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Thread: Congruence Problems

  1. #1
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    Congruence Problems

    a) Let p be an odd prime and let
    a = $\displaystyle \prod$ (2j-1) = (1)(3)(5)....(p-2)
    Prove that $\displaystyle a^2$ = $\displaystyle (-1)^{(p+1)/2}$ mod p

    (Product is from j=1 to (p-1)/2)

    b) Let p be a prime. Prove that
    $\displaystyle \left(\begin{array}{cc}2p\\p\end{array}\right)$ = 2 mod p

    $\displaystyle \left(\begin{array}{cc}2p\\p\end{array}\right)$ is 2p choose p, as in statistics. so $\displaystyle (2p)!/p!(2p-p)!$
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Hint for a:

    Wilson theorem.
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  3. #3
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Hint for a:

    Wilson theorem.
    Wilson's crossed my mind but I don't see how to work (p-1)! into the proof
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Another hint(or solution)...

    $\displaystyle t\equiv -(p-t)(\mod p)$, so

    $\displaystyle 2\cdot4\cdot6 ... (p-1)\equiv (-1)^{\frac{p-1}{2}}1\cdot3\cdot5...(p-2)(\mod p)$

    ...
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