1. Elementary Number Theory and Methods of Proof

2. Basic Definitions An integer n is an even number if there exists an integer k such that n = 2k. An integer n is an odd number if there exists an integer k such that n = 2k+1. An integer n is a prime number if and only if n>1 and if n=rs for some positive integers r and s then r=1 or s=1.

3. Simple Exercises The sum of two even numbers is even. The product of two odd numbers is odd. direct proof.

4. Rational Number R is rational  there are integers a and b such that and b ≠ 0. numerator denominator Is 0.281 a rational number? Is 0 a rational number? If m and n are non-zero integers, is (m+n)/mn a rational number? Is the sum of two rational numbers a rational number? Is 0.12121212…… a rational number?

5. a “divides” b (a|b): b = ak for some integer k Divisibility 5|15 because 15 = 3  5 n|0 because 0 = n  0 1|n because n = 1  n n|n because n = n  1 A number p > 1 with no positive integer divisors other than 1 and itself is called a prime. Every other number greater than 1 is called composite. 2, 3, 5, 7, 11, and 13 are prime, 4, 6, 8, and 9 are composite.

6. 1. If a | b, then a | bc for all c. 2. If a | b and b | c, then a | c. 3. If a | b and a | c, then a | sb + tc for all s and t. 4. For all c ≠ 0, a | b if and only if ca | cb. Simple Divisibility Facts Proof of (1) direct proof.

7. 1. If a | b, then a | bc for all c. 2. If a | b and b | c, then a | c. 3. If a | b and a | c, then a | sb + tc for all s and t. 4. For all c ≠ 0, a | b if and only if ca | cb. Simple Divisibility Facts Proof of (2) direct proof.

8. Divisibility by a Prime Theorem. Any integer n > 1 is divisible by a prime number. Idea of induction.

9. Every integer, n>1, has a unique factorization into primes: p 0 ≤ p 1 ≤ ··· ≤ p k p 0 p 1 ··· p k = n Fundamental Theorem of Arithmetic Example: 61394323221 = 3·3·3·7·11·11·37·37·37·53

10. Claim: Every integer > 1 is a product of primes. Prime Products Proof: (by contradiction) Suppose not. Then set of non-products is nonempty. There is a smallest integer n > 1 that is not a product of primes. In particular, n is not prime. So n = k·m for integers k, m where n > k,m >1. Since k,m smaller than the least nonproduct, both are prime products, eg., k = p 1  p 2    p 94 m = q 1  q 2    q 214

11. Prime Products …So n = k  m = p 1  p 2    p 94  q 1  q 2    q 214 is a prime product, a contradiction.  The set of nonproducts > 1 must be empty. QED Claim: Every integer > 1 is a product of primes. (The proof of the fundamental theorem will be given later.) Idea of induction (or smallest counterexample).

12. For b > 0 and any a, there are unique numbers q ::= quotient(a,b), r ::= remainder(a,b), such that a = qb + r and 0  r < b. The Quotient-Reminder Theorem When b=2, this says that for any a, there is a unique q such that a=2q or a=2q+1. When b=3, this says that for any a, there is a unique q such that a=3q or a=3q+1 or a=3q+2. We also say q = a div b and r = a mod b.

13. For b > 0 and any a, there are unique numbers q ::= quotient(a,b), r ::= remainder(a,b), such that a = qb + r and 0  r < b. The Division Theorem 0 b 2b kb (k+1)b Given any b, we can divide the integers into many blocks of b numbers. For any a, there is a unique “position” for a in this line. q = the block where a is in a r = the offset in this block Clearly, given a and b, q and r are uniquely defined. -b

14. The Square of an Odd Integer 3 2 = 9 = 8+1, 5 2 = 25 = 3x8+1 …… 131 2 = 17161 = 2145x8 + 1, ……… Idea 1: prove that n 2 – 1 is divisible by 8. Idea 2: consider (2k+1) 2 Idea 0: find counterexample.

15. The Square of an Odd Integer Idea 3: Use quotient-remainder theorem. Proof by cases.

16. Trial and Error Won’t Work! Euler conjecture: has no solution for a,b,c,d positive integers. Open for 218 years, until Noam Elkies found Fermat (1637): If an integer n is greater than 2, then the equation a n + b n = c n has no solutions in non-zero integers a, b, and c. Claim: has no solutions in non-zero integers a, b, and c. False. But smallest counterexample has more than 1000 digits.

17. Since m is an odd number, m = 2l+1 for some natural number l. So m 2 is an odd number. The Square Root of an Even Square Statement: If m 2 is even, then m is even Contrapositive: If m is odd, then m 2 is odd. So m 2 = (2l+1) 2 = (2l) 2 + 2(2l) + 1 Proof (the contrapositive): Proof by contrapositive.

18. Suppose was rational. Choose m, n integers without common prime factors (always possible) such that Show that m and n are both even, thus having a common factor 2, a contradiction! Theorem: is irrational. Proof (by contradiction): Irrational Number

19. so can assume so n is even. so m is even. Theorem: is irrational. Proof (by contradiction):Want to prove both m and n are even. Proof by contradiction. Irrational Number

20. Infinitude of the Primes Theorem. There are infinitely many prime numbers. Claim: if p divides a, then p does not divide a+1. Let p 1, p 2, …, p N be all the primes. Proof by contradiction. Consider p 1 p 2 …p N + 1.

21. Greatest Common Divisors Given a and b, how to compute gcd(a,b)? Can try every number, but can we do it more efficiently? Let’s say a>b. 1.If a=kb, then gcd(a,b)=b, and we are done. 2.Otherwise, by the Division Theorem, a = qb + r for r>0.

22. Greatest Common Divisors Let’s say a>b. 1.If a=kb, then gcd(a,b)=b, and we are done. 2.Otherwise, by the Division Theorem, a = qb + r for r>0. Euclid: gcd(a,b) = gcd(b,r)! a=12, b=8 => 12 = 8 + 4 gcd(12,8) = 4 a=21, b=9 => 21 = 2x9 + 3gcd(21,9) = 3 a=99, b=27 => 99 = 3x27 + 18 gcd(99,27) = 9 gcd(8,4) = 4 gcd(9,3) = 3 gcd(27,18) = 9

23. Euclid’s GCD Algorithm Euclid: gcd(a,b) = gcd(b,r) gcd(a,b) if b = 0, then answer = a. else write a = qb + r answer = gcd(b,r) a = qb + r

24. gcd(a,b) if b = 0, then answer = a. else write a = qb + r answer = gcd(b,r) Example 1 GCD(102, 70) 102 = 70 + 32 = GCD(70, 32) 70 = 2x32 + 6 = GCD(32, 6) 32 = 5x6 + 2 = GCD(6, 2) 6 = 3x2 + 0 = GCD(2, 0) Return value: 2.

25. gcd(a,b) if b = 0, then answer = a. else write a = qb + r answer = gcd(b,r) Example 2 GCD(252, 189) 252 = 1x189 + 63 = GCD(189, 63) 189 = 3x63 + 0 = GCD(63, 0) Return value: 63.

26. gcd(a,b) if b = 0, then answer = a. else write a = qb + r answer = gcd(b,r) Example 3 GCD(662, 414) 662 = 1x414 + 248 = GCD(414, 248) 414 = 1x248 + 166 = GCD(248, 166) 248 = 1x166 + 82 = GCD(166, 82) 166 = 2x82 + 2 = GCD(82, 2) 82 = 41x2 + 0 = GCD(2, 0) Return value: 2.

27. Euclid: gcd(a,b) = gcd(b,r) a = qb + r Correctness of Euclid’s GCD Algorithm When r = 0:

28. Euclid: gcd(a,b) = gcd(b,r) a = qb + r Correctness of Euclid’s GCD Algorithm When r = 0: Then gcd(b, r) = gcd(b, 0) = b since every number divides 0. But a = qb so gcd(a, b) = b = gcd(b, r), and we are done.

29. Euclid: gcd(a,b) = gcd(b,r)a = qb + r Correctness of Euclid’s GCD Algorithm Let d be a common divisor of b, r When r > 0:

30. Euclid: gcd(a,b) = gcd(b,r)a = qb + r Correctness of Euclid’s GCD Algorithm Let d be a common divisor of a, b. When r > 0:

31. Euclid: gcd(a,b) = gcd(b,r)a = qb + r Correctness of Euclid’s GCD Algorithm Let d be a common divisor of b, r  b = k 1 d and r = k 2 d for some k 1, k 2.  a = qb + r = qk 1 d + k 2 d = (qk 1 + k 2 )d => d is a divisor of a Let d be a common divisor of a, b  a = k 3 d and b = k 1 d for some k 1, k 3.  r = a – qb = k 3 d – qk 1 d = (k 3 – qk 1 )d => d is a divisor of r So d is a common factor of a, b iff d is a common factor of b, r  d = gcd(a, b) iff d = gcd(b, r) When r > 0: