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

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  • Kevin Acres
    I did indeed look through all of the interesting links that you sent. I m just wondering if I can get the time to work through the information and see if it
    Message 1 of 18 , Jun 1, 2007
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      I did indeed look through all of the interesting links that you sent. I'm
      just wondering if I can get the time to work through the information and see
      if it holds true that p^2 always divides the minimal base^(p-1/x)-1 that p
      does.

      Kevin.

      -----Original Message-----
      From: Jens Kruse Andersen [mailto:jens.k.a@...]
      Sent: 02 June 2007 01:12

      [...]

      Did you look at the first link I gave you:
      http://www.fermatquotient.com/FermatQuotienten/FermQ_Sort
      It includes those bases searched to 5.074*10^12 (with no odd solutions),
      and much more.
      It was updated 3 days ago and appears to be the current records.
      http://www.fermatquotient.com/ says:
      "http://www.fermatquotient.com/FermatQuotienten/News
      Neues und Rekorde / News and records / Update 29.05.2007"
    • elevensmooth
      ... I suspect it s provable, but I can t locate my copy of Hardy and Wright tonight to skim for inspiration. The minimal value is the column called order at
      Message 2 of 18 , Jun 2, 2007
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        --- In primenumbers@yahoogroups.com, "Kevin Acres" <research@...> wrote:

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

        I suspect it's provable, but I can't locate my copy of Hardy and
        Wright tonight to skim for inspiration.

        The minimal value is the column called "order" at
        http://oddperfect.org/FermatQuotients.html , so it was easy to verify
        that p^2 divided the minimal value for all of those.

        William
        Poohbah of oddperfect.org
      • Phil Carmody
        ... Subscirbe! Phil () ASCII ribbon campaign () Hopeless ribbon campaign / against HTML mail / against gratuitous bloodshed [stolen with
        Message 3 of 18 , Jun 3, 2007
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          --- 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 4 of 18 , Jun 3, 2007
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            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 5 of 18 , Jun 5, 2007
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              --- 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 6 of 18 , Jun 14, 2007
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                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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