History of Mathematics Paul Yiu Department of Mathematics Florida Atlantic University Summer 2011 6A: Fermat: Beginning of modern number theory (2)
Fermat s little theorem Given a prime p, and any geometric progression 1, a, a 2,..., p must divide some number a n 1 for which n divides p 1. If N is any multiple of the smallest n for which this is so, p divides also a N 1. 1
Fermat s little theorem Given a prime p, and any geometric progression 1, a, a 2,..., p must divide some number a n 1 for which n divides p 1. If N is any multiple of the smallest n for which this is so, p divides also a N 1. This can only be true when a is not divisible by p. Consider the mod p residues of 1, a, a 2,..., Each of these residues is a positive integer < p. Therefore, two of them must be the same. If h is the smallest positive integer for which there is a greater integer k satisfying a k and a h have the same residue, then a h (a k h 1) is divisible by p, and a k h 1 is divisible by p. 2
Fermat s little theorem Given a prime p, and any geometric progression 1, a, a 2,..., p must divide some number a n 1 for which n divides p 1. If N is any multiple of the smallest n for which this is so, p divides also a N 1. 3
Fermat s little theorem Given a prime p, and any geometric progression 1, a, a 2,..., p must divide some number a n 1 for which n divides p 1. If N is any multiple of the smallest n for which this is so, p divides also a N 1. This can only be true when a is not divisible by p. Consider the mod p residues of 1, a, a 2,..., Each of these residues is a positive integer < p. Therefore, two of them must be the same. If h is the smallest positive integer for which there is a greater integer k satisfying a k and a h have the same residue, then a h (a k h 1) is divisible by p, and a k h 1 is divisible by p. Next, we show that a p 1 1 is divisible by p. We offer two proofs. 4
Fermat s little theorem (modern proof) If p is a prime, then for every integer a not divisible by p, a p 1 1 mod p. Proof. If p is prime, the nonzero residues 1, 2,..., p 1 form a group under multiplication. Therefore, if a is an integer not divisible by p, the residues a 1, a 2,..., a(p 1) form a permutation of 1, 2,..., p 1. Therefore, (a 1)(a 2) (a(p 1)) 1 2 (p 1) mod p, Since each of 1, 2,..., p 1 is prime to p, canceling these common factors, we have a p 1 1 mod p. 5
Fermat s little theorem (Inductive proof) Given a prime p, and any geometric progression 1, a, a 2,..., p must divide some number a n 1 for which n divides p 1. If N is any multiple of the smallest n for which this is so, p divides also a N 1. This is true for a = 2: 2 p = (1 + 1) p = 1 + ( ) p + 1 ( ) ( ) p p + + + 1. 2 p 1 Each of the binomial coefficients ( ) p p(p 1) (p k + 1) = k 1 2 k is an integer divisible by p. Therefore, 2 p 2 is divisible by p, so is 2 p 1 1. 6
Fermat s little theorem (inductive proof) For every integer a, (a + 1) p = a p + ( ) p a p 1 + 1 Therefore, (a + 1) p a p 1 is divisible by p, or ((a + 1) p (a + 1)) (a p a) is divisible by p. ( ) ( ) p p a p 2 + + a + 1. 2 p 1 Since 2 p 2 is divisible by p, it follows that a p a is divisible by p for every positive integer a 2, (certainly including a = 1 as well). 7
Fermat s little theorem (inductive proof) For every integer a, (a + 1) p = a p + ( ) p a p 1 + 1 Therefore, (a + 1) p a p 1 is divisible by p, or ((a + 1) p (a + 1)) (a p a) is divisible by p. ( ) ( ) p p a p 2 + + a + 1. 2 p 1 Since 2 p 2 is divisible by p, it follows that a p a is divisible by p for every positive integer a 2, (certainly including a = 1 as well). Therefore, if a is not divisible by p, then a p 1 1 mod p. 8
Fermat s little theorem Given a prime p, and any geometric progression 1, a, a 2,..., p must divide some number a n 1 for which n divides p 1. If N is any multiple of the smallest n for which this is so, p divides also a N 1. Now, if n is the smallest number for which a n 1 is divisible by p, and if a N 1 is also divisible by p, then N must be divisible by n. 9
Fermat s little theorem Given a prime p, and any geometric progression 1, a, a 2,..., p must divide some number a n 1 for which n divides p 1. If N is any multiple of the smallest n for which this is so, p divides also a N 1. Now, if n is the smallest number for which a n 1 is divisible by p, and if a N 1 is also divisible by p, then N must be divisible by n. Suppose N is not divisible by n. Write N = nb + r for 0 < r < n. a N 1 = a nb+r 1 = (a n ) b a r 1 a r 1 mod p. Since a N 1 is divisible by p, so is a r 1 for 0 < r < n. This contradicts the minimality of n. 10
Application to primality test Let k be an odd prime. Every prime factor of M k := 2 k 1 is of the form 2kr + 1. 11
Application to primality test Let k be an odd prime. Every prime factor of M k := 2 k 1 is of the form 2kr + 1. Proof. Let p be a prime factor of 2 k 1. Since k is prime, it is the smallest n for which 2 n 1 is divisible by p. Now, 2 p 1 1 is also divisible by p. Therefore, p 1 is divisible by k, and indeed by 2k since p 1 is even. Therefore, there is an integer r for which p 1 = 2kr, and p = 2kr + 1. 12
Testing Mersenne primes Let k be an odd prime. Every prime factor of M k := 2 k 1 is of the form 2kr + 1. Example 1: M 11 = 2 11 1 = 2047. The only prime divisors are of the form 22r + 1. We need only test prime divisors < 2047 45. It is enough to check p = 23 (with r = 1). Now, indeed, M 11 = 2047 = 23 89. Note that the other prime factor is 89 = 2 11 4 + 1. 13
Testing Mersenne primes Let k be an odd prime. Every prime factor of M k := 2 k 1 is of the form 2kr + 1. Example 1: M 11 = 2 11 1 = 2047. The only prime divisors are of the form 22r + 1. We need only test prime divisors < 2047 45. It is enough to check p = 23 (with r = 1). Now, indeed, M 11 = 2047 = 23 89. Note that the other prime factor is 89 = 2 11 4 + 1. Example 2: M 17 = 2 17 1 = 131071. We need only check prime divisor < 13107 < 363 which are of the form 34r + 1. There are only 4 such primes: 103, 137, 239, and 307. None of these divides 131071, which is therefore a prime. 14
Infinite descent: area of Pythagorean triangle There is no right angled triangle in numbers whose area is a square. Infinite descent: If there is a triangle whose area is a square, then from such a triangle, it is possible to construct a smaller triangle whose area is also a square, and so on, which is impossible. 15
Antecedent of Fermat s theorem There is no right angled triangle in numbers whose area is a square. Bachet s edition of Diophantus Arithmetica contains a supplement to Book VI, in which Bachet gives a necessary and sufficient condition for a rational number A to be the area of a right triangle: (2A) 2 + K 4 = square for some integer K. Bachet asks whether it is possible for a right triangle to have an area equal to a square. In the margin at the end of his copy of Arithmetica, Fermat gave his proof. 16
Fermat s proof by infinite descent There is no right angled triangle in numbers whose area is a square. Proof. Suppose there is one such integer right triangle. We may assume it primitive, i.e., its sidelengths are a = m 2 n 2, n = 2mn, c = m 2 + n 2 for relatively prime integers m, n of different parity. Its area A = mn(m 2 n 2 ) is a square. 17
Fermat s proof by infinite descent There is no right angled triangle in numbers whose area is a square. Proof. Suppose there is one such integer right triangle. We may assume it primitive, i.e., its sidelengths are a = m 2 n 2, n = 2mn, c = m 2 + n 2 for relatively prime integers m, n of different parity. Its area A = mn(m 2 n 2 ) is a square. Now, no two of m, n, m 2 n 2 share common divisors. Therefore, each of these is a square. Write m = p 2 and n = q 2. Then m 2 n 2 = p 4 q 4 is also a square. 18
Fermat s proof by infinite descent There is no right angled triangle in numbers whose area is a square. Proof. Suppose there is one such integer right triangle. We may assume it primitive, i.e., its sidelengths are a = m 2 n 2, n = 2mn, c = m 2 + n 2 for relatively prime integers m, n of different parity. Its area A = mn(m 2 n 2 ) is a square. Now, no two of m, n, m 2 n 2 share common divisors. Therefore, each of these is a square. Write m = p 2 and n = q 2. Then m 2 n 2 = p 4 q 4 is also a square. Note that p 4 q 4 = (p 2 + q 2 )(p 2 q 2 ) and that p 2 + q 2 and p 2 q 2 do not have common divisors, we must have p 2 + q 2 = r 2 and p 2 q 2 = s 2 for some integers r and s. 19
Fermat s proof by infinite descent There is no right angled triangle in numbers whose area is a square. Proof. Suppose there is one such integer right triangle. We may assume it primitive, i.e., its sidelengths are a = m 2 n 2, n = 2mn, c = m 2 + n 2 for relatively prime integers m, n of different parity. Its area A = mn(m 2 n 2 ) is a square. Now, no two of m, n, m 2 n 2 share common divisors. Therefore, each of these is a square. Write m = p 2 and n = q 2. Then m 2 n 2 = p 4 q 4 is also a square. Note that p 4 q 4 = (p 2 + q 2 )(p 2 q 2 ) and that p 2 + q 2 and p 2 q 2 do not have common divisors, we must have p 2 + q 2 = r 2 and p 2 q 2 = s 2 for some integers r and s. From these, 2p 2 = r 2 + s 2 and (2p) 2 = 2(r 2 + s 2 ) = (r + s) 2 + (r s) 2. In other words, (r s, r + s, 2p) is an integer right triangle, and its area is 1 2 (r + s)(r s) = 1 2 (r2 s 2 ) = q 2. This is a smaller triangle since q 2 = n < mn(m 2 n 2 ) = A. 20
Fermat s Last Theorem If n 3, the equation x n + y n = z n has no solution in nonzero integers x, y, z. Origin: Fermat s remark on Diophantus II.8. To divide a given square number into two squares. On the other hand it is impossible to separate a cube into two cubes, or a biquadrate into two biquadrates, or generally any power except a square into two powers of the same exponent. I have discovered a truly marvelous proof of this, which, however, the margin is not large enough to contain. 21
FLT for n = 4 There is no solution of x 4 + y 4 = z 4 in nonzero integers. Suppose, to the contrary, that there are positive integers x, y, z satisfying x 4 + y 4 = z 4. Then z 4 y 4 = x 4. Consider the right triangle with sides z 4 y 4, 2z 2 y 2 and z 4 + y 4. This has area z 2 y 2 (z 4 y 4 ) = x 4 y 2 z 2 = (x 2 yz) 2, contradicting the fact that the area of a right triangle cannot be a square. 22
Brief history of the Fermat Last Theorem Fermat s proof for n = 4. n = 3: Euler gave two proofs in 1760 and 1770. n = 5: P. G. Lejeune Dirichlet (1805-1859) and A. M. Legendre (1752-1833) in 1825. By 1985, known to be true for n 125, 000. G. Faltings (1954 ) proved in 1983 that for each exponent n 3, the equation x n + y n = z n has at most a finite number of solution in x, y, z in which x, y, z have no common divisors. 23
Brief history of the Fermat Last Theorem Fermat s proof for n = 4. n = 3: Euler gave two proofs in 1760 and 1770. n = 5: P. G. Lejeune Dirichlet (1805-1859) and A. M. Legendre (1752-1833) in 1825. By 1985, known to be true for n 125, 000. G. Faltings (1954 ) proved in 1983 that for each exponent n 3, the equation x n + y n = z n has at most a finite number of solution in x, y, z in which x, y, z have no common divisors. In June 1993, Andrew Wiles (1953 ) of Princeton University announced a proof of Fermat s Last Theorem. But a gap was discovered later in the proof. With the help of Richard Taylor, Wiles proof of the Fermat Last Theorem was finally completed in September 1994, and published in the Annals of Mathematics, 1995. 24
Further contributions and challenges Fermat s two-square theorem (proved by Euler): Every prime number of the form 4k + 1 can be written uniquely as a sum of two squares. For examples, 5 = 1 2 + 2 2, 13 = 2 2 + 3 2, 17 = 1 2 + 4 2, 29 = 2 2 + 5 2. Every prime number of the form 3k + 1 can be written as x 2 + 3y 2. Every prime number of the form 8k + 1 or 8k + 3 can be written as x 2 + 2y 2. Find a cube which, added to its proper divisors, gives a square. For example: 7 3 + 7 2 + 7 + 1 = 400 = 20 2. Find a square which, added to its proper divisors, gives a cube: σ(x 2 ) x 2 = y 3. Example given by Frenicle (1612-1675): x = 2 2 5 7 11 37 67 163 191 263 439 499, y = 3 2 7 3 13 19 31 2 67 109. 25