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

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

Educational Program for Gifted Youth (EPGY)

Number Theory

Dana Paquin, Ph.D

paquin@math.stanford.edu

Summer 2010

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 The Four Numbers Game 5Problem Set 7

12 Primitive Roots 82Problem Set 89

28 Pick’s Theorem 182Problem Set 185

Problem Set 200

Problem Set 221

36 Number Theory Problems from AMC, AHSME, AIME, USAMO,

The Four Numbers Game

Choose 4 numbers and place them at the corners of a square At the midpoint ofeach edge, write the difference of the two adjacent numbers, subtracting the smallerone from the larger This produces a new list of 4 numbers, written on a smallersquare Now repeat this process The game ends if/when a square with 0 at everyvertex is achieved Here’s an example starting with the four numbers 1,5,3,2 We’llcall this the (1, 5, 3, 2) game; note that the first number (1) is placed in the upperleft-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 whether or not all games must end infinitely many steps Once it’s clear how the game works, it’s easier if we display thegame more compactly as follows:

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 ofthe 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 thelength 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 thelength of the game for each configuration Are the lengths all the same? Canyou make any observations/conjectures?

Example 1.3 What is the greatest length of games using 4 integers between 0 and9?

Example 1.4 Work out a few examples of the Four Numbers Game with rationalnumbers at the vertices Does the game always end?

Observation 1.1 What happens if you multiply the 4 start numbers by a positiveinteger m? Is the length of the game changed? Once you’ve made and formallystated a conjecture, can you prove it?

Observation 1.2 Find several games with length at least 4 What do you observeabout the numbers that appear after Step 4?

Theorem 1.1 Every Four Numbers Game played with nonnegative integers hasfinite length More precisely, if we let A denote the largest of the 4 nonnegativeintegers and if k is the least integer such that A

2k < 1, then the length of the game

is at most 4k

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 thiscase?

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 thiscase?

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:

t0 = 0, t1 = 1, t2 = 1, t3 = 2, t4 = 4, t5 = 7,

In general,

tn = tn−3+ tn−2+ tn−1.We’ll define the n-th Tribonacci game as follows:

T1 = (t2, t1, t0, 0) = (1, 1, 0, 0)

Tn = (tn, tn−1, tn−2, tn−3)

Can you find an equation for the length of Tn? Begin this problem by doingsome 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 ?

9 Numerous mathematical research papers have been written about the FourNumbers Game(and related games) The sequence of numbers that appear

in the games are also called Ducci sequences after the Italian mathematicianEnrico Ducci Investigate Ducci sequences and their properties, extensions ofthe 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?

Theorem 2.1 Properties of Divisibility

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,

at − b = c

at − as = ca(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) | (n2+ 1)

Solution: n2+ 1 = n2− 1 + 2 = (n − 1)(n + 1) + 2 Thus, if (n + 1) | (n2+ 1), wemust have (n + 1) | 2 since (n + 1) | (n − 1)(n + 1) Thus, n + 1 = 1 or n + 1 = 2.Now, n + 1 6= 1 since n ≥ 1 We conclude that n + 1 = 2, so the only n such that(n + 1) | (n2 + 1) is n = 1

Example 2.2 If 7 | (3x + 2) prove that 7 | (15x2− 11x − 14.)

Solution: Observe that 15x2− 11x − 14 = (3x + 2)(5x − 7) We have 7s = (3x + 2)for some integer s, so

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 withyour answer to the previous problem

3 Show that if d 6= 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 aredivisible by 5?

(a) n2 − 4

(b) n2 + 8n + 7

(c) n4 − 1

(d) n2 − 2n

6 Find all integers n ≥ 1 so that n3−1 is prime Hint: n3−1 = (n2+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 n4+ 4 is prime

10 Find all integers n ≥ 1 so that n4+ 4n 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 n2+ 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 Provethat (3, 5, 7) is the only prime triplet

14 Prove that if 3 | (a2+ b2), then 3 | a and 3 | b Hint: If 3 - a and 3 - b, what arethe possible remainders upon division by 3?

15 Let n be an integer greater than 1 Prove that if one of the numbers

2n− 1, 2n+ 1

is prime, then the other is composite

16 Suppose that p is an odd prime and that

17 Find, with proof, the unique square which is the product of four consecutiveodd numbers

18 Suppose that a is an integer greater than 1 and that n is a positive integer.Prove that if an+ 1 is prime, then a is even and n is a power of 2 Primes ofthe form 22k+ 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 an− 1 is prime, then a = 2 and n is a prime Primes of the form

2n− 1 are called Mersenne primes

20 Prove that the product of four consecutive natural numbers is never a perfectsquare

21 Can you find an integer n > 1 such that the sum

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(a2− b2)(a2+ b2)

is divisible by 30

Proof by Contradiction

In a proof by contradiction (or reductio ad absurdum), we assume, along with thehypotheses, the logical negation of the statement that we are trying to prove, andthen reach some kind of contradiction Upon reaching a contradiction, we concludethat the original assumption (i.e the negation of the statement we are trying toprove) 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 concludethat 6 −√

35 < 1

10.Example 3.2 Let a1, a2, , anbe an arbitrary permutation of the numbers 1, 2, , n,where n is an odd number Prove that the product

since the ak’s are a reordering of 1, 2, , n S is an odd number of summands ofodd integers adding to the even integer 0 This is a contradiction, so our initialassumption that all the ak − k are odd is thus false, so one of the terms ak− k iseven, and hence the product is even.

Example 3.3 Prove that there are no positive integer solutions to the equation

y = 0, which is again a contradiction Thus, there are no positive integer solutions

to the equation x2− y2 = 1

Example 3.4 If a, b, c are odd integers, prove that ax2 + 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 pand q have no prime factors in common, so either p and q are both odd, or one isodd and the other even Now

a pq

2

+ b pq

+ c = 0 =⇒ ap2+ bpq + cq2 = 0

If both p and p were odd, then ap2+ bpq + cq2 is also odd and hence 6= 0 Similarly

if one of them is even and the other odd then either ap2 + bpq or bpq + cq2 is evenand ap2+ bpq + cq2 is odd This contradiction proves that the equation cannot have

2 = r

2

s2, so 2s2 = r2

This means that r2 must be even, so r must be even, say r = 2c Then

2s2 = (2c)2 = 4c2,so

s2 = 2c2,

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 important results

Theorem 3.1 If n is an integer greater than 1, then n can be written as a finiteproduct of primes

Proof Proof by contradiction Assume that the theorem is false Then there arecomposite 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,

N = pp1p2· · · pk

is a finite product of primes This is a contradiction, so we conclude that any integergreater 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 toEuclid Assume that there are only finitely many (say, n) prime numbers Then{p1, p2, , pn} is a list that exhausts all the primes Consider the number

N = p1p2· · · pn+ 1

This is a positive integer, clearly greater than 1 Observe that none of the primes

on the list {p1, p2, , pn} divides N , since division by any of these primes leaves aremainder 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 theassumption that any finite list of primes leads to the existence of a prime outsidethis list, so we have reached a contradiction This implies that the number of primes

is infinite

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 a1, a2, , a2000 be natural numbers such that

Prove that at least one of the ak’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 positivepalindrome which is divisible by 10

5 Let 0 < α < 1 Prove that √

α > α

6 In 4ABC, ∠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

13 Let a, b, c be integers satisfying a2+ b2 = c2 Show that abc must be even

Mathematical Induction

Mathematical induction is a powerful method for proving statements that are dexed” by the integers For example, induction can be used to prove the following:

“in-• The sum of the interior angles of any n-gon is 180(n − 2) degrees

• The inequality n! > 2n is true for all integers n ≥ 4

• 7n− 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 ≥ n0,where P (n) is a statement involving the integer n, and n0 is the starting point, orbase case For example, for the third assertion, P (n) is the statement 7n − 1 isdivisible by 6, and the base case is n0 = 1 Here’s how induction works:

1 Base case First, prove that P (n0) 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 ≥ n0, as P (n0) is true by step (1), and step (2) then implies that P (n0+ 1) is true,which implies that P (n0+ 2) is true, etc

You can think of induction in the following way Suppose that you have arrangedinfinitely 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 dominoswill 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 Showingthat 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)

Example 4.2 Prove that n! > 2n for all integers n ≥ 4.

Solution: P (n) is the statement n! > 2n The base case is n0 = 4

(i) Base case

4! = 24 > 24 = 16,

so the base case P (4) is true

(ii) Inductive hypothesis Assume that n! > 2n We must use this assumption toprove that (n + 1)! > 2n+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 sides of the inductive hypothesis

by (n + 1)

n! > 2n(n + 1)n! > (n + 1)2n(n + 1)! > (n + 1)2n

Finally, note that (n + 1) > 2, so

(n + 1)! > (n + 1)2n > 2 · 2n> 2n+1,

so we conclude that

(n + 1)! > 2n+1,

as needed

Thus, n! > 2n for all integers n ≥ 4

Example 4.3 Prove that the expression 33n+3− 26n − 27 is a multiple of 169 for allnatural numbers n

Solution: P (n) is the assertion 33n+3− 26n − 27 is a multiple of 169, and the basecase is n0 = 1

(i) Base case Observe that 33(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 aninteger M such that

33n+3− 26n − 27 = 169M

We must prove that there is an integer K so that

33(n+1)+3− 26(n + 1) − 27 = 169K

We have:

Thus, 33n+3− 26n − 27 is a multiple of 169 for all natural numbers n.

Example 4.4 Prove that if k is odd, then 2n+2 divides

k2n − 1for all natural numbers n

Solution: Let k be odd P (n) is the statement that 2n+2 is a divisor of k2 n

− 1, andthe base case is n0 = 1

(i) Base case

k2− 1 = (k − 1)(k + 1)

is divisible by 21+2= 8 for any odd natural number k since k − 1 and k + 1 are

consecutive even integers

(ii) Inductive hypothesis Assume that 2n+2 is a divisor of k2 n

Example 4.5 The Fibonacci Numbers are given by

F0 = 0, F1 = 1, Fn+1 = Fn+ Fn−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,

Prove that for all integers n ≥ 1,

Fn−1Fn+1 = Fn2 + (−1)n+1.Solution: P (n) is the statement that

Fn−1Fn+1 = Fn2+ (−1)nand the base case is n0 = 1

(i) Base case If n = 1, then 0 = F0F2 = 12+ (−1)1

(ii) Inductive hypothesis Assume that Fn−1Fn+1 = Fn2 + (−1)n Then, using thefact that Fn+2 = Fn+ Fn+1, we have

which establishes the assertion by induction

Example 4.6 Prove that

is an integer for all integers n ≥ 0

Solution: P (n) is the statement that

is an integer and the base case is n0 = 0

(i) Base case Since 0 is an integer, the statement is clearly true when n = 0.(ii) Inductive hypothesis Assume that

is an integer We must show that

is also an integer We have:

+n4 + 2n3 + 2n2 + n + 2n3+ 3n2+ 2n + n2+ n + 1 ,which is an integer by the inductive hypothesis and since the second grouping

is a sum of integers

3 Prove that n5− 5n3+ 4n is divisible by 120 for all integers n ≥ 1.

4 Prove that n9− 6n7+ 9n5− 4n3 is divisible by 8640 for all integers n ≥ 1

5 Prove that

n2 | ((n + 1)n− 1)for all integers n ≥ 1

6 Show that

(x − y) | (xn− yn)for all integers n ≥ 1

7 Use the result of the previous problem to show that

87672345− 81012345

is divisible by 666

8 Show that

2903n− 803n− 464n+ 261n

is divisible by 1897 for all integers n ≥ 1

9 Prove that if n is an even natural number, then the number 13n+ 6 is divisible

by 7

10 Prove that n! ≥ 3n for all integers n ≥ 7

11 Prove that 2n ≥ n2 for all integers n ≥ 4

12 Prove that for every integer n ≥ 2, n3− n is a multiple of 6

13 Consider the sequence defined by a1 = 1 and an =√

2an−1 Prove that an < 2for all integers n ≥ 1

14 Prove that the equation

x2+ y2 = znhas a solution in positive integers x, y, z for all integers n ≥ 1

15 Prove that n3+ (n + 1)3+ (n + 2)3 is divisible by 9 for all integers n ≥ 1

17 Prove that

4n

n + 1 ≤ (2n)!

(n!)2

for all integers n ≥ 1

18 Show that 7n− 1 is divisible by 6 for all integers n ≥ 0

19 Consider the Fibonacci sequence {Fn} defined by F0 = 0, F1 = 1, Fn+1 =

Fn + Fn−1, n ≥ 1 Prove that each of the following statements is true for allintegers n ≥ 1

!n

− 1 +

√52

!n#

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

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 forcomputing the greatest common divisor of two integers We’ll begin by illustratingthe method with an example

Example 5.2 Compute (54, 21)

Solution: The first step is to divide 54 by 21, which gives a quotient of 2 and aremainder of 12 We write this as

Example 5.3 Compute (36, 132), and use your computation to find integers x and

y such that (36, 132) = 36x + 132y

We conclude that (53, 77) = 1, so 53 and 77 are relatively prime.

Working backwards, we have:

Theorem 5.1 Let a and b be integers, not both zero Then (a, b) can be written as

a linear combination of a and b, i.e there exist integers x and y such that

(a, b) = ax + by,and these integers can be found by the Euclidean algorithm method illustrated inthe examples

Note that since (a, b) | a and (a, b) | b,

(a, b) | ax + byfor all integers x and y

By Theorem 5.1, we can always find integers x and y so that (a, b) = ax + by Ingeneral, let’s consider the possible values that we can obtain from numbers of theform

ax + bywhen we substitute all possible integers for x and y For example, consider the case

a = 42 and b = 30 Note that (42, 30) = 6 Complete the table of values of 42x + 30ybelow for the given values of x and y

Observe that (42, 30) = 6 appears in the table, and is the smallest positive value

of ax + by In general, this is always true (and can be proven via the Euclideanalgorithm)

Theorem 5.2 Let a and b be integers, not both zero Then the smallest positivevalue of ax + by (taken over all integers x and y) is (a, b)

Suppose that a, b, c are integers and that a | bc When is it true that a is also adivisor of c? For example, 8 | 4 · 10 = 40, but 8 - 4 and 8 - 10 We can use Theorem5.1 to answer this question

Lemma 5.3 If a | bc and if (a, b) = 1, then a | c

Proof Since (a, b) = 1, there are integers x and y such that

ax + by = 1,and since a | bc, there is an integers k such that ak = bc Then

2 Express (17, 37) as a linear combination of 17 and 37.

3 Express (399, 703) as a linear combination of 399 and 703

4 Find integers r and s such that 547r + 632s = 1

5 Find integers r and s such that 398r + 600s = 2

6 Find integers r and s such that 922r + 2163s = 7

7 Use the Euclidean algorithm to find (29, 11), and show that



= 1

Hint: use the theorem on linear combinations

10 Show that if there is no prime p such that p | a and p | b, then (a, b) = 1

11 Show that if p is a prime and a is an integer, then either (a, p) = 1 or (a, p) = p

12 Prove that (a, b)n= (an, bn) for all natural numbers n

13 Suppose that (a, b) = 1 Show that (a + b, a2− ab + b2) = 1 or 3

14 A number L is called a common multiple of m and n if both m and n divide

L The smallest such L is called the least common multiple of m and n and isdenoted by lcm(m, n) For example, lcm(3, 7) = 21 and lcm(12, 66) = 132.(a) Find each of the following

i lcm(8, 12)

(d) Suppose that (m, n) = 18 and lcm(m, n) = 720 Find m and n Is theremore than one possibility? If so, find all of them.

(e) Suppose that (a, b) = 1 Show that for every integer c, the equation

ax + by = chas a solution in integers x and y

(f) Find integers x and y such that 37x + 47y = 103

15 Find two positive integers a and b such that a2+b2 = 85113 and lcm(a, b) = 1764

16 For all integers n ≥ 0, define

Fn= 22n + 1

Fn is called the n-th Fermat number Find (Fn, Fm)

17 Let a be an integer greater than or equal to 1 Find all integers b ≥ 1 such that(2b− 1) | (2a− 1)

18 Show that

(n3 + 3n + 1, 7n3+ 18n2− n − 2) = 1for all integers n ≥ 1

19 Let the integers an and bn be defined by the relationship

an+ bn√

2 = (1 +√

2)nfor all integers n ≥ 1 Prove that (an, bn) = 1 for all integers n ≥ 1

20 Find integers x, y, z that satisfy the equation

6x + 15y + 20z = 1

21 Under what conditions on a, b, c is it true that the equation

ax + by + cz = 1has a solution? Describe a general method for finding a solution when oneexists

Prime Factorization and the

Fundamental Theorem of

Arithmetic

Theorem 6.1 Let p be a prime number, and suppose that p | ab Then either p | a

or p | b (or p divides both a and b)

Proof Suppose that p is a prime number that divides the product ab If p | a, then

we have nothing to prove, so let’s assume that p - a Consider the greatest commondivisor (a, p) We know that

Theorem 6.2 Let p be a prime number, and suppose that p divides the product

a1a2· · · ar Then p divides at least one of the factors a1, a2, , ar

Proof If p | a1, then we have nothing to prove, so let’s assume that p - a1 ApplyingTheorem 6.1 to the product

a1(a2a3· · · ar),

we conclude that

p | a2a3· · · ar.Now, if p | a2, then we are finished, so let’s assume that p - a2 Applying Theorem6.1 to the product

a2(a3a4· · · ar),

we conclude that

p | a3a4· · · ar.Continuing, we eventually find some ak so that p | ak

Our goal now is to prove that every integer n ≥ 2 can be factored uniquely into

a product of primes p1p2· · · pn Before we prove this result (which seems naturaland, perhaps, obvious), let’s look at an example that should illustrate that uniquefactorization into primes is, in fact, not obvious

Thus, 60 has two completely different prime factorizations in E

Although this example is somewhat contrived, it should convince you that there isreal mathematical content to unique prime factorization Certain number systemshave unique factorization, and others do not The set Z of integers has importantproperties that make the unique factorization theorem true

Theorem 6.3 Fundamental Theorem of Arithmetic (FTA) Every integer n ≥

2 can be factored into a product of primes

n = p1p2· · · pn

in exactly one way

Proof Notice that the FTA actually contains two separate assertions that we mustprove:

1 We must prove that every integer n ≥ 2 can be factored into a product ofprimes

2 We must prove that there is only one such factorization.

We’ll begin by proving the first assertion We’ll construct a proof by contradiction.Suppose that there exist integers greater than 2 that cannot be written as a product

of primes There must be a smallest such integer Call the smallest such integer N Since N cannot be written as a product of primes, we can conclude that N is notprime Thus, there exist integers b and c such that

N = bc,with b, c > 1 and b, c < N Since N is the smallest integer that cannot be written as

a product of primes, b and c can both be written as products of primes:

b = p1p2· · · pk, c = q1q2· · · ql,where all of the pi and qi are prime Then

N = bc = p1p2· · · pkq1q2· · · qkcan also be written as a product of primes This is a contradiction, so we concludethat no such integers exist Thus, every integer n ≥ 2 can be factored into a product

of primes

Next, we’ll prove the second assertion Suppose that there exists an integer n that

we can factor as a product of primes in two ways, say

n = p1p2· · · pk = q1q2· · · ql

We must show that these two factorizations are the same, possibly after rearrangingthe order of the factors First, observe that

p1 | n = q1q2· · · ql,

so by Theorem 6.2, p1 must divide one of the qi We can rearrange the qi’s so that

p1 | q1 But q1 is also a prime number, so its only divisors are 1 and q1 Thus, weconclude that

p1 = q1.Now we cancel p1 = q1 from both sides of the equation to obtain

p2p3· · · pk = q2q3· · · ql.Repeating the same argument as before, we note that

p2 | q1q2· · · ql,

so by Theorem 6.2, p2 must divide one of the qi’s, and after rearranging, we concludethat p2 | q2, so

p2 = q2

since q2 is prime Canceling p2 = q2 from both sides of the equation, we obtain

p1 = q1, p2 = q2, p3 = q3, , pk= qk.Thus, there is only one way to write an integer n ≥ 2 as a product of primes.Applications of the Fundamental Theorem of Arithmetic

Example 6.2 Show that √

2 is irrational

Solution: Proof by contradiction Suppose that √

2 is rational Then there exist

2 = r

s.Then

2 = r

2

s2,so

2s2 = r2.Let n denote the number of prime factors in the prime factorization of s Then thereare 2n prime factors in the prime factorization of s2, and since 2 is prime, there are2n + 1 prime factors in the prime factorization of 2s2, so in particular, 2s2 has an oddnumber of prime factors Next, let m denote the number of prime factors in the primefactorization of r Then there are 2m prime factors in the prime factorization of r2,

so in particular, r2 has an even number of prime factors However, this contradictsthe FTA since

2s2 = r2.Thus, we conclude that√

2 is irrational

Example 6.3 Suppose that a and n are positive integers and that √n

a is rational.Prove that √n

asn= rn.Without loss of generality, we may assume that (r, s) = 1 (otherwise, divide thenumerator and denominator by (r, s) so that the fraction is in lowest terms) Wewill use proof by contradiction to show that s = 1 Suppose that s > 1 Then there

is a prime p that divides s, so

p | asn= rn.Thus, by Theorem 6.2,

2 < 2, √

2 is not an integer, so it is not rational by the result of thisexample

Example 6.4 Show that log102 is irrational

Solution: Proof by contradiction Suppose that log102 is rational Then there existintegers r, s such that

log102 = r

s.Then

10r/s = 2,so

10r = 2s,or

5r2r = 2s,which contradicts the FTA Thus, log102 is irrational

Example 6.5 Prove that if the polynomial

for some polynomial q(x) with integer coefficients Then we have

14 − 7 = 7 = p(m) − 7 = (m − a1)(m − a2)(m − a3)(m − a4)q(m)

Since the factors m − ak are all distinct, we have decomposed the integer 7 into atleast four different factors However, by the FTA, the integer 7 can be written as aproduct of at most 3 different integers: 7 = (−7)(1)(−1) Thus, we have reached acontradiction, so we conclude that the polynomial cannot take the value 14 for anyintegral value of x

Example 6.6 Prove that m5 + 3m4n − 5m3n2 − 15m2n3 + 4mn4 + 12n5 is neverequal to 33

Solution: Observe that

m5+ 3m4n − 5m3n2− 15m2n3+ 4mn4+ 12n5

= (m − 2n)(m − n)(m + n)(m + 2n)(m + 3n)

Now, 33 can be decomposed as the product of at most four different integers: 33 =(−11)(3)(1)(−1) or 33 = (−3)(11)(1)(−1) If n 6= 0, the factors in the above productare all different By the FTA, they cannot multiply to 33, since 33 is the product

of at most 4 different factors and the expression above is the product of 5 differentfactors for n 6= 0 If n = 0, the product of the factors is m5, and 33 is clearly not afifth power Thus, m5+ 3m4n − 5m3n2− 15m2n3+ 4mn4+ 12n5 is never equal to 33

Example 6.7 Prove that there is exactly one natural number n such that 28+211+2n

is a perfect square

Solution: Suppose that k is an integer such that

k2 = 28+ 211+ 2n = 2304 + 2n= 482 + 2n.Then

k2− 482 = (k − 48)(k + 48) = 2n

By the FTA,

k − 48 = 2s and k + 48 = 2t,where s + t = n But then

2t− 2s= 48 − (−48) = 96 = 3 · 25, so

2s(2t−s− 1) = 3 · 25

By the FTA, s = 5 and t − s = 2, so s + t = n = 12 Thus, the only natural number

n such that 28 + 211+ 2n is a perfect square is n = 12

5 Prove that if n ≥ 2, then √n

n is irrational Hint: show that if n > 2, then

2n> n

6 Prove that log107 is irrational

7 Prove that log 3

log 2 is irrational.

8 Find the smallest positive integer such that n/2 is a square and n/3 is a cube

9 In this exercise, you will continue your investigation of the set E, the set of evennumbers

(a) Classify all primes in E We will refer to such integers as E-primes

(b) We have seen that 60 has two different factorizations as a product of primes Show that 180 has three different factorizations as a product ofE-primes

E-(c) Find the smallest number with four different factorizations in E

(d) The number 12 has only one factorization as a product of primes in E:

12 = 2 · 6 Describe all even numbers that have only one factorization as aproduct of E-primes

10 Let M denote the set of positive integers that leave a remainder of 1 whendivided by 4, i.e

M = {1, 5, 9, 13, 17, 21, }

Note that all numbers in M are numbers of the form 4k + 1 for k = 0, 1, 2, (a) Show that the product of two numbers in M is also in M, i.e if a and bboth leave a remainder of 1 when divided by 4, then ab does as well.(b) Find the first six M-primes in M An integer is an M-prime if its onlydivisors in M are 1 and itself

(c) Find a number in M that has two different factorizations as a product ofM-primes Conclude that M does not have unique factorization

11 Consider the set

F = {a + b

√

−6},where a and b are integers

(a) A prime in F is an element of F which has no factors in F other than 1 anditself Show that 2 and 5 are F-primes.

(b) Show that 7 and 31 are not F-primes

(c) Find two different factorizations of the number 10 in F

(d) Conclude that F does not have unique factorization

12 Show that if p is a prime and p | an, then pn | an

13 How many zeros are there at the end of 100!?

14 Prove that the sum

1/3 + 1/5 + 1/7 + · · · + 1/(2n + 1)

is never an integer Hint: Look at the largest power of 3 ≤ n

15 Find the number of ways of factoring 1332 as the product of two positive tively prime factors each greater than 1

rela-16 Let p1, p2, , pt be different primes and a1, a2, at be natural numbers Findthe number of ways of factoring pa1

Introduction to Congruences and Modular Arithmetic

Definition 7.1 We say that a is congruent to b modulo m, and write

a ≡ b mod m,

if m divides a − b

Equivalently, a ≡ b mod m if a and b leave the same remainder upon division by m

By the Division Algorithm, we observe that a ≡ b mod m if and only if there exists

an integer k such that a = b + km

Example 7.1 7 ≡ 2 mod 5 since 5 | (7−2) Note that 7 and 2 both leave remainder

2 upon division by 5

Example 7.2 47 ≡ 35 ≡ 5 mod 6 since 6|(47 − 35) and 6|(35 − 5) Note that 47,

35, and 5 all leave remainder 5 upon division by 6

Example 7.3 9 ≡ 0 mod 3 since 3 | 9 Note that 9 leaves a remainder of 0 upondivision by 3

Example 7.4 15 ≡ 7 ≡ −1 mod 8 since 8 | (15 − 7) and 8 | (7 − −1)

Example 7.5 Construct an addition table and a multiplication table for arithmeticmodulo 5

Solution: