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Re: [PrimeNumbers] Wolstenholme's Theorem

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  • Ignacio Larrosa Cañestro
    Monday, September 04, 2006 6:21 PM [GMT+1=CET], ... I think there are two or more versions of the theorem, related I suppose. In Introduction to Analytic
    Message 1 of 3 , Sep 4, 2006
      Monday, September 04, 2006 6:21 PM [GMT+1=CET],
      Phil Carmody <thefatphil@...> escribió:

      > I don't suppose anyone can sketch a proof of it, could they?
      >
      > A quick search on the interweb failed to yield anything, and I don't
      > remember seeing it in any of the books on my limited bookshelf at
      > home. I suspect it's in H&W. (DB - can you issue a challenge that's
      > worth a book to the winner where only I can solve it, please ;-) )
      >

      I think there are two or more versions of the theorem, related I suppose. In
      "Introduction to Analytic Number Theory" of T. M. Apostol, there is this:

      If p >= 5 is prime, then

      S_{p-2} = Sum((p - 1)!/k, k, 1, p-1) = 0 (mod p^2)

      Dem.:

      That sum is the coefficient of -x in the polynome

      g(x) = (x - 1)(x - 2)...(x - (p-1))

      = x^(p-1) - S_1*x^(p-2) + S_2*x^(p-3) - ...

      + S_{p-3}*x^2 - S_{p-2}*x + (p-1)!

      The roots of g(x) are 1, 2, ..., p-1. They also are roots of the congruence
      h(x) = x^(p-1) - 1.

      But then, the difference f(x) = g(x) - h(x) has degree p-2 and p-1 roots
      module p. Then, for Lagrange's th., all of coefficients of f(x) are zero
      (mod p) (Wilson's th. is a corolary).

      Therefore, S_k, k = 1, .. ,p-2 are zero (mod p) and

      g(p-1) = (p-1)! = p^(p-1) - S_1*p^(p-2) + ... S_{p-3}*p^2 - S_{p-2}*p +
      (p-1)!

      (here p > 3 is important)

      ===> S_{p-2}*p = 0 (mod p^3)

      ===> S_{p-2} = 0 (mod p^2) (q.e.d.)

      Best regards,

      Ignacio Larrosa Cañestro
      A Coruña (España)
      ilarrosa@...
    • Phil Carmody
      ... Many thanks Ignacio! The formulation of the theorem I have seen was different, but closely related (namely the C(2p-1,p-1) one), and hopefully not too much
      Message 2 of 3 , Sep 4, 2006
        --- Ignacio Larrosa Cañestro <ilarrosa@...> wrote:
        > Monday, September 04, 2006 6:21 PM [GMT+1=CET],
        > Phil Carmody <thefatphil@...> escribió:
        > > I don't suppose anyone can sketch a proof of it, could they?
        >
        > I think there are two or more versions of the theorem, related I suppose. In
        > "Introduction to Analytic Number Theory" of T. M. Apostol, there is this:
        >
        > If p >= 5 is prime, then
        >
        > S_{p-2} = Sum((p - 1)!/k, k, 1, p-1) = 0 (mod p^2)
        ...
        > ===> S_{p-2} = 0 (mod p^2) (q.e.d.)

        Many thanks Ignacio! The formulation of the theorem I have seen was different,
        but closely related (namely the C(2p-1,p-1) one), and hopefully not too much of
        a leap from the above.

        In particular, I think I noticed that on Dave Rusin's archive of useful
        sci.math.* posts, there's a discussion which seems to relate the two
        formulations.

        I thought I spotted a generalisation (of the other formulation), and wished to
        prove it before relying on its truth!

        Thanks again,
        Phil

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