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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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  • $\begingroup$ You have confused $(2n)!$ with $(2n!)$. Also the square of $(2n)!$ isn't $4n!$ (or even $(4n)!$). $\endgroup$ Commented Oct 6, 2012 at 6:02

2 Answers 2

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If $p=4n+1$, then the nonzero elements of $\mathbb{Z}_p$ are $\{1,2, \dots, 4n \}$. You can write this as $\{1,2, \dots, 2n, -2n, \dots, -2,-1 \}$. The product of all of these elements is $-1$ from Wilson's Theorem. But this is the same thing as $(1 \cdot 2 \cdots 2n)^{2}(-1)^{2n} = (1 \cdot 2 \cdots 2n)^{2} = ((2n!))^{2}$.

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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.

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