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Math Help - Prove variation of Wilson

  1. #1
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    Prove variation of Wilson

    I'm trying to prove this for all n \geq 1

    (p-1)!^{p^{n-1}}\equiv -1\bmod p^n

    I think it's just a variation of Wilson's theorem but I can't seem to prove. I've tried induction but I can't change from \bmod p^n to \bmod p^{n+1}

    Any help would be great?
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  2. #2
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    Quote Originally Posted by alan4cult View Post
    I'm trying to prove this for all n \geq 1

    (p-1)!^{p^{n-1}}\equiv -1\bmod p^n

    I think it's just a variation of Wilson's theorem but I can't seem to prove. I've tried induction but I can't change from \bmod p^n to \bmod p^{n+1}

    Any help would be great?
    Hint: If a\equiv b(\bmod p^k)\implies a^p \equiv b^p (\bmod p^{k+1}).
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Hint: If a\equiv b(\bmod p^k)\implies a^p \equiv b^p (\bmod p^{k+1}).
    How can I prove this fact?
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  4. #4
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    Quote Originally Posted by alan4cult View Post
    How can I prove this fact?
    Write a = b + cp^k \implies a^p = (b+cp^k)^p \implies a^p = b^p + pb^{p-1}cp^k + N where p^{k+1}|N.
    Thus, a^p \equiv b^p (\bmod p^{k+1}).
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    Write a = b + cp^k \implies a^p = (b+cp^k)^p \implies a^p = b^p + pb^{p-1}cp^k + N where p^{k+1}|N.
    Thus, a^p \equiv b^p (\bmod p^{k+1}).
    Excellent that solves my problem. Thank you.
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