# Why does $(\frac{p-1}{2}!)^2 = (-1)^{\frac{p+1}{2}}$ mod $p$?

This question already has an answer here:

• Show for prime numbers of the form $p=4n+1$, $x=(2n)!$ solves the congruence $x^2\equiv-1 \pmod p$. $p$ is therefore not a gaussian prime.

#### Solutions Collecting From Web of "Why does $(\frac{p-1}{2}!)^2 = (-1)^{\frac{p+1}{2}}$ mod $p$?"
$$(p-1)!=1\cdot2\cdots\frac{p-1}{2}\cdot\frac{p+1}{2}\cdots(p-2)(p-1)$$ We have the congruences
\begin{align*} p-1&\equiv -1\pmod p\\ p-2&\equiv -2\pmod p\\ &\vdots\\ \frac{p+1}{2}&\equiv -\frac{p-1}{2}\pmod{p}\end{align*}
$$(p-1)!\equiv 1\cdot(-1)\cdot2\cdot(-2)\cdots\frac{p-1}{2}\cdot\left(-\frac{p-1}{2}\right)\pmod p.$$
$\therefore (p-1)!\equiv (-1)^{\frac{p-1}{2}}\left(1\cdot2\cdots\frac{p-1}{2}\right)^{2}\pmod p.$
From Wilson’s theorem, $(p-1)!\equiv -1\pmod{p}$
Thus $$-1\equiv (-1)^{\frac{p-1}{2}}\left[\left(\frac{p-1}{2}\right)!\right]^{2}\pmod{p}$$
It follows that $\left[\left(\frac{p-1}{2}\right)!\right]^{2}\equiv (-1)^{\frac{p+1}{2}}\pmod{p}$.