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Math Help - calculate this by wilson's Theorem

  1. #1
    Junior Member
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    calculate this by wilson's Theorem

    express 1^2 * 3^2 * 5^2 * .... * (p-2)^2 (mod p) as a power of -1 .

    suppose p is an odd prime.
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  2. #2
    Super Member PaulRS's Avatar
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    Note that:
    1\equiv{-(p-1)}(\bmod.p); 3\equiv{-(p-3)}(\bmod.p);... p-4\equiv{-4}(\bmod.p); p-2\equiv{-2}(\bmod.p)

    Thus: (p-1)!\equiv{(-1)^{\left(\frac{p-1}{2}\right)}\cdot{1^2\cdot{3^2}\cdot{...\cdot{(p-2)^2}}}}(\bmod.p)

    But (p-1)!\equiv{-1}(\bmod.p) by Wilson's Theorem

    Thus: 1^2\cdot{3^2}\cdot{...\cdot{(p-2)^2}}\equiv{(-1)^{1-\left(\frac{p-1}{2}\right)}}(\bmod.p)
    Last edited by PaulRS; June 1st 2008 at 06:17 PM.
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