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Twin polynomials contest

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  • henridelille
    I ask about triple primes (p(d),q(d),r(d)) with r(d)=32d^4+8d^2+1 Does somebody find a solution ???
    Message 1 of 19 , Jan 24, 2006
      I ask about triple primes (p(d),q(d),r(d)) with

      r(d)=32d^4+8d^2+1

      Does somebody find a solution ???
    • henridelille
      I have done a mistake : of couse Jacques is the leader !
      Message 2 of 19 , Jan 24, 2006
        I have done a mistake : of couse Jacques is the leader !
      • kradenken
        If we are having a contest we need to at least seed it with a titanic prime d=253!+85052 yields a 1001 digit prime 4*((253!+85052)^2)+1 (along with its 1501
        Message 3 of 19 , Jan 24, 2006
          If we are having a contest we need to at least seed it with a titanic
          prime
          d=253!+85052 yields a 1001 digit prime
          4*((253!+85052)^2)+1
          (along with its 1501 digit companion)
          4*(253!+85052)*(2*(253!+85052)^2+1)+1
          Cheers
          Ken
          --- In primenumbers@yahoogroups.com, "henridelille"
          <henridelille@y...> wrote:
          >
          > Dear all,
          >
          > I propose the following contest:
          >
          > Consider these two polynomials:
          >
          > p(d)=4d^2+1
          > q(d)=4d(2d^2+1)+1
          >
          > The polynomials are often prime for the same parameter d.
          > I conjecture that there are an infinty of prime pairs (p,q).
          >
          > My own record is: d=502 -> p=1008017 and q=1012050073
          >
          > Contest : find very high parameters d with (p,q) prime pairs
          >
        • henridelille
          Wonderfull Ken ! I follow up my explanation : Theorem (Fermat) - Every prime p=4k+1 write p=a^2+b^2 where a,b are natural numbers. Theorem - A prime number is
          Message 4 of 19 , Jan 24, 2006
            Wonderfull Ken !

            I follow up my explanation :

            Theorem (Fermat) - Every prime p=4k+1 write p=a^2+b^2 where a,b are
            natural numbers.

            Theorem - A prime number is well approximated by the square of
            rationnal numbers p=(u/v)^2+e/v^2 and e small, v as large as we want.
            (it is related to continued fractions)

            Then a prime 4k+1 is related to a quasi "congruent number" d=ab/2 (see
            previous message) in the sense:

            p=a^2+b^2=(u/v)^2+e/v^2 and e is small compared to v.

            to be continued...
          • Jens Kruse Andersen
            ... Let x=2d. Then the goal is primes x^2+1 and x^3+2x+1. x^2+1 would be a generalized Fermat prime and many of them are known. I got no hit on x^3+2x+1 from
            Message 5 of 19 , Jan 24, 2006
              henridelille wrote:

              > p(d)=4d^2+1
              > q(d)=4d(2d^2+1)+1
              >
              > Contest : find very high parameters d with (p,q) prime pairs

              Let x=2d. Then the goal is primes x^2+1 and x^3+2x+1.
              x^2+1 would be a generalized Fermat prime and many of them are known.
              I got no hit on x^3+2x+1 from the tested GF primes in
              The Prime Pages database.

              http://perso.wanadoo.fr/yves.gallot/primes/results.html has more
              small GF primes.
              b^1024+1 and b^2048+1 gave nothing.
              b^512+1 gave several hits corresponding to these x:
              65710^256
              177536^256
              248578^256
              1072214^256
              4561880^256

              PrimeForm/GW has proven all primes.

              --
              Jens Kruse Andersen
            • thefatphil
              ... 4*(2^1024+73006634)^2+1 8*(2^1024+73006634)^3+4*(2^1024+73006634)+1 32*(2^1024+73006634)^4+8*(2^1024+73006634)^2+1 4*(2^1024+403799789)^2+1
              Message 6 of 19 , Jan 25, 2006
                --- In primenumbers@yahoogroups.com, "henridelille"
                <henridelille@y...> wrote:
                >
                > I ask about triple primes (p(d),q(d),r(d)) with
                >
                > r(d)=32d^4+8d^2+1
                >
                > Does somebody find a solution ???

                4*(2^1024+73006634)^2+1
                8*(2^1024+73006634)^3+4*(2^1024+73006634)+1
                32*(2^1024+73006634)^4+8*(2^1024+73006634)^2+1

                4*(2^1024+403799789)^2+1
                8*(2^1024+403799789)^3+4*(2^1024+403799789)+1
                32*(2^1024+403799789)^4+8*(2^1024+403799789)^2+1

                4*(2^1024+693904189)^2+1
                8*(2^1024+693904189)^3+4*(2^1024+693904189)+1
                32*(2^1024+693904189)^4+8*(2^1024+693904189)^2+1

                The largest of each triplet is titanic. 10 GHz hours, but 3 was twice
                the yield I expected in that time. Assuming O(D^5) growth, a
                1000-digit smallest prime should be doable in <1GHz week. I'll not do
                that myself, but can donate sieve-technology to anyone who wants to.
                (My guess is that PFGW -f would be less than half speed.)

                Can we have a complete explanation now both the pairs and the triplets
                have reached titanic sizes, please?

                Phil
              • henridelille
                I am impressed by your all results. I think yhat everyone has won. I continue my explanation. (see my 2 previous mails) As a prime 4k+1 can be considered
                Message 7 of 19 , Jan 25, 2006
                  I am impressed by your all results. I think yhat everyone has won.

                  I continue my explanation. (see my 2 previous mails)

                  As a prime 4k+1 can be considered connected to a "quasi-congruent"
                  number d=ab/2 it is natural to ask whether there is a elliptic curve
                  attached y^2=x^3-d^2x.

                  In fact as this not really a congruent number the elliptic curve
                  shall be modified. For some heuristic reasons there are two families
                  which hit : E1 y^2=8u(x^3-d^2x)+v and E2 y^2=8u(x^3+d^2x)-v, the
                  parameters u and v depending of the prime p.

                  Now, in the particular case of primes p(d)=4d^2+1, the parameters
                  v=1 is good for both E1 and E2 but u for E2 is u=1 and for E1 is u=
                  (2e)^2 where e is a natural number depending of the continued
                  fraction of sqrt(p). It is too long to explain the details...

                  Now put x=p/2 in E1 and E2 and you get:

                  E1 z1=32d^4+8d^2+1
                  E2 z2=4d(2d^2+1)

                  Here is our polynomials:

                  p(d)=4d^2+1
                  q(d)=z2+1=4d(2d^2+1)+1
                  r(d)=z1=32d^4+8d^2+1

                  I have immediately remarked that for small parameters d the numbers
                  p,q,r are often prime together.

                  I don't know exactly why but now you know that it is connected to:
                  - congruent numbers
                  - continued fractions
                  - elliptic curves with rational solutions

                  It requests further deep investigations to undersand this phenomena.

                  You can find an more detailed overview (in french) on my website
                  http://perso.wanadoo.fr/hh-mouvement.com/

                  Note number 4

                  Enjoy primes
                • kradenken
                  Hi, My final entry in contest d=450!+42455 4*((450!+42455)^2)+1 (digits:2002) 4*(450!+42455)*(2*(450!+42455)^2+1)+1 (digits:3002) Both certified prime using
                  Message 8 of 19 , Jan 25, 2006
                    Hi,
                    My final entry in contest

                    d=450!+42455
                    4*((450!+42455)^2)+1 (digits:2002)
                    4*(450!+42455)*(2*(450!+42455)^2+1)+1 (digits:3002)

                    Both certified prime using primo (Copyright © 2003-2004, Marcel
                    Martin).
                    The first taking 1h 18mn 22s the second 8h 5mn 2s

                    Primes found in a couple of hours using a pfgw script file
                    (source available on request)
                    Cheers
                    Ken
                    --- In primenumbers@yahoogroups.com, "henridelille"
                    <henridelille@y...> wrote:
                    >
                    > I am impressed by your all results. I think yhat everyone has won.
                    >
                    > I continue my explanation. (see my 2 previous mails)
                    >
                    > As a prime 4k+1 can be considered connected to a "quasi-congruent"
                    > number d=ab/2 it is natural to ask whether there is a elliptic
                    curve
                    > attached y^2=x^3-d^2x.
                    >
                    > In fact as this not really a congruent number the elliptic curve
                    > shall be modified. For some heuristic reasons there are two
                    families
                    > which hit : E1 y^2=8u(x^3-d^2x)+v and E2 y^2=8u(x^3+d^2x)-v, the
                    > parameters u and v depending of the prime p.
                    >
                    > Now, in the particular case of primes p(d)=4d^2+1, the parameters
                    > v=1 is good for both E1 and E2 but u for E2 is u=1 and for E1 is u=
                    > (2e)^2 where e is a natural number depending of the continued
                    > fraction of sqrt(p). It is too long to explain the details...
                    >
                    > Now put x=p/2 in E1 and E2 and you get:
                    >
                    > E1 z1=32d^4+8d^2+1
                    > E2 z2=4d(2d^2+1)
                    >
                    > Here is our polynomials:
                    >
                    > p(d)=4d^2+1
                    > q(d)=z2+1=4d(2d^2+1)+1
                    > r(d)=z1=32d^4+8d^2+1
                    >
                    > I have immediately remarked that for small parameters d the
                    numbers
                    > p,q,r are often prime together.
                    >
                    > I don't know exactly why but now you know that it is connected to:
                    > - congruent numbers
                    > - continued fractions
                    > - elliptic curves with rational solutions
                    >
                    > It requests further deep investigations to undersand this
                    phenomena.
                    >
                    > You can find an more detailed overview (in french) on my website
                    > http://perso.wanadoo.fr/hh-mouvement.com/
                    >
                    > Note number 4
                    >
                    > Enjoy primes
                    >
                  • henridelille
                    Jens is the winner for the twin contest and Phil wins the triple contest with impressive solutions. I hope that you are not deceived by my explanations. An
                    Message 9 of 19 , Jan 26, 2006
                      Jens is the winner for the twin contest and Phil wins the triple
                      contest with impressive solutions.

                      I hope that you are not deceived by my explanations.

                      An interesting point is to look at randomly choosen twin or triple
                      polynomials and compare their performance in giving primes to our
                      p,q,r.
                    • henridelille
                      An add-on about my explanations: If someone is able to prove that for all primes there is a set of parameters u,v in order that p/2 belongs to the elliptic
                      Message 10 of 19 , Jan 26, 2006
                        An add-on about my explanations:

                        If someone is able to prove that for all primes there is a set of
                        parameters u,v in order that p/2 belongs to the elliptic curves E1 or
                        E2, then he would prove the longstanding problem : there is an
                        infinity of primes x^2+1
                      • Werner D. Sand
                        Hello henridelille, I think it was Euler, not Fermat, who said p = 4k+1 = a²+b². And this is a quality not only of prime numbers, f.e. 25 or 85. Werner
                        Message 11 of 19 , Jan 26, 2006
                          Hello henridelille,

                          I think it was Euler, not Fermat, who said p = 4k+1 = a²+b². And this
                          is a quality not only of prime numbers, f.e. 25 or 85.

                          Werner
                        • henridelille
                          Indded it is Fermat Euler s theorem states p=6n+1 - p=x^2+3y^2
                          Message 12 of 19 , Jan 26, 2006
                            Indded it is Fermat

                            Euler's theorem states p=6n+1 -> p=x^2+3y^2
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