Prove that if $p > 3$ is a prime then $2(p-3)! \pmod{p} =-1 \pmod{p}$.

Question

Prove that if $p > 3$ is a prime then $2(p-3)! \pmod{p} =-1 \pmod{p}$..

I am totally lost; at first I thought this could be done by induction, but unfortunately this is not possible (at least I believe) because $p + 1$ won't be a prime so $P(n) =/> P(n+1)$.

I tried using modular arithmetic but I haven't gotten far.

Can anybody hint me here?

Answer 1

You could start with Wilsons's theorem $(p-1)!\equiv -1 \mod p$, and then try to make $(p-3)!$ appears in this identity...