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Math Help - 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: (p-1)!\equiv{-1}(\bmod.p)

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

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

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

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

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

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