MA3412S2_Hil2014.pdf

# Proof the factorial p 1 is the product of the

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• 38

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Proof The factorial ( p - 1)! is the product of the integers k ( p - k ) for k = 1 , 2 , . . . , m . Moreover k ( p - k ) ≡ - k 2 (mod p ) for k = 1 , 2 , . . . , m . Therefore ( p - 1)! = m Y k =1 k ( p - k ) m Y k =1 ( - k 2 ) = ( - 1) m ( m !) 2 (mod p ) . Thus ( m !) 2 + ( - 1) m ( - 1) m (( p - 1)! + 1) (mod p ) . It follows from Wilson’s Theorem (Theorem 2.10) that ( m !) 2 + ( - 1) m is divisible by p , as required. We now prove Fermat’s Two Squares Theorem using a method based on properties of the ring Z [ - 1] of Gaussian integers published by Richard Dedekind in 1894. Theorem 2.12 (Fermat’s Two Squares Theorem) Let p be an odd prime number. Then there exist integers x and y such that p = x 2 + y 2 if and only if p 1 (mod 4) . 18

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Proof If x is an even integer then x 2 0 (mod 4). If x is an odd integer then x 2 1 (mod 4). It follows that if x and y are integers, and if x 2 + y 2 is an odd integer then x 2 + y 2 1 (mod 4). (It is not possible to represent an integer congruent to 3 modulo 4 as a sum of two squares.) Thus only odd primes p satisfying p 1 (mod 4) can be represented as the sum of two squares. Now let p be an odd prime satisfying p 1 (mod 4), and let m = 1 2 ( p - 1). Then m is an even integer. It follows from Corollary 2.11 that p divides ( m !) 2 + 1. We now consider the nature of the prime number p , considered as an element of the ring Z [ - 1] of Gaussian integers, where Z [ - 1] = { x + y - 1 : x, y Z } . Now ( m !) 2 + 1 factorizes as the product ( m !) 2 + 1 = ( m ! + - 1)( m ! - - 1) of Gaussian integers m ! + - 1 and m ! - - 1. But neither m ! + - 1 nor m ! - - 1 is divisible by the prime number p in the ring Z [ - 1] of Gaussian integers, despite the fact that p divides the product of these two Gaussian integers. It follows that if the prime number p satisfies p 1 (mod 4) then p is not a prime element of the ring Z [ - 1]. But Z [ - 1] is a Euclidean domain, and is thus a principal ideal domain (Proposition 2.6), and therefore every irreducible element of Z [ - 1] is prime (Theorem 2.9). Because p is not a prime element of Z [ - 1], it cannot be an irreducible element of Z [ - 1], and therefore there must exist Gaussian integers ω, θ Z [ - 1] such that p = ωθ , where neither ω nor θ is a unit of Z [ - 1]. Now the norm x 2 + y 2 of any non-zero Gaussian integer x + y - 1 is a positive integer, and has the value one if and only if x + y - 1 is a unit of Z [ - 1]. It follows that | ω | 2 and | θ | 2 are positive integers satisfying | ω | 2 > 1 and | θ | 2 > 1. But | ω | 2 | θ | 2 = p 2 , and the only factors of p 2 are 1, p and p 2 . It follows that | ω | 2 = | θ | 2 = p . Let ω = x + y - 1. Then p = | ω | 2 = x 2 + y 2 . Thus a prime number p satisfying p 1 (mod 4) can be represented in the form p = x 2 + y 2 for some integers x and y , as required. Remark The above proof of Fermat’s Two Squares theorem uses the fact that if p is a prime number satisfying p 1 (mod 4) then there exists an integer w satisfying the congruence w 2 ≡ - 1 (mod p ). The existence of such an integer shows that the number - 1 is a quadratic residue of p when p 1 (mod 4), and can be proved in various ways. One of these ways involves the use of Wilson’s Theorem, as explained above, to show that if p 1 (mod 4) then - 1 ( m !) 2 (mod p ), where m = ( p
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• Fall '16
• Jhon Smith
• Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

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