Question

If p is prime, then (p-1)! + 1 =0 mod p

Prove it

Answer 1

Wow interestingly all your questions are the ones I have only just studied a few months earlier! :) Here you go: Wilson's Theorem! I suggest you read up on this theorem.

I think this is a good proof:

(taken from http://primes.utm.edu/notes/proofs/Wilsons.html)

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.

There you go! Hope that helped. Have a great day!

Answer 2

Thankyou