If prime $p \equiv 1\pmod 4$ and $b = ((p-1)/2)!$ then show that $b^2 \equiv -1\pmod p$.
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.