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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!

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    You have confused $(2n)!$ with $(2n!)$. Also the square of $(2n)!$ isn't $4n!$ (or even $(4n)!$).2012-10-06

2 Answers 2