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proof of wilson's theorem by differentiation on fermat's little theorem? watch

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    Does it work? I'm trying to do it, but keep getting to (p-1)! (three horizontal lines) 0 (mod p), which obviously isn't right.

    Could you show how? Bearing in mind I'm an A level student with very little further maths knowledge
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    You simply cannot differentiate like that with modulo arithmetic. Take for example:

    x^5 \equiv 1\mathrm{mod\ 31}

    x\equiv 2\mathrm{mod\ 31} is a solution but differentiating the equation as you seem to have done gives

    5x^4 \equiv 0\mathrm{mod\ 31} for which the solution doesn't work. Moduluar arithmetic is that little bit more tricky, you need to turn a modular equivalence into an equality for differnetiating
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    Basically, the set your function operates on, \mathbb{F}_p, doesn't have enough structure to support differentiation. The proof of Wilson's theorem is actually quite easy, and boils down to the observation that every number other than \pm 1 has a unique multiplicative inverse which is not the number itself.
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    I read somewhere that Wilson's Theorem "follows directly from Fermat's Little Theorem".
    If differentiation like that isn't a correct approach using the Little Theorem as starting point, what is?
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    I'm not saying there isn't a link, but I'm not really seeing it. The normal approach is as Zhen describes (and in fact, Wilson's theorem is in many ways "simpler" than Fermat's Little Theorem).
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    http://users.broadsurf.co.uk/johnbul...ninduction.pdf
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    (Original post by DFranklin)
    I'm not saying there isn't a link, but I'm not really seeing it. The normal approach is as Zhen describes (and in fact, Wilson's theorem is in many ways "simpler" than Fermat's Little Theorem).
    The first proof I learnt was more complicated, but does indeed start with Fermat's Little Theorem.

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    We know that x^{p-1} - 1 \equiv 0 \pmod p for all x \in \mathbb{F}_p^*, so by various properties of \mathbb{F}_p which I cannot recall, x^{p-1} - 1 \equiv (x-1)(x-2)\cdots(x-p+1) \pmod p. Now consider x = 0.
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    (Original post by Zhen Lin)
    The first proof I learnt was more complicated, but does indeed start with Fermat's Little Theorem.

    Spoiler:
    Show
    We know that x^{p-1} - 1 \equiv 0 \pmod p for all x \in \mathbb{F}_p^*, so by various properties of \mathbb{F}_p which I cannot recall, x^{p-1} - 1 \equiv (x-1)(x-2)\cdots(x-p+1) \pmod p. Now consider x = 0.
    Oh yes, thanks. Though you could argue it starts with observations about roots of polynomials in \mathbb{F}_p.
 
 
 
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