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Twin polynomials contest

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  • henridelille
    Excuse me Bernard for the spelling of your name. A first step about the origin of (p,q). I start with the congruent numbers : Definition - A number d is said
    Message 1 of 19 , Jan 24, 2006
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      Excuse me Bernard for the spelling of your name.

      A first step about the origin of (p,q).

      I start with the "congruent numbers":

      Definition - A number d is said congruent if it is equal to the area
      of a rationnal rectangle triangle : d =ab/2 , c^2=a^2+b^2 with a,b,c
      belonging to the set of rationnals.
      The first examples are d= 5,6,7,13,14,15,..157

      Therorem - A number d is congruent if the elliptic curve
      y^2=x^3-(d^2)x has a point with rationnal coordinates (x,y).

      To be continued...
    • 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 2 of 19 , Jan 24, 2006
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        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 3 of 19 , Jan 24, 2006
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          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 4 of 19 , Jan 24, 2006
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            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 5 of 19 , Jan 24, 2006
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              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 6 of 19 , Jan 24, 2006
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                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 7 of 19 , Jan 25, 2006
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                  --- 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 8 of 19 , Jan 25, 2006
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                    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 9 of 19 , Jan 25, 2006
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                      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 10 of 19 , Jan 26, 2006
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                        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 11 of 19 , Jan 26, 2006
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                          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 12 of 19 , Jan 26, 2006
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                            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 13 of 19 , Jan 26, 2006
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                              Indded it is Fermat

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