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Number Theory: Can someone explain this to me please?

  1. Jun 4, 2009 #1
    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. jcsd
  3. Jun 4, 2009 #2
    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.
     
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