- #1

cwatki14

- 57

- 0

My train of thought...:

If m is composite, which has a prime factors r and s such that r does not equal s, then m divides (m-1)! then (m-1)! is congruent to 0 (mod m).

Now consider the case of m=p^2 where p is a prime.

If m> 2p then px2p divides (m-1)! therefore (m-1)! is congruent to 0 (mod m).

If m<= 2p then 2p>= p^2 by dividing each side by p we show that 2 >=p. The only case that exists when p=2. We can note the 3! is congruent to 2 (mod4).

Thus we have proven that (m-1)! is not congruent to -1 (mod m) is m is composite. In fact, when m is any composite besides 4, (m-1)! is congruent to 0 (mod m)

Also, we can note if m is composite that it has prime factors such that p<m. Hence if n divides (m-1)!+1 then p also divides (m-1)! +1. This is impossible because p divides (m-1)! and it can not divide (m-1)!+1

Does this suffice for the proof?