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

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  • Jens Kruse Andersen
    ... 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
    Message 1 of 18 , Jun 1, 2007
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      Kevin Acres wrote:
      > A little bit of pertinent information just arrived in my inbox:
      >
      > Wilfrid Keller and Jörg Richstein searched the bases
      > b = 29, 47, 61
      > and found no example of
      > b^(p-1) = 1 mod p^2
      > for any prime for p < 10^11.

      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"

      --
      Jens Kruse Andersen
    • 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 2 of 18 , Jun 1, 2007
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        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 3 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 4 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 5 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 6 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 7 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 8 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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