Prove is $n \in \mathbb{N}$ and is $p$ is prime such that $p|(n!)^2+1$ then $(p-1)/2$ is even? Can anyone help me prove if $n \in \mathbb{N}$ and is $p$ is prime such that $p|(n!)^2+1$ then $(p-1)/2$ is even?
I'm attempting to use Fermats little theorem, so far I have only shown $p$ is odd.
I want to show that $p \equiv 1 \pmod 4$

If I am not mistaken this is just a special case of the first supplement to the quadratic reciprocity law:
$p|x^2+1\Rightarrow x^2 \equiv -1 \pmod p$. This is solvable if and only if $p \equiv 1 \pmod 4$. And you have a given solution.
