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Math Help - Wilson's Theorem

  1. #1
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    Wilson's Theorem

    So I've been working on this problem for too long now and I need some advice as to how to proceed. I've tried using induction and I've also tried plugging in concrete values to try and discern a pattern to no avail. I DO know that Wilson's Theorem applies.

    Wilson's Theorem
    If p is a prime, then (p-1)! \equiv -1 (mod p)

    Problem
    Show that if p is an odd prime and a and b are non negative integers with a+b=p-1, then a!b! + (-1)^a \equiv 0 (mod p)
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  2. #2
    Super Member PaulRS's Avatar
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    Re: Wilson's Theorem

    Note that: a! = 1\cdot ... \cdot a \equiv_p \left( (-1) \cdot (p-1) \right)\cdot .. \cdot \left( (-1) \cdot (p-a) \right) and that p-a = b + 1

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  3. #3
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    Re: Wilson's Theorem

    Quote Originally Posted by PaulRS View Post
    Note that: a! = 1\cdot ... \cdot a \equiv_p \left( (-1) \cdot (p-1) \right)\cdot .. \cdot \left( (-1) \cdot (p-a) \right) and that p-a = b + 1

    I tried factoring out the (-1) getting a! congruent to (-1)^a[(p-1)...(p-a)]. Then I tried to multiply by b! considering that p-a=b+1, but I'm getting nowhere. Man I feel stupid!
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: Wilson's Theorem

    Quote Originally Posted by PaulRS View Post
    Note that: a! = 1\cdot ... \cdot a \equiv_p \left( (-1) \cdot (p-1) \right)\cdot .. \cdot \left( (-1) \cdot (p-a) \right) and that p-a = b + 1
    I'm going to try to complete this.

    a!\equiv (-1)^a(p-1)\cdots(p-a)\pmod{p}

    a!b!\equiv (-1)^a(p-1)\cdots(p-a)b!\pmod{p}

    \equiv (-1)^a(p-1)!\pmod{p} (Note that p-a = b + 1.)

    \equiv -(-1)^a\pmod{p} (Using Wilson's theorem)

    a!b!+(-1)^a\equiv 0\pmod{p}
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  5. #5
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    Re: Wilson's Theorem

    Quote Originally Posted by alexmahone View Post

    (-1)^a(p-1)\cdots(p-a)b!\pmod{p}

    \equiv (-1)^a(p-1)!\pmod{p} (Note that p-a = b + 1.)
    I don't understand how you got this congruence.
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  6. #6
    MHF Contributor alexmahone's Avatar
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    Re: Wilson's Theorem

    Quote Originally Posted by UNLVRich View Post
    I don't understand how you got this congruence.
    (p-1)\cdots(p-a)b!=(p-1)\cdots(p-a)(p-a-1)!=(p-1)!
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  7. #7
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    Re: Wilson's Theorem

    Thank you!
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