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Thread: h+k=p-1 ==> h!k!+[(-1)^h]=0 (mod p)

  1. #1
    Newbie demdirbu's Avatar
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    h+k=p-1 => h!k!+[(-1)^h]=0 (mod p)

    $\displaystyle p$, prime odd number


    $\displaystyle h \geq 0 ,k \geq 0 $

    Show that:

    $\displaystyle h+k=p-1 \Rightarrow h!k!+(-1)^{h}=0 (mod p)$
    Last edited by demdirbu; Nov 26th 2008 at 10:36 PM.
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  2. #2
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    Quote Originally Posted by demdirbu View Post
    $\displaystyle p$, prime odd number


    $\displaystyle h \geq 0 ,k \geq 0 $

    Show that:

    $\displaystyle h+k=p-1 \Rightarrow h!k!+(-1)^{h}=0 \pmod p$
    This is a lightly disguised form of Fermat's theorem. Notice that $\displaystyle (-1)^hh! = (-1)(-2)\cdots(-h)$. Modulo p, this is equivalent to $\displaystyle (p-1)(p-2)\cdots(p-h) = (p-1)(p-2)\cdots(k+1)$. When you multiply this by k! you get (p-1)!.
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