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Thread: prove by wilson theorem

  1. #1
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    prove by wilson theorem

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  2. #2
    Super Member PaulRS's Avatar
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    (i) By Wilson's Theorem: $\displaystyle (p-1)!\equiv{-1}(\bmod.p)$

    $\displaystyle (p-1)(p-2)...\left(\frac{p+1}{2}\right)\cdot{\left(\frac{p-1}{2}\right)!}\equiv{-1}(\bmod.p)$

    But note that $\displaystyle p-1\equiv{-1}(\bmod.p)$ $\displaystyle p-2\equiv{-2}(\bmod.p)$ and so on... $\displaystyle \frac{p+1}{2}\equiv{-\frac{p-1}{2}}(\bmod.p)$

    Thus: $\displaystyle (-1)^{\left(\frac{p-1}{2}\right)}\left[\left(\frac{p-1}{2}\right)!\right]^2\equiv{-1}(\bmod.p)$

    Now if $\displaystyle p\equiv{1}(\bmod.4)$ then $\displaystyle (-1)^{\left(\frac{p-1}{2}\right)}=1$

    Thus: $\displaystyle \left[\left(\frac{p-1}{2}\right)!\right]^2\equiv{-1}(\bmod.p)$

    (iii) This shows that you have at least 2 solutions if $\displaystyle p\equiv{1}(\bmod.4)$ ( a=(-a) )
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  3. #3
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    (Let $\displaystyle p$ be odd).
    Define $\displaystyle f(x) \in \mathbb{Z}_p[x]$ as $\displaystyle f(x) = x^{p-1} -1 - (x-1)(x-2)...(x-(p-1))$.
    This polynomial has degree $\displaystyle p-2$. But every $\displaystyle a\in \mathbb{Z}_p^{\text{x}}$ solves this polynomial.
    Thus, $\displaystyle \deg f(x) = p-2$ has value of zero for $\displaystyle p-1$ distinct values, and so it must be a zero polynomial.
    So let $\displaystyle x=0$ and we get: $\displaystyle -1 - (p-1)! \equiv 0 (\bmod p)\implies (p-1)! \equiv -1(\bmod p)$
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