If $ p \equiv 1 \pmod{4}$, prove $((\frac{p-1}{2})!)^2 \equiv -1 \pmod {p}$ where p is prime. Characteristics: The fields where $ p \equiv 1 \pmod{4}$ has half the number from 1 to $\frac{p-1}{2}$ both in positive and the negative. There can be paired up such that when multiplied together, they equal $ -(p-1) \equiv +1 \pmod p$. Only one pair equals -1 and that is $\pm 1$.
Why does that happen?

If we start from Wilson's theorem, $(p-1)! \equiv -1 \pmod{p}$ for a prime $p$, then we can write
$$\begin{align}
-1 &\equiv (p-1)! = \prod_{k=1}^{\frac{p-1}{2}} k \cdot \prod_{m = 1}^{\frac{p-1}{2}} (p-m)\\
&\equiv \left(\frac{p-1}{2}\right)! \cdot (-1)^{\frac{p-1}{2}}\prod_{m = 1}^{\frac{p-1}{2}} (m-p)\\
&\equiv (-1)^{\frac{p-1}{2}} \left(\left(\frac{p-1}{2}\right)!\right)^2
\end{align}$$
for odd primes $p$. Now if $p \equiv 1 \pmod{4}$, the factor $(-1)^{\frac{p-1}{2}}$ is $1$, hence
$$\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv -1 \pmod{p}$$
then.
It is Wilson's theorem that follows from pairing up each $1 < k < p-1$ with its inverse, leaving $(p-1)! \equiv 1\cdot(p-1) \equiv -1 \pmod{p}$.
