume very little mathematical knowledge beyond
Algebra and Trigonometry. Here and there some of the problems
might use certain properties of the complex numbers.
A note on the topic selection. I tried to cover most Number The-
ory that is useful in contests. I also wrote notes (which I have not
transcribed) dealing with primitive roots, quadratic reciprocity, dio-
phantine equations, and the geometry of numbers. I shall finish writ-
ing them when laziness leaves my weary soul.
I would be very glad to hear any comments, and please forward
me any corrections or remarks on the material herein.
David A. Santos
Chapter 1
Preliminaries
1.1 Introduction
We can say that no history of mankind would ever be complete
without a history of Mathematics. For ages numbers have fasci-
nated Man, who has been drawn to them either for their utility at
solving practical problems (like those of measuring, counting sheep,
etc.) or as a fountain of solace.
Number Theory is one of the oldest and most beautiful branches
of Mathematics. It abounds in problems that yet simple to state, are
very hard to solve. Some number-theoretic problems that are yet
unsolved are:
1. (Goldbach’s Conjecture) Is every even integer greater than 2
the sum of distinct primes?
2. (Twin Prime Problem) Are there infinitely many primes p such
that p + 2 is also a prime?
3. Are there infinitely many primes that are 1 more than the square
of an integer?
4. Is there always a prime between two consecutive squares of
integers?
In this chapter we cover some preliminary tools we need before
embarking into the core of Number Theory.
1
2 Chapter 1
1.2 Well-Ordering
The set N = {0, 1, 2, 3, 4, . . .} of natural numbers is endowed with two
operations, addition and multiplication, that satisfy the following prop-
erties for natural numbers a, b, and c:
1. Closure: a + b and ab are also natural numbers.
2. Associative laws: (a + b) + c = a + (b + c) and a(bc) = (ab)c.
3. Distributive law: a(b + c) = ab + ac.
4. Additive Identity: 0 + a = a + 0 = a
5. Multiplicative Identity: 1a = a1 = a.
One further property of the natural numbers is the following.
1 Axiom Well-Ordering Axiom Every non-empty subset S of the nat-
ural numbers has a least element.
As an example of the use of the Well-Ordering Axiom, let us prove
that there is no integer between 0 and 1.
2
Example Prove that there is no integer in the interval ]0; 1[.
Solution: Assume to the contrary that the set S of integer s in ]0; 1[ is
non-empty. Being a set of positive integers, it must contain a least
element, say m. Now, 0 < m
2
< m < 1, and so m
2
∈ S . But this is
saying that S has a positive integer m
2
which is smaller than its least
positive integer m. This is a contradiction and so S = ∅.
We denote the set of all integers by Z, i.e.,
Z = {. . . − 3, −2, −1, 0, 1, 2, 3, . . .}.
A rational number is a number which can be expressed as the ratio
a
b
of two integers a, b, where b = 0. We denote the set of rational
numbers by Q. An irrational number is a number which cannot be
expressed as the ratio of two integers. Let us give an example of an
irrational number.
Well-Ordering 3
3 Example Prove that
√
2 is irrational.
Solution: The proof is by contradiction. Suppose that
√
2 were ra-
tional, i.e., that
√
2 =
a
b
for some integers a, b. This implies that the
set
A = {n
√
2 : both n and n
√
2 positive integers}
is nonempty since it contains a. By Well-Ordering A has a smallest
element, say j = k
√
2. As
√
2 − 1 > 0,
j(
√
2 − 1) = j
√
2 − k
√
2 = (j − k)
√
2
is a positive integer. Since 2 < 2
√
2 implies 2 −
√
2 <
√
2 and also
j
√
2 = 2k, we see that
(j − k)
√
2 = k(2 −
√
2) < k(
√
2) = j.
Thus (j − k)
√
2 is a positive integer in A which is smaller than j. This
contradicts the choice of j as the smallest integer in A and hence,
finishes the proof.
4 Example Let a, b, c be integers such that a
6
+ 2b
6
= 4c
6
. Show that
a = b = c = 0.
Solution: Clearly we can restrict ourselves to nonnegative numbers.
Choose a triplet of nonnegative integers a, b, c satisfying this equa-
tion and with
max(a, b, c) > 0
as small as possible. If a
6
+ 2b
6
= 4c
6
then a must be even, a = 2a
1
.
This leads to 32a
6
1
+ b
6
= 2c
6
. Hence b = 2b
1
and so 16a
6
1
+ 32b
6
1
= c
6
.
This gives c = 2c
1
, and so a
6
1
+ 2b
6
1
= 4c
6
1
. But clearly max(a
1
, b
1
, c
1
) <
max(a, b, c). This means that all of these must be zero.
5 Example (IMO 1988) If a, b are positive integers such that
a
2
+ b
2
1 + ab
is
an integer, then
a
2
+ b
2
1 + ab
is a perfect square.
4 Chapter 1
Solution: Suppose that
a
2
+ b
2
1 + ab
= k is a counterexample of an integer
which is not a perfect square, with max(a, b) as small as possible. We
may assume without loss of generality that a < b for if a = b then
0 < k =
2a
2
a
2
+ 1
< 2,
which forces k = 1, a perfect square.
Now, a
2
+b
2
−k(ab+1) = 0 is a quadratic in b with sum of the roots
ka and product of the roots a
2
− k. Let b
1
, b be its roots, so b
1
+ b = ka
and b
1
b = a
2
− k.
As a, k are positive integers, supposing b
1
< 0 is incompatible with
a
2
+ b
2
1
= k(ab
1
+ 1). As k is not a perfect square, supposing b
1
= 0 is
incompatible with a
2
+ 0
2
= k(0 · a + 1). Also
b
1
=
a
2
− k
b
<
b
2
− k
b
< b.
Thus we have found another positive integer b
1
for which
a
2
+ b
2
1
1 + ab
1
= k
and which is smaller than the smallest max(a, b). This is a contradic-
tion. It must be the case, then, that k is a perfect square.
Ad Pleniorem Scientiam
6 APS Find all integer solutions of a
3
+ 2b
3
= 4c
3
.
7 APS Prove that the equality x
2
+ y
2
+ z
2
= 2xyz can hold for whole
numbers x, y, z only when x = y = z = 0.
1.3 Mathematical Induction
The Principle of Mathematical Induction is based on the following
fairly intuitive observation. Suppose that we are to perform a task
that involves a certain number of steps. Suppose that these steps
must be followed in strict numerical order. Finally, suppose that we
know how to perform the n-th task provided we have accomplished
[...]... Cauchy to prove the Arithmetic-Mean-Geometric Mean Inequality It consists in proving a statement first for powers of 2 and then interpolating between powers of 2 9 Mathematical Induction 15 Theorem (Arithmetic-Mean-Geometric-Mean Inequality) Let a1, a2, , an be nonnegative real numbers Then √ n a1a2 · · · an ≤ a1 + a2 + · · · + a n n Proof Since the square of any real number is nonnegative, we have... number is nonnegative, we have √ √ ( x1 − x2)2 ≥ 0 Upon expanding, x1 + x2 √ ≥ x1x2, (1.2) 2 which is the Arithmetic-Mean-Geometric-Mean Inequality for n = 2 Assume that the Arithmetic-Mean-Geometric-Mean Inequality holds true for n = 2k−1, k > 2, that is, assume that nonnegative real numbers w1, w2, , w2k−1 satisfy w1 + w2 + · · · + w2k−1 k−1 ≥ (w1w2 · · · w2k−1 )1/2 k−1 2 Using (1.2) with x1 =... Applying (1.3) to both factors on the right hand side of the above , we obtain k y1 + y2 + · · · + y2k ≥ (y1y2 · · · y2k )1/2 (1.4) 2k u k−1 k This means that the 2 -th step implies the 2 -th step, and so we have proved the Arithmetic-Mean-Geometric-Mean Inequality for powers of 2 1/ 10 Chapter 1 Now, assume that 2k−1 < n < 2k Let y1 = a1, y2 = a2, , yn = an, and yn+1 = yn+2 = · · · = y2k = a1 + a2 + ·... at least three who know one another, or at least three who do not know one another 80 APS Show that in any sum of non-negative real numbers there is always one number which is at least the average of the numbers and that there is always one member that it is at most the average of the numbers 81 APS We call a set “sum free” if no two elements of the set add up to a third element of the set What is the... 22n−1 Fn(x) dx = 2n 2 −1 0 (Hint: Let x = sin2 θ.) 1.4 Binomial Coefficients 1.5 Vi`te’s Formulæ e 1.6 Fibonacci Numbers The Fibonacci numbers fn are given by the recurrence f0 = 0, f1 = 1, fn+1 = fn−1 + fn, n ≥ 1 (1.5) Thus the first few Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, A number of interesting algebraic identities can be proved using the above recursion 42 Example Prove that f1 +... real numbers Prove the HarmonicMean- Geometric-Mean Inequality: n 1 1 1 + + ··· + y1 y2 yn ≤ √ n y1y2 · · · yn 7 Let a1, , an be positive real numbers, all different Set s = a1 + a2 + · · · + an (a) Prove that (n − 1) 1 1 < s − ar 1≤r≤n ar 1≤r≤n (b) Deduce that 4n n 1 1 n+1 n+2 2n 24 for all natural numbers n > 1 39 APS In how many regions will a sphere be divided by n planes passing through its centre if no three planes pass through one... the ArithmeticMean-Geometric-Mean Inequality 3 Prove that if n > 1, then 1 · 3 · 5 · · · (2n − 1) < nn 4 Prove that if n > 1 then n (n + 1)1/n − 1 < 1 + 1 1 1 1 + ···+ < n 1 − + 1/n 2 n (n + 1) n+1 5 Given that u, v, w are positive, 0 < a ≤ 1, and that u + v + w = 1, prove that 1 −a u 1 −a v 1 −a w ≥ 27 − 27a + 9a2 − a3 15 Mathematical Induction 6 Let y1, y2, , yn be positive real numbers Prove the... Solution: Split the numbers {1, 2, 3, , 126} into the six sets {1, 2}, {3, 4, 5, 6}, {7, 8, , 13, 14}, {15, 16, , 29, 30}, {31, 32, , 61, 62} and {63, 64, , 126} 24 Chapter 1 By the Pigeonhole Principle, two of the seven numbers must lie in one of the six sets, and obviously, any such two will satisfy the stated inequality 71 Example Given any set of ten natural numbers between 1 and... 1 27 APS Prove that the sum of the cubes of three consecutive positive integers is . Arithmetic-Mean-Geometric-Mean Inequality for n = 2. Assume that the Arithmetic-Mean-Geometric-Mean Inequality holds true for n = 2 k−1 , k > 2, that is, assume that nonnegative real numbers. of the natural numbers is the following. 1 Axiom Well-Ordering Axiom Every non-empty subset S of the nat- ural numbers has a least element. As an example of the use of the Well-Ordering Axiom,. Elementary Number Theory Notes c David A. Santos January 15, 2004 ii Contents Preface v 1 Preliminaries 1 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Well-Ordering