The question is \"Compute ((p -1)/2)! modulo p\" I know Wilson\'s theorem that (

Question

The question is "Compute ((p -1)/2)! modulo p"
I know Wilson's theorem that (p-1)! -1 modulo p.

By experimenting with different primes, it looks like ((p -1)/2)! -1 mod p, but I need a formal proof.

Explanation / Answer

If p is composite, then it has divisors among the integers 1, 2, 3, 4, ... , p-1 and gcd ((p-1)!,p) > 1, so we cannot 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). 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) which is the same as (p-2)! = 1 (mod p)

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