Tài liệu lý thuyết số cho học sinh năng khiếu tại Đại Học Stanford

Stanford University
Educational Program for Gifted Youth (EPGY)
Number Theory
Dana Paquin, Ph.D.

Summer 2010
Stanford University EPGY Number Theory
Note: These lecture notes are adapted from the following sources:
1. Ivan Niven, Herbert S. Zuckerman, and Hugh L. Montgomery, An Introduction
to Number Theory, Fifth Edition, John Wiley & Sons, Inc., 1991.
2. Joseph H. Silverman, A Friendly Introduction to Number Theory, Third Edition,
Prentice Hall, 2006.
3. Harold M. Stark, An Introduction to Number Theory, The MIT Press, 1987.
1
Contents
1 The Four Numbers Game 5
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2 Elementary Properties of Divisibility 9
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3 Proof by Contradiction 13
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
4 Mathematical Induction 17
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
5 The Greatest Common Divisor (GCD) 24
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
6 Prime Factorizatio n and the Fundamental Theorem of Arithmetic 31
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
7 Introduction to Congruences and Modular Arithmetic 39
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
8 Applications of Congruences and Modular Arithmetic 46
Problem Set 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Problem Set 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
9 Linear Congruence Equations 56
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
10 Fermat’s Little Theorem 66
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
11 Eul er’ s Phi-Function and The Euler-Fermat Theorem 74
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
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Stanford University EPGY Number Theory
12 Prim it ive Roots 82
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
13 Squares Modulo p and Quadratic Residues 92
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
14 Intro duct io n to Quadratic Reciprocity 102
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
15 The Law of Quadratic Reciprocity 110
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
16 Dio phantine Equations 115
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
17 Fibonacci Numbers and Linear Recurrences 120
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
Fibonacci Nim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
Unsolved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
18 Mer senne Pri mes and Perfect Numbers 131
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
19 Powers Modulo m and Successive Squaring 138
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
20 Co mput ing k-th Roots Modulo m 141
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
21 RSA Public Key Cr yptog raphy 145

Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
22 Pythagor ean Triples 151
23 Which Primes are Sums of Two Squares? 153
24 Lagr ange’ s Theo rem 157
25 Co ntinued Fractions 159
26 Geo me tr ic Numbe rs 164
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
27 Square-Triangular Numbers and Pell’s Equation 170
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
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Stanford University EPGY Number Theory
28 Pick’s Theorem 182
Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
29 Farey Sequences and Ford Circles 193
Problem Set. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
30 The C ard G ame SE T 202
31 Magi c Squares 207
32 Mathe mati cal G ames 212
33 The 5 Card Trick of Fitch Cheney 215
34 Co nway’s Rational Tangles 217
35 Invariants and Monovariants 219
Problem Set. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
36 Number Theory Problems from AMC, AHSME, AIME, USAMO,
and IMO Mathematics Contests 223
37 Chal leng e C ontest Problems 228
4
Chapter 1
The Four Numbers Game
Choose 4 numbers and place them at the corners of a square. At the midpoint of
each edge, write the difference of the two adjacent numbers, subtracting the smaller

one from the larger. This produces a new list of 4 numbers, written on a smaller
square. Now repeat this process. The game ends if/when a square with 0 at every
vertex is achieved. Here’s an example starting with the four numbers 1,5,3,2. We’ll
call this the (1, 5, 3, 2) game; note that the first number (1) is placed in the upper
left-hand corner.
The (1, 5, 3, 2) game ends after 7 steps. We’ll call this the length of the (1, 5, 3, 2)
game. We’ll be interested in determining whe ther or not all games must end in
finitely many steps. Once it’s clear how the game works, it’s easier if we display the
game more compactly as follows:
1 5 3 2
4 2 1 1
2 1 0 3
1 1 3 1
0 2 2 0
2 0 2 0
2 2 2 2
0 0 0 0
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Stanford University EPGY Number Theory
Example 1.1 1. Find the length of the (1, 3, 8, 17) game.
2. Find the length of the (1, 2, 2, 5) game.
3. Find the length of the (0, 1, 6, π) game.
Example 1.2 Is the length of the game affected by rotations and/or reflections of
the square?
1. Find the length of the (9, 7, 5, 1) game.
2. Find the length of the (7, 5, 1, 9) game.
3. More generally, there are 4 total ways to “rotate” the (9, 7, 5, 1) game. Find the
length of each one.
4. Find the length of the (5, 9, 7, 1) game (vertical reflection).
5. Find the length of the (1, 7, 5, 9) game (horizontal reflection).

6. Find the length of the (9, 1, 5, 7) game (major diagonal reflection).
7. Find the length of the (7, 5, 9, 1) game (minor diagonal reflection).
8. There are 24 possible ways to arrange the numbers 9,7,5,1 on the vertices of
a square–only 8 of them can be achieved by rotation and reflection. Find the
length of the game for each configuration. Are the lengths all the same? Can
you make any observations/conjectures?
Example 1.3 What is the greatest length of games using 4 integers between 0 and
9?
Example 1.4 Work out a few examples of the Four Numbers Game with rational
numbers at the vertices. Does the game always end?
Observation 1.1 What happens if you multiply the 4 start numbers by a positive
integer m? Is the length of the game changed? Once you’ve made and formally
stated a conjecture, can you prove it?
Observation 1.2 Find several games with length at least 4. What do you observe
about the numbers that appear after Step 4?
Theorem 1.1 Every Four Numbers Game played with nonnegative integers has
finite length. More precisely, if we let A denote the largest of the 4 nonnegative
integers and if k is the least integer such that
A
2
k
< 1, then the length of the game
is at most 4k.
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Stanford University EPGY Number Theory
Problem Set
1. Play the Three Numbers Game shown below using the same rules as the Four
Numbers Game, and determine its length.
2. Experiment with examples of the k-Numbers Game for k = 5, 6, 7, 8. For each k,
can you find examples of k-Numbers Games with finite length? Infinite length?

Do you observe any patterns?
3. How does the length of the (a, b, c, d) game compare to the length of the (ma +
e, mb + e, mc + e, md + e) game?
4. Let a, b, c, d be nonnegative real numbers, and suppose that a ≥ c ≥ b ≥ d.
What is the maximum length of the Four Numbers Game (a, b, c, d) in this
case?
5. Let a, b, c, d be nonnegative real numbers, and suppose that a ≥ b ≥ d ≥ c.
What is the maximum length of the Four Numbers Game (a, b, c, d) in this
case?
6. Let a, b, c, d be nonnegative real numbers, and suppose that any 2 of the numbers
a, b, c, d are equal. What is the maximum length of the Four Numbers Game
(a, b, c, d) in this case?
7. The Tribonacci numbers are defined as follows:
t
0
= 0, t
1
= 1, t
2
= 1, t
3
= 2, t
4
= 4, t
5
= 7, . . . .
In general,
t
n
= t

n−3
+ t
n−2
+ t
n−1
.
We’ll define the n-th Tribonacci game as follows:
T
1
= (t
2
, t
1
, t
0
, 0) = (1, 1, 0, 0)
T
n
= (t
n
, t
n−1
, t
n−2
, t
n−3
)
Can you find an equation for the length of T
n
? Begin this problem by doing

some experiments, and try to make a conjecture based on your observations.
Then try to prove your conjecture.
8. Can you find a Four Numbers Game of length 20? Length 100? More generally,
for a given integer N (possibly very large), can you find a Four Numbers Game
of length N?
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Stanford University EPGY Number Theory
9. Numerous mathematical research papers have been written about the Four
Numbers Game(and related games). The sequence of numbers that appear
in the games are also called Ducci sequences after the Italian mathematician
Enrico Ducci. Investigate Ducci sequences and their properties, extensions of
the Four Numbers Game, the Four Real Numbers Game, k-Numbers Games,
and/or other related topics. For example, if 4 nonnegative integers are picked
at random, what’s the probability that the game ends in 8 or fewer steps?
8
Chapter 2
Elementary Properties of
Divisibility
One of the most fundamental ideas in elementary number theory is the notion of
divisibility:
Definition 2.1 If a and b are integers, with a = 0, and if there is an integer c such
that ac = b, then we say that a divides b, and we write a | b. If a does not divide
b, then we write a  b.
For example,
2 | 18, 1 | 42, 3 | (−6), − 7 | 49, 9  80, − 6  31.
Theorem 2.1 Propertie s of Divi sibil ity
1. If a, b, c, m, n are integers such that c | a and c | b, then c | (am + nb ).
2. If x, y, z are integers such that x | y and y | z, then x|z.
Proof. Since c | a and c | b, there are integers s, t such that sc = a, tc = b. Thus
am + nb = c(sm + tn),

so c | (am + bn). Similarly, since x | y and y | z, there are integers u, v with
xu = y, yv = z. Hence xuv = z, so x | z.
Theorem 2.2 If a | b and a | (b + c), then a | c.
Proof. Since a | b, there is an integer s such that as = b. Since a | (b + c), there is
an integer t such that at = b + c. Thus,
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Stanford University EPGY Number Theory
at − b = c
at − as = c
a(t − s) = c.
Since t and s are both integers, t − s is also an integer, so a | c.
Example 2.1 Find all positive integers n ≥ 1 for which
(n + 1) | (n
2
+ 1).
Solution: n
2
+ 1 = n
2
− 1 + 2 = (n − 1)(n + 1) + 2. Thus, if (n + 1) | (n
2
+ 1), we
must have (n + 1) | 2 since (n + 1) | (n −1)(n + 1). Thus, n + 1 = 1 or n + 1 = 2.
Now, n + 1 = 1 since n ≥ 1. We conclude that n + 1 = 2, so the only n such that
(n + 1) | (n
2
+ 1) is n = 1.
Example 2.2 If 7 | (3x + 2) prove that 7 | (15x
2
− 11x − 14.).

Solution: Observe that 15x
2
−11x −14 = (3x + 2)(5x −7). We have 7s = (3x + 2)
for some integer s, so
(15x
2
− 11x − 14) = 7s(5x − 7).
Thus, 7 | (15x
2
− 11x − 14).
Theorem 2.3 The Division Algorithm: If a and b are positive integers, then
there are unique integers q and r such that
a = bq + r, 0 ≤ r < b.
We refer to this theorem as an algorithm because we can find the quotient q and the
remainder r by using ordinary long division to divide a by b. We observe that b | aif
and only if r = 0.
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Stanford University EPGY Number Theory
Problem Set
1. List all the divisors of the integer 12.
2. List all the numbers which divide both 24 and 36. Compare your answer with
your answer to the previous problem.
3. Show that if d = 0 and d | a, then d | (−a) and −d | a.
4. Show that if a | b and b | a, then a = b or a = −b.
5. Suppose that n is an integer such that 5|(n + 2). Which of the following are
divisible by 5?
(a) n
2
− 4
(b) n

2
+ 8n + 7
(c) n
4
− 1
(d) n
2
− 2n
6. Find all integers n ≥ 1 so that n
3
−1 is prime. Hint: n
3
−1 = (n
2
+n+1)(n−1).
7. Show that if ac | bc, then a | b.
8. (a) Prove that the product of three consecutive integers is divisible by 6.
(b) Prove that the product of four consecutive integers is divisible by 24.
(c) Prove that the product of n consecutive integers is divisible by n!.
9. Find all integers n ≥ 1 so that n
4
+ 4 is prime.
10. Find all integers n ≥ 1 so that n
4
+ 4
n
is prime.
11. Prove that the square of any integer of the form 5k + 1 is of the same form.
12. Prove that 3 is not a divisor of n
2

+ 1 for all integers n ≥ 1.
13. A prime triplet is a triple of numbers of the form (p, p + 2, p + 4), for which p,
p + 2, and p + 4 are all prime. For example, (3, 5, 7) is a prime triplet. Prove
that (3, 5, 7) is the only prime triplet.
14. Prove that if 3 | (a
2
+ b
2
), then 3 | a and 3 | b. Hint: If 3  a and 3  b, what are
the possible remainders upon division by 3?
15. Let n be an integer greater than 1. Prove that if one of the numbers
2
n
− 1, 2
n
+ 1
is prime, then the other is composite.
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Stanford University EPGY Number Theory
16. Suppose that p is an odd prime and that
a
b
= 1 +
1
2
+
1
3
+ ··· +
1

p − 1
.
Show that p | a.
17. Find, with proof, the unique square which is the product of four consecutive
odd numbers.
18. Suppose that a is an integer greater than 1 and that n is a positive integer.
Prove that if a
n
+ 1 is prime, then a is even and n is a power of 2. Primes of
the form 2
2
k
+ 1 are called Fermat primes.
19. Suppose that a is an integer greater than 1 and that n is a positive integer.
Prove that if a
n
− 1 is prime, then a = 2 and n is a prime. Primes of the form
2
n
− 1 are called Mersenne primes.
20. Prove that the product of four consecutive natural numbers is never a perfect
square.
21. Can you find an integer n > 1 such that the sum
1 +
1
2
+
1
3
+ ··· +

1
n
is an integer?
22. Show that every integer of the form
4 · 14
k
+ 1, k ≥ 1
is composite. Hint: show that there is a factor of 3 when k is odd and a factor
of 5 when k is even.
23. Show that every integer of the form
521 · 12
k
+ 1, k ≥ 1
is composite. Hint: show that there is a factor of 13 when k is odd, a factor of
5 when k ≡ 2 mod 4, and a factor of 29 when 4 | k.
24. Show that for all integers a and b,
ab(a
2
− b
2
)(a
2
+ b
2
)
is divisible by 30.
12
Chapter 3
Proof by Contradiction
In a proof by contradiction (or reductio ad absurdum), we assume, along with the

hypotheses, the logical negation of the statement that we are trying to prove, and
then reach some kind of contradiction. Upon reaching a contradiction, we conclude
that the original assumption (i.e. the negation of the statement we are trying to
prove) is false, and thus the statement that we are trying to prove must be true.
Example 3.1 Show, without using a calculator, that 6 −
√
35 <
1
10
.
Solution: Assume that 6 −
√
35 ≥
1
10
. Then
6 −
1
10
≥
√
35,
so
59 ≥ 10
√
35.
Squaring both sides we obtain
3481 ≥ 3500,
which is a contradiction. Thus our original assumption must be false, so we conclude
that 6 −

√
35 <
1
10
.
Example 3.2 Let a
1
, a
2
, . . . , a
n
be an arbitrary permutation of the numbers 1, 2, . . . , n,
where n is an odd number. Prove that the product
(a
1
− 1)(a
2
− 2) ···(a
n
− n)
is even.
Solution: It is enough to prove that some difference a
k
− k is even. Assume that
all the differences a
k
− k are odd. Clearly
S = (a
1
− 1) + (a

2
− 2) + ··· + (a
n
− n) = 0,
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Stanford University EPGY Number Theory
since the a
k
’s are a reordering of 1, 2, . . . , n. S is an odd number of summands of
odd integers adding to the even integer 0. This is a contradiction, so our initial
assumption that all the a
k
− k are odd is thus false, so one of the terms a
k
− k is
even, and hence the product is even.
Example 3.3 Prove that there are no positive integer solutions to the equation
x
2
− y
2
= 1.
Solution: Assume that there is a solution (x, y) where x and y are positive integers.
Then we can factor the left-hand side of the equation to obtain
(x − y)(x + y) = 1.
Since x and y are both positive integers, x−y and x+y are integers. Thus, x−y = 1
and x + y = 1 or x − y = −1 and x + y = −1. In the first case, we add the two
equations to obtain x = 1 and y = 0, which contradicts the assumption that x and y
are both positive. In the second case, we add the two equations to obtain x = −1 and
y = 0, which is again a contradiction. Thus, there are no positive integer solutions

to the equation x
2
− y
2
= 1.
Example 3.4 If a, b, c are odd integers, prove that ax
2
+ bx + c = 0 does not have
a rational number solution.
Solution: Suppose
p
q
is a rational solution to the equation. We may assume that p
and q have no prime factors in common, so either p and q are both odd, or one is
odd and the other even. Now
a

p
q

2
+ b

p
q

+ c = 0 =⇒ ap
2
+ bpq + cq
2

= 0.
If both p and p were odd, then ap
2
+ bpq + cq
2
is also odd and hence = 0. Similarly
if one of them is even and the other odd then either ap
2
+ bpq or bpq + cq
2
is even
and ap
2
+ bpq + cq
2
is odd. This contradiction proves that the equation cannot have
a rational root.
Example 3.5 Show that
√
2 is irrational.
Solution: Proof by contradiction. Suppose that
√
2 is irrational, i.e.
√
2 =
r
s
,
where r and s have no common factors (i.e. the fraction is in lowest terms). Then
2 =

r
2
s
2
, so 2s
2
= r
2
.
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Stanford University EPGY Number Theory
This means that r
2
must be even, so r must be even, say r = 2c. Then
2s
2
= (2c)
2
= 4c
2
,
so
s
2
= 2c
2
,
so s is also even. This is a contradiction since r and s have no common factors.
Thus,
√

2 must be irrational.
We conclude with two imp ortant results.
Theorem 3.1 If n is an integer greater than 1, then n can be written as a finite
product of primes.
Proof. Proof by contradiction. Assume that the theorem is false. Then there are
composite numbers which cannot be represented as a finite product of primes. Let
N be the smallest such number. Since N is the smallest such number, if 1 < n < N,
then the theorem is true for n. Let p be a prime divisor of N. Since N is composite,
1 <
N
p
< N,
so the theorem is true for
N
p
. Thus, there are primes p
1
, p
2
···p
k
such that
N
p
= p
1
p
2
···p
k

.
Thus,
N = pp
1
p
2
···p
k
is a finite product of primes. This is a contradiction, so we conclude that any integer
greater than 1 can be written as a finite product of primes.
Theorem 3.2 There are infinitely many prime numbers.
Proof. Proof by contradiction. The following beautiful proof is attributed to
Euclid. Ass ume that there are only finitely many (say, n) prime numbers. Then
{p
1
, p
2
, . . . , p
n
} is a list that exhausts all the primes. Consider the number
N = p
1
p
2
···p
n
+ 1.
This is a positive integer, clearly greater than 1. Observe that none of the primes
on the list {p
1

, p
2
, . . . , p
n
} divides N, since division by any of these primes leaves a
remainder of 1. Since N is larger than any of the primes on this list, it is either
a prime or divisible by a prime outside this list. Thus we have shown that the
assumption that any finite list of primes leads to the existence of a prime outside
this list, so we have reached a contradiction. This implies that the number of primes
is infinite.
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Stanford University EPGY Number Theory
Problem Set
1. The product of 34 integers is equal to 1. Show that their sum cannot be 0.
2. Prove that the sum of two odd squares cannot be a square.
3. Let a
1
, a
2
, . . . , a
2000
be natural numbers such that
1
a
1
+
1
a
2
+ ··· +

1
a
2000
= 1.
Prove that at least one of the a
k
’s is even. Hint: clear the denominators.
4. A palindrome is an integer whose decimal expansion is symmetric, e.g. 1, 2, 11, 121,
15677651 (but not 010, 0110) are palindromes. Prove that there is no positive
palindrome which is divisible by 10.
5. Let 0 < α < 1. Prove that
√
α > α.
6. In ABC, ∠A > ∠B. Prove that BC > AC.
7. Show that if a is rational and b is irrational, then a + b is irrational.
8. Prove that there is no smallest positive real number.
9. Prove that there are no positive integer solutions to the equation
x
2
− y
2
= 10.
10. Given that a, b, c are odd integers, prove that the equation ax
2
+ bx + c = 0
cannot have a rational root.
11. Prove that there do not exist positive integers a, b, c and n such that
a
2
+ b

2
+ c
2
= 2
n
abc.
12. Show that the equation
b
2
+ b + 1 = a
2
has no positive integer solutions a, b.
13. Let a, b, c be integers satisfying a
2
+ b
2
= c
2
. Show that abc must be even.
16
Chapter 4
Mathematical Induction
Mathematical induction is a powerful method for proving statements that are “in-
dexed” by the integers. For example, induction can be used to prove the following:
• The sum of the interior angles of any n-gon is 180(n −2) degrees.
• The inequality n! > 2
n
is true for all integers n ≥ 4.
• 7
n

− 1 is divisible by 6 for all integers n ≥ 1.
Each assertion can be put in the form:
P(n) is true for all integers n ≥ n
0
,
where P (n) is a statement involving the integer n, and n
0
is the starting point, or
base case. For example, for the third assertion, P (n) is the statement 7
n
− 1 is
divisible by 6, and the base case is n
0
= 1. Here’s how induction works:
1. Base case. First, prove that P (n
0
) is true.
2. Inductive step. Next, show that if P(k) is true, then P(k +1) must also be true.
Observe that these two steps are sufficient to prove that P (n) is true for all integers
n ≥ n
0
, as P(n
0
) is true by step (1), and step (2) then implies that P (n
0
+ 1) is true,
which implies that P (n
0
+ 2) is true, etc.
You can think of induction in the following way. Suppose that you have arranged

infinitely many dominos in a line, corresponding to statements P (1), P (2), P (3),
. . If you make the first domino fall, then you can be sure that all of the dominos
will fall, provided that whenever one domino falls, it will knock down its neighbor.
Knocking the first domino down is analogous to establishing the base case. Showing
that each falling domino knocks down its neighbor is equivalent to showing that P (n)
implies P (n + 1).
Example 4.1 Prove that for any integer n ≥ 1,
1 + 2 + 3 + ··· + n =
n(n + 1)
2
.
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Stanford University EPGY Number Theory
Example 4.2 Prove that n! > 2
n
for all integers n ≥ 4.
Solution: P (n) is the statement n! > 2
n
. The base case is n
0
= 4.
(i) Base case.
4! = 24 > 2
4
= 16,
so the base case P (4) is true.
(ii) Inductive hypothesis. Assume that n! > 2
n
. We must use this assumption to
prove that (n + 1)! > 2

n+1
. The left-hand side of the inductive hypothesis is n!,
and the left-hand side of the statement that we want to prove is (n + 1)! = (n +
1)n!. Thus, it seems natural to multiply both side s of the inductive hypothesis
by (n + 1).
n! > 2
n
(n + 1)n! > (n + 1)2
n
(n + 1)! > (n + 1)2
n
.
Finally, note that (n + 1) > 2, so
(n + 1)! > (n + 1)2
n
> 2 · 2
n
> 2
n+1
,
so we conclude that
(n + 1)! > 2
n+1
,
as needed.
Thus, n! > 2
n
for all integers n ≥ 4.
Example 4.3 Prove that the expression 3
3n+3

−26n −27 is a multiple of 169 for all
natural numbers n.
Solution: P (n) is the assertion 3
3n+3
− 26n − 27 is a multiple of 169, and the base
case is n
0
= 1.
(i) Base case. Observe that 3
3(1)+3
− 26(1) − 27 = 676 = 4(169) so P (1) is true.
(ii) Inductive hypothesis. Assume that P (n) is true, i.e. that is, that there is an
integer M such that
3
3n+3
− 26n − 27 = 169M.
We must prove that there is an integer K so that
3
3(n+1)+3
− 26(n + 1) − 27 = 169K.
We have:
18
Stanford University EPGY Number Theory
3
3(n+1)+3
− 26(n + 1) − 27 = 3
3n+3+3
− 26n − 26 − 27
= 27(3
3n+3

) − 26n − 27 − 26
= 27(3
3n+3
) − 26n − 26(26n) + 26(26n)
−27 − 26(27) + 26(27) − 26
= 27(3
3n+3
) − 27(26n) − 27(27) + 26(26n) + 26(27) − 26
= 27(3
3n+3
− 26n − 27) + 676n + 676
= 27(169M) + 169 · 4n + 169 · 4
= 169(27M + 4n + 4).
Thus, 3
3n+3
− 26n − 27 is a multiple of 169 for all natural numbers n.
Example 4.4 Prove that if k is odd, then 2
n+2
divides
k
2
n
− 1
for all natural numbers n.
Solution: Let k be odd. P (n) is the statement that 2
n+2
is a divisor of k
2
n
−1, and

the base case is n
0
= 1.
(i) Base case.
k
2
− 1 = (k −1)(k + 1)
is divisible by 2
1+2
= 8 for any odd natural number k since k − 1 and k + 1 are
consecutive even integers.
(ii) Inductive hypothesis. Assume that 2
n+2
is a divisor of k
2
n
− 1. Then there is
an integer a such that 2
n+2
a = k
2
n
− 1. Then
k
2
n+1
− 1 = (k
2
n
− 1)(k

2
n
+ 1) = 2
n+2
a(k
2
n
+ 1).
Since k is odd, k
2
n
+ 1 is even and so k
2
n
+ 1 = 2b for some integer b. This gives
k
2
n+1
− 1 = 2
n+2
a(k
2
n
+ 1) = 2
n+3
ab,
and so the assertion follows by induction.
Example 4.5 The Fibonacci Numbers are given by
F
0

= 0, F
1
= 1, F
n+1
= F
n
+ F
n−1
, n ≥ 1,
i.e. every number after the second one is the sum of the preceding two.
The first several terms of the Fibonacci sequence are
0, 1, 1, 2, 3, 5, 8, 13, 21, . . . .
19
Stanford University EPGY Number Theory
Prove that for all integers n ≥ 1,
F
n−1
F
n+1
= F
2
n
+ (−1)
n+1
.
Solution: P (n) is the statement that
F
n−1
F
n+1

= F
2
n
+ (−1)
n
and the base case is n
0
= 1.
(i) Base case. If n = 1, then 0 = F
0
F
2
= 1
2
+ (−1)
1
.
(ii) Inductive hypothesis. Assume that F
n−1
F
n+1
= F
2
n
+ (−1)
n
. Then, using the
fact that F
n+2
= F

n
+ F
n+1
, we have
F
n
F
n+2
= F
n
(F
n
+ F
n+1
)
= F
2
n
+ F
n
F
n+1
= F
n−1
F
n+1
− (−1)
n
+ F
n

F
n+1
= F
n+1
(F
n−1
+ F
n
) + (−1)
n+1
= F
2
n+1
+ (−1)
n+1
,
which establishes the assertion by induction.
Example 4.6 Prove that
n
5
5
+
n
4
2
+
n
3
3
−

n
30
is an integer for all integers n ≥ 0.
Solution: P (n) is the statement that
n
5
5
+
n
4
2
+
n
3
3
−
n
30
is an integer and the base case is n
0
= 0.
(i) Base case. Since 0 is an integer, the statement is clearly true when n = 0.
(ii) Inductive hypothesis. Assume that
n
5
5
+
n
4
2

+
n
3
3
−
n
30
is an integer. We must show that
(n + 1)
5
5
+
(n + 1)
4
2
+
(n + 1)
3
3
−
n + 1
30
20
Stanford University EPGY Number Theory
is also an integer. We have:
(n + 1)
5
5
+
(n + 1)

4
2
+
(n + 1)
3
3
−
n + 1
30
=
n
5
+ 5n
4
+ 10n
3
+ 10n
2
+ 5n + 1
5
+
n
4
+ 4n
3
+ 6n
2
+ 4n + 1
2
+

n
3
+ 3n
2
+ 3n + 1
3
−
n + 1
30
=

n
5
5
+
n
4
2
+
n
3
3
−
n
30

+

n
4

+ 2n
3
+ 2n
2
+ n + 2n
3
+ 3n
2
+ 2n + n
2
+ n + 1

,
which is an integer by the inductive hypothesis and since the second grouping
is a sum of integers.
21
Stanford University EPGY Number Theory
Problem Set
1. Prove that for any integer n ≥ 1,
2
0
+ 2
1
+ ··· + 2
n−1
= 2
n
− 1.
2. Prove that for any integer n ≥ 1, n
2

is the sum of the first n odd integers. (For
example, 3
2
= 1 + 3 + 5.)
3. Prove that n
5
− 5n
3
+ 4n is divisible by 120 for all integers n ≥ 1.
4. Prove that n
9
− 6n
7
+ 9n
5
− 4n
3
is divisible by 8640 for all integers n ≥ 1.
5. Prove that
n
2
| ((n + 1)
n
− 1)
for all integers n ≥ 1.
6. Show that
(x − y) | (x
n
− y
n

)
for all integers n ≥ 1.
7. Use the result of the previous problem to show that
8767
2345
− 8101
2345
is divisible by 666.
8. Show that
2903
n
− 803
n
− 464
n
+ 261
n
is divisible by 1897 for all integers n ≥ 1.
9. Prove that if n is an even natural number, then the number 13
n
+ 6 is divisible
by 7.
10. Prove that n! ≥ 3
n
for all integers n ≥ 7.
11. Prove that 2
n
≥ n
2
for all integers n ≥ 4.

12. Prove that for every integer n ≥ 2, n
3
− n is a multiple of 6.
13. Consider the sequence defined by a
1
= 1 and a
n
=
√
2a
n−1
. Prove that a
n
< 2
for all integers n ≥ 1.
14. Prove that the equation
x
2
+ y
2
= z
n
has a solution in positive integers x, y, z for all integers n ≥ 1.
15. Prove that n
3
+ (n + 1)
3
+ (n + 2)
3
is divisible by 9 for all integers n ≥ 1.

22
Stanford University EPGY Number Theory
16. Prove that
1
n + 1
+
1
n + 2
+ ··· +
1
3n + 1
> 1
for all integers n ≥ 1.
17. Prove that
4
n
n + 1
≤
(2n)!
(n!)
2
for all integers n ≥ 1.
18. Show that 7
n
− 1 is divisible by 6 for all integers n ≥ 0.
19. Consider the Fibonacci sequence {F
n
} defined by F
0
= 0, F

1
= 1, F
n+1
=
F
n
+ F
n−1
, n ≥ 1. Prove that each of the following statements is true for all
integers n ≥ 1.
(a) F
1
+ F
3
+ F
5
+ ··· + F
2n−1
= F
2n
(b) f
2
+ F
4
+ F
6
+ ··· + F
2n
= F
2n+1

− 1
(c) F
n
< 2
n
(d) F
n
=
1
√
5

1 +
√
5
2

n
−

1 +
√
5
2

n

23
Chapter 5
The Greatest Common Divisor

(GCD)
Definition 5.1 Let a and b be integers, not both zero. Let d be the largest number
in the set of common divisors of a and b. We call d the greatest common divisor
of a and b, and we write
d = gcd(a, b),
or, more simply,
d = (a, b).
Example 5.1 We compute some simple gcd’s.
• (6, 4) = 2
• (3, 5) = 1
• (16, 24) = 8
• (4, 0) = 4
• (5, 5) = 5
• (3, 12) = 3
Definition 5.2 If (a, b) = 1, we say that a and b are relatively prime.
In general, we’d like to be able to compute (a, b) without listing all of the factors
of a and b. The Euclidean Algorithm is the most efficient method known for
computing the greatest common divisor of two integers. We’ll begin by illustrating
the method with an example.
Example 5.2 Compute (54, 21).
24