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Wolstenholme's Theorem

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  • Phil Carmody
    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
    Message 1 of 3 , Sep 4, 2006
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      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 ;-) )

      Phil

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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 2 of 3 , Sep 4, 2006
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        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 3 of 3 , Sep 4, 2006
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          --- 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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