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Re: [PrimeNumbers] Re: Thanks to all. Proposed binary Prime number test fails.

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  • Phil Carmody
    ... Subscirbe! Phil () ASCII ribbon campaign () Hopeless ribbon campaign / against HTML mail / against gratuitous bloodshed [stolen with
    Message 1 of 18 , Jun 3, 2007
      --- Mark Rodenkirch <mgrogue@...> wrote:
      > This was meant to be sent to the group and I sent it to Jens by
      > accident.
      >
      > If anyone else here has a PowerPC G5, I have a program that could
      > search a range of about 1e12 in a day (for a single base).

      Subscirbe!

      Phil

      () ASCII ribbon campaign () Hopeless ribbon campaign
      /\ against HTML mail /\ against gratuitous bloodshed

      [stolen with permission from Daniel B. Cristofani]



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    • Mark Rodenkirch
      ... YGM. --Mark [Non-text portions of this message have been removed]
      Message 2 of 18 , Jun 3, 2007
        On Jun 3, 2007, at 9:21 AM, Phil Carmody wrote:

        > > This was meant to be sent to the group and I sent it to Jens by
        > > accident.
        > >
        > > If anyone else here has a PowerPC G5, I have a program that could
        > > search a range of about 1e12 in a day (for a single base).
        >
        > Subscirbe!

        YGM.

        --Mark

        [Non-text portions of this message have been removed]
      • elevensmooth
        ... Yes. Let d be the smallest value so that p divides b^d-1. First observe that if b^x-1 and b^y-1 and both divisible by p^2 (or any other number), then
        Message 3 of 18 , Jun 5, 2007
          --- In primenumbers@yahoogroups.com, "Kevin Acres" <research@...> wrote:

          > just wondering ...
          > if it holds true that p^2 always divides the
          > minimal base^(p-1/x)-1 that p does.

          Yes. Let d be the smallest value so that p divides b^d-1.

          First observe that if b^x-1 and b^y-1 and both divisible by p^2 (or
          any other number), then b^(gcd(x,y))-1 is also divisible by p^2 (or
          any other number).

          Second, observe that b^(pd)-1 is divisible by p^2. To see this,
          express b^d as (ps+1), then expand (ps+1)^p with the binomial theorem.

          Third, observe that every case of divisibility by p is of the form
          b^(kd)-1, and every case of divisibility by p^2 is also a case of
          divisiblity by p, so the minimal case of divisibility by p^2 must be
          of the form b^(kd)-1.

          Finally, if there is any k<p such that b^(kd)-1 is divisible by p^2,
          then b^gcd(kd,pd) must also be divisible by p^2, but gcd(kd,pd)=d.

          Therefore, in all cases of Vanishing Fermat Quotients, p^2 divides the
          same primitive term that p divides.

          William
          Poohbah of OddPerfect.org

          P.S. We still need a few large factors of Vanishing Fermat Quotients
          at http://oddperfect.org/FermatQuotients.html
        • mistermac39
          Earlier, with reference to 29, and 47 the number 61 came up. It is a factor of the 15th Fibonacci number 610 Ring any bells?
          Message 4 of 18 , Jun 14, 2007
            Earlier, with reference to 29, and 47 the number 61 came up.

            It is a factor of the 15th Fibonacci number 610

            Ring any bells?
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