Let p be a prime larger than 2. Prove that 2(p − 3)! ≡ −1 mod p

Answer 1

(p-1)(p-2) = p^2 - 3p + 2, so (p-1)(p-2) ≡ 2 mod p for p> 2

Thus, 2(p - 3!) ≡ (p-1)(p-2)(p - 3!) ≡ p-1! mod p

From Wilson's theorem, p-1! mod p ≡ -1 mod p

Thus, 2(p - 3!) ≡ -1 mod p