Let p be a prime Show that if x2 1 in Zopf p then either x

Let p be a prime. Show that if x^2 = 1 in Zopf _p. then either x = 1 or x = - 1. Use this result to prove Wilson\'s Theorem: (p - 1)! = - 1 (mod p).

Solution

1, 2, 3, 4, ... , p-1

2^.3^.4^....^.(p-2) = 1 (mod p)

(p-1) (p-2)! = p-1 (modp)

(p-1)! = -1 (mod p)

It is easy to check the result when p is 2 or 3, so let us assume p > 3. If p is composite, then its positive divisors are among the integers 1, 2, 3, 4, ... , p-1 and it is clear that gcd((p-1)!,p) > 1, so we can not have (p-1)! = -1 (mod p). However if p is prime, then each of the above integers are relatively prime to p. So for each of these integers a there is another b such that ab = 1 (mod p). It is important to note that this b is unique modulo p, and that since p is prime, a = b if and only if a is 1 or p-1. Now if we omit 1 and p-1, then the others can be grouped into pairs whose product is one showing 2.3.4.....(p-2) = 1 (mod p) (or more simply (p-2)! = 1 (mod p)). Finally, multiply this equality by p-1 to complete the proof.