TITLE: 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. 
QUESTION [7 upvotes]: I need to show that for prime numbers of the form $p=4n+1$, $x=(2n)!$ solves the congruence $x^2 \equiv-1\pmod p$.
I then need to show this implies p isn't a gaussian prime. 
I have started to solve this using Wilson's theorem that a number $z$ is prime iff $(z-1)!\equiv-1\pmod z$. Therefore the endpoint of my proof should be that $(p-1)!\equiv-1\pmod p$.
As $p$ is of the form $p=4n+1$, I only need to prove that $4n!$ is congruent to $-1$ modulo $p$. 
Here is my working so far: Starting with the congruence $x^2 \equiv -1\pmod p$:  
$$\eqalign{x^2 = -1\pmod p&\implies 
 x^2 + 1 = kp
\implies (x-i)(x+i) = kp\cr
&\implies ((2n)!-i)((2n!)+i))=kp
\implies 4n!- 1 =kp\cr}$$ 
This is where I start to run out of any ideas that seem to get me anywhere.
Any tips would be greatly appreciated!

REPLY [3 votes]: I think the part about $p$ not being a Gaussian prime has not been addressed. You have arrived at $p\mid(x+i)(x-i)$ (with $x=(2n)!$). Now $p$ divides neither $x+i$ nor $x-i$, since $${x\pm i\over p}={x\over p}\pm{1\over p}i$$ is not a Gaussian integer. But the Gaussian integers are a unique factorization domain, and in such a domain a prime that divides a product divides (at least) one of the terms being multiplied. Thus, $p$ can't be a prime in the Gaussians.