If $p$ is prime and congruent to $1$, then show $((\frac{p-1}{2})!)^2 \equiv -1 \pmod p$ I got another one.
Quadratic residues are completely new to me...
Thanks!

Take the congruences, $p-1 \equiv -1 \pmod p$, $\text{ }\text{ } p-2 \equiv -2 \pmod p$
and so on upto, $\frac{p+1}2 \equiv -\frac{p-1}2 \pmod p$.
Multiplying and rearranging, $$(p-1)!\equiv 1\cdot (-1)\cdot 2 \cdot (-2) ...\frac{p-1}2 \cdot (-\frac{p-1}2) \equiv (-1)^{\frac{p-1}2}[(\frac{p-1}{2})!]^2 \pmod p$$
Thus, by Wilson's theorem, $ -1 \equiv (-1)^{\frac{p-1}2}[(\frac{p-1}{2})!]^2 \pmod p$. 
If by "$p$ is congruent to $1$", you meant that $p \equiv 1 \pmod 4$, we are done as $(-1)^{\frac{p-1}2} =1$.
