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Math Help - Variation of Wilson's Theorem

  1. #1
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    Variation of Wilson's Theorem

    I am having troubles with this problem, I tried using a proof similar to that of Wilson's theorem but I can't get it to work! Any help would be greatly appreciated.

    Show that if p is prime and 2 \leq k < p then (p-k)!(k-1)! \equiv (-1)^k \pmod{p}
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by RonyPony View Post
    I am having troubles with this problem, I tried using a proof similar to that of Wilson's theorem but I can't get it to work! Any help would be greatly appreciated.

    Show that if p is prime and 2 \leq k < p then (p-k)!(k-1)! \equiv (-1)^k \pmod{p}
     (p-k)!(k-1)!\equiv (p-k)!(k-1-p)(k-2-p)\cdots(1-p)\equiv (p-k)!(p-k+1)(p-k+2)\cdots(p-1)(-1)^{k+1}=(-1)^{k-1}(p-1)!\equiv (-1)^k \bmod{p}
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Champaign, Illinois
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    Quote Originally Posted by RonyPony View Post
    I am having troubles with this problem, I tried using a proof similar to that of Wilson's theorem but I can't get it to work! Any help would be greatly appreciated.

    Show that if p is prime and 2 \leq k < p then (p-k)!(k-1)! \equiv (-1)^k \pmod{p}
     (p-k)!(k-1)!\equiv (p-k)!(k-1-p)(k-2-p)\cdots(1-p)

     \equiv(p-k)!(p-k+1)(p-k+2)\cdots(p-1)(-1)^{k-1}=(-1)^{k-1}(p-1)!\equiv (-1)^k \bmod{p}
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