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

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  • Mark Rodenkirch
    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
    Message 1 of 18 , Jun 1, 2007
      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).

      --Mark
    • 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 2 of 18 , Jun 1, 2007
        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 3 of 18 , Jun 2, 2007
          --- 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 4 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 5 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 6 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 7 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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