Question

Let p be an odd prime number, so that p = 2k + 1 for some positive integer k. Prove that
(k!)2 ≡ (−1)k+1 mod p.
Hint: Try to see how to group the terms in the product
(p − 1)! = (2k)! = 1 ∗ 2 ∗ 3 · · ·(2k − 2) ∗ (2k − 1) ∗ (2k)
to get two products, each equal to k! modulo p.

Answer 1

Consider the product $(p-1)! = 1\cdot(p-1)\cdot 2\cdot (p-2)\cdots k\cdot (p-k)$

$$1\cdot(p-1)\cdot 2\cdot (p-2)\cdots k\cdot (p-k)\ \equiv\ 1\cdot (-1)\cdot 2\cdot (-2)\cdots k\cdot (-k)\ \equiv\ -1 \pmod{p}$$

(See Wilson's theorem).

$$1\cdot (-1)\cdot 2\cdot (-2)\cdots k\cdot (-k) = (-1)^k (k!)^2$$

Therefore:

$$(-1)^k (k!)^2 \ \equiv\ -1 \pmod{p}$$ $$(k!)^2 \ \equiv\ (-1)^{k+1} \pmod{p}$$