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Math Help - Wilson's theorem proofing questions?

  1. #1
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    Question Wilson's theorem proofing questions?

    I've been asked to prove these 2 things :

    1. (-1)^{\frac{n-1}{2}}(1\cdot2\cdot\cdot\cdot(\frac{n-1}{2}))^{2} = -1

    2. if n\equiv1 (mod 4) then x2=-1 has a solution.

    I haven't manage to come by enough examples of Wilson's theorem, so I'm really in the dark here.
    Any help will be much appreciated.

    Thanks.
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  2. #2
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    Re: Wilson's theorem proofing questions?

    I will prove when 'n' is an odd prime (so we can invoke Wilson's Theorem).

    Let p be an odd prime. Then by Wilson's Theorem:

    -1 \equiv (p-1)! mod p
    -1 \equiv 1 \cdot 2 \cdot ... \cdot \frac{p-1}{2} \cdot \frac{p+1}{2} \cdot ... \cdot (p-1) mod p
    Since p \equiv 0 mod p we can freely subtract p from each term in the second half of the previous equation. This gives us:
    -1 \equiv 1 \cdot 2 \cdot ... \cdot \frac{p-1}{2} \cdot (-\frac{p-1}{2} \cdot ... \cdot (-1)) mod p
    Now note that in the second half of the right hand side there are \frac{p-1}{2} '-1's' (pardon my notation abuse). Pulling out the '-1's' we have the result.

    If p \equiv 1 mod 4 then \frac{p-1}{2} \equiv 0 mod 2. Hence, by our previous result, -1 \equiv [(\frac{p-1}{2})!]^2 mod p. Therefore x^2 \equiv -1 mod p has a solution.
    Thanks from Laban
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  3. #3
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    Re: Wilson's theorem proofing questions?

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  4. #4
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    Re: Wilson's theorem proofing questions?

    Quote Originally Posted by wsldam View Post
    I will prove when 'n' is an odd prime (so we can invoke Wilson's Theorem).

    Let p be an odd prime. Then by Wilson's Theorem:

    -1 \equiv (p-1)! mod p
    -1 \equiv 1 \cdot 2 \cdot ... \cdot \frac{p-1}{2} \cdot \frac{p+1}{2} \cdot ... \cdot (p-1) mod p
    Since p \equiv 0 mod p we can freely subtract p from each term in the second half of the previous equation. This gives us:
    -1 \equiv 1 \cdot 2 \cdot ... \cdot \frac{p-1}{2} \cdot (-\frac{p-1}{2} \cdot ... \cdot (-1)) mod p
    Now note that in the second half of the right hand side there are \frac{p-1}{2} '-1's' (pardon my notation abuse). Pulling out the '-1's' we have the result.

    If p \equiv 1 mod 4 then \frac{p-1}{2} \equiv 0 mod 2. Hence, by our previous result, -1 \equiv [(\frac{p-1}{2})!]^2 mod p. Therefore x^2 \equiv -1 mod p has a solution.
    Thanks a lot! I would have never figured it out on my own. I'm still baffled as to why we're getting these questions in a linear algebra course.
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