Homework Help: Number Theory: Can someone explain this to me please?

1. Jun 4, 2009

Firepanda

Wilson's Theorem:

(p-1)! ≡ -1 mod p

Statement:

As an immediate deduction from wilson's theorem we see that if p is prime with p ≡ 1 mod 4 then the congruence x2 ≡ -1 mod p has solutions
x = +-(r!), where r = (p-1)/2.

How do I plug in p ≡ 1 mod 4 into wilsoms theorem so I can see this? I'm missing something here an I'd be grateful if someone could explain...

Thanks.

Last edited: Jun 4, 2009
2. Jun 4, 2009

Count Iblis

If a is a primitive element, then

(p-1)! = a^[(p-1)(p)/2] = -1

We know that p - 1 is divisible by 4, so we can put

p - 1 = 4 n in the exponent:

a^(2 n p) = -1

By Fermat's theorem a^p = a, so we have:

a^(2 n) = -1

Then, it follows that a^n will be a solution of x^2 = -1.