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Prime Generating Polynomial contest

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  • Ed Pegg Jr
    $250 and numerous smaller prizes will go to the best prime generating polynomials, in the latest Al Zimmermann programming contest. Details at
    Message 1 of 23 , Mar 14, 2006
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      $250 and numerous smaller prizes will go to the best prime generating
      polynomials, in the latest Al Zimmermann programming contest.

      Details at
      http://recmath.org/contest/description.php

      Ed Pegg Jr
    • Phil Carmody
      ... Can you tell us how you discovered that polynomial? The discriminant is very smooth, being 5^3*11^2, and I suspect that that s an essential ingredient to a
      Message 2 of 23 , Mar 14, 2006
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        --- aldrich617 <aldrich617@...> wrote:
        > I am searching for a way to prove the following Theorem:
        >
        > For any integer 'B', the value 'A' of the equation
        > A = 5B^4 -10B^3 + 20B^2 -15B +11 will have as factors only
        > integers that end in a one, excluding all others.
        >
        > This seems to be true at least up to 10^18. I think that proving
        > it could give us new insights into primality testing,
        > and factoring. Moreover there are similar equations, vast in
        > number, apparently with the same property, that could then probably
        > be verified to be similar threads of pure one. Together these
        > would form an infinite interconnecting web.

        Can you tell us how you discovered that polynomial?
        The discriminant is very smooth, being 5^3*11^2, and I suspect that that's an
        essential ingredient to a proof.

        It seems that if p== +/-1 mod 10, then your polynomial splits at least into 2
        quadratics, and if p== +1 mod 10, then at least one of those quadratics splits.

        This property should be easily explainable, but alas it's late and my brain's
        on holiday. (I wrote this last night, but forgot to send).

        > Also, I am looking for a good factorizing program.

        Dario Alpern's Java ECM/QS Applet is hard to beat, and very convenient.
        For tougher nuts, GMP-ECM and Jason Papadopoulos' msieve are pretty much
        impossible to beat.

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      • Jack Brennen
        ... It can also be rewritten as: A = (((B*2-1)^2+5)^2*5-4)/16 Solving for A == 0 (modulo p), we can easily eliminate p == 2 and p == 5 directly from the
        Message 3 of 23 , Mar 15, 2006
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          Phil Carmody wrote:
          >>
          >>For any integer 'B', the value 'A' of the equation
          >>A = 5B^4 -10B^3 + 20B^2 -15B +11 will have as factors only
          >>integers that end in a one, excluding all others.
          >
          >
          > Can you tell us how you discovered that polynomial?
          > The discriminant is very smooth, being 5^3*11^2, and I suspect that that's an
          > essential ingredient to a proof.
          >

          It can also be rewritten as:

          A = (((B*2-1)^2+5)^2*5-4)/16


          Solving for A == 0 (modulo p), we can easily eliminate p == 2
          and p == 5 directly from the original expression of A. Then
          you get:

          (((B*2-1)^2+5)^2*5-4)/16 == 0 (modulo p)

          Multiply through by 16 (permissible because we eliminated p == 2),
          and rearrange slightly:

          ((B*2-1)^2+5)^2*5 == 4 (modulo p)

          In order for this to have solutions, 5 must be a quadratic
          residue modulo p, which eliminates primes p ending in 3 or 7.


          All that remains is to show that this last equation has no
          solutions when p is a prime ending in 9. I couldn't see any
          immediate way to do that, although I'm convinced that it's
          true.
        • Mark Underwood
          To get those creative juices flowing, here s a reminder of some *really* simple (two term) prime generating polynomials :) 6x^2 + 17 is prime from x=0 to x=16
          Message 4 of 23 , Mar 15, 2006
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            To get those creative juices flowing, here's a reminder of some
            *really* simple (two term) prime generating polynomials :)

            6x^2 + 17 is prime from x=0 to x=16

            10x^2 + 19 is prime from x=0 to x=18.

            2x^2 + 29 is prime from x=0 to x=28.

            (notice that the coefficients in each sum to a prime :) )

            Mark



            --- In primenumbers@yahoogroups.com, Ed Pegg Jr <ed@...> wrote:
            >
            > $250 and numerous smaller prizes will go to the best prime
            generating
            > polynomials, in the latest Al Zimmermann programming contest.
            >
            > Details at
            > http://recmath.org/contest/description.php
            >
            > Ed Pegg Jr
            >
          • Werner D. Sand
            You know Euler s x^2 + x + 41 prime from x=0 to 40 ? or: x^2 - 79x + 1601 prime from x=0 to 79 ? Werner
            Message 5 of 23 , Mar 16, 2006
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              You know Euler's

              x^2 + x + 41 prime from x=0 to 40 ?

              or:

              x^2 - 79x + 1601 prime from x=0 to 79 ?

              Werner



              --- In primenumbers@yahoogroups.com, "Mark Underwood"
              <mark.underwood@...> wrote:
              >
              >
              > To get those creative juices flowing, here's a reminder of some
              > *really* simple (two term) prime generating polynomials :)
              >
              > 6x^2 + 17 is prime from x=0 to x=16
              >
              > 10x^2 + 19 is prime from x=0 to x=18.
              >
              > 2x^2 + 29 is prime from x=0 to x=28.
              >
              > (notice that the coefficients in each sum to a prime :) )
              >
              > Mark
              >
              >
              >
              > --- In primenumbers@yahoogroups.com, Ed Pegg Jr <ed@> wrote:
              > >
              > > $250 and numerous smaller prizes will go to the best prime
              > generating
              > > polynomials, in the latest Al Zimmermann programming contest.
              > >
              > > Details at
              > > http://recmath.org/contest/description.php
              > >
              > > Ed Pegg Jr
              > >
              >
            • Lelio sknet
              * You know Euler s ... Or all these previously found http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html [Non-text portions of this message have
              Message 6 of 23 , Mar 16, 2006
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                * You know Euler's
                >
                > x^2 + x + 41 prime from x=0 to 40 ?
                >
                > or:
                >
                > x^2 - 79x + 1601 prime from x=0 to 79 ?
                >
                > Werner



                Or all these previously found

                http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html











                [Non-text portions of this message have been removed]
              • Mark Underwood
                Oh yes. I was thinking about polynomials with just two terms. Speaking of which, 2x^2 - 181 is prime from x=0 to x=28. (Not bad for being plagued with factors
                Message 7 of 23 , Mar 16, 2006
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                  Oh yes. I was thinking about polynomials with just two terms.

                  Speaking of which, 2x^2 - 181 is prime from x=0 to x=28.

                  (Not bad for being plagued with factors of 19!)

                  Also, from x=0 to 1000 it is prime 525 times. (The amazing x^2+x+41
                  is prime 582 times.)


                  Mark



                  --- In primenumbers@yahoogroups.com, "Lelio sknet" <lelio@...> wrote:
                  >
                  >
                  >
                  > * You know Euler's
                  > >
                  > > x^2 + x + 41 prime from x=0 to 40 ?
                  > >
                  > > or:
                  > >
                  > > x^2 - 79x + 1601 prime from x=0 to 79 ?
                  > >
                  > > Werner
                  >
                  >
                  >
                  > Or all these previously found
                  >
                  > http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html
                  >
                  >
                  >
                  >
                  >
                  >
                  >
                  >
                  >
                  >
                  >
                  > [Non-text portions of this message have been removed]
                  >
                • Mark Underwood
                  Oops, I meant 2x^2 - 181 is prime from x=0 to x=27.
                  Message 8 of 23 , Mar 16, 2006
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                    Oops, I meant 2x^2 - 181 is prime from x=0 to x=27.

                    --- In primenumbers@yahoogroups.com, "Mark Underwood"
                    <mark.underwood@...> wrote:
                    >
                    >
                    > Oh yes. I was thinking about polynomials with just two terms.
                    >
                    > Speaking of which, 2x^2 - 181 is prime from x=0 to x=28.
                    >
                    > (Not bad for being plagued with factors of 19!)
                    >
                    > Also, from x=0 to 1000 it is prime 525 times. (The amazing x^2+x+41
                    > is prime 582 times.)
                    >
                    >
                    > Mark
                    >
                    >
                  • Werner D. Sand
                    ... Fine site, thanks.
                    Message 9 of 23 , Mar 16, 2006
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                      --- In primenumbers@yahoogroups.com, "Lelio sknet" <lelio@...> wrote:
                      >
                      >
                      >
                      > * You know Euler's
                      > >
                      > > x^2 + x + 41 prime from x=0 to 40 ?
                      > >
                      > > or:
                      > >
                      > > x^2 - 79x + 1601 prime from x=0 to 79 ?
                      > >
                      > > Werner
                      >
                      >
                      >
                      > Or all these previously found
                      >
                      > http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      > [Non-text portions of this message have been removed]
                      >


                      Fine site, thanks.
                    • Lelio sknet
                      ... * You can submit your binomial 2 x^2 – 181 to Eric Weisstein in the site bellow. He knows about only 12 better than yours and that includes Euler and
                      Message 10 of 23 , Mar 16, 2006
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                        > Oh yes. I was thinking about polynomials with just two terms.
                        >
                        > Speaking of which, 2x^2 - 181 is prime from x=0 to x=28.
                        >
                        > Mark
                        >

                        *


                        You can submit your binomial 2 x^2 – 181 to Eric Weisstein in the site
                        bellow.

                        He knows about only 12 better than yours and that includes Euler and
                        Legendre.



                        http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html






                        Lélio





                        [Non-text portions of this message have been removed]
                      • Mark Underwood
                        Just checked the prime sequence for 2x^2 - 181 at Encylopedia of Integer Sequences and it isn t there. And a google didn t find anything. I *assumed* it has
                        Message 11 of 23 , Mar 17, 2006
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                          Just checked the prime sequence for 2x^2 - 181 at Encylopedia of
                          Integer Sequences and it isn't there. And a google didn't find
                          anything. I *assumed* it has been known for a long time...but who
                          knows? So maybe I will submit it to Mr. Weisstein. I wonder if there
                          is a site which has a more or less comprehensive listing of prime
                          generating polynomials.


                          Mark


                          --- In primenumbers@yahoogroups.com, "Lelio sknet" <lelio@...> wrote:
                          >
                          > > Oh yes. I was thinking about polynomials with just two terms.
                          > >
                          > > Speaking of which, 2x^2 - 181 is prime from x=0 to x=28.
                          > >
                          > > Mark
                          > >
                          >
                          > *
                          >
                          >
                          > You can submit your binomial 2 x^2 – 181 to Eric Weisstein in the
                          site
                          > bellow.
                          >
                          > He knows about only 12 better than yours and that includes Euler and
                          > Legendre.
                          >
                          >
                          >
                          > http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html
                          >
                          >
                          >
                          >
                          >
                          >
                          > Lélio
                          >
                          >
                          >
                          >
                          >
                          > [Non-text portions of this message have been removed]
                          >
                        • Ignacio Larrosa Cañestro
                          ... that that s an ... Why? ((2*B - 1)^2 + 5)^2*5 = 4 (mod p) === ((2*B - 1)^2 + 5)^2 = 4*5^(-1) (mod p) Then 4*5^(-1) must be a quadratic residue modulo p,
                          Message 12 of 23 , Mar 17, 2006
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                            --- In primenumbers@yahoogroups.com, Jack Brennen <jb@...> wrote:
                            >
                            > Phil Carmody wrote:
                            > >>
                            > >>For any integer 'B', the value 'A' of the equation
                            > >>A = 5B^4 -10B^3 + 20B^2 -15B +11 will have as factors only
                            > >>integers that end in a one, excluding all others.
                            > >
                            > >
                            > > Can you tell us how you discovered that polynomial?
                            > > The discriminant is very smooth, being 5^3*11^2, and I suspect
                            that that's an
                            > > essential ingredient to a proof.
                            > >
                            >
                            > It can also be rewritten as:
                            >
                            > A = (((B*2-1)^2+5)^2*5-4)/16
                            >
                            >
                            > Solving for A == 0 (modulo p), we can easily eliminate p == 2
                            > and p == 5 directly from the original expression of A. Then
                            > you get:
                            >
                            > (((B*2-1)^2+5)^2*5-4)/16 == 0 (modulo p)
                            >
                            > Multiply through by 16 (permissible because we eliminated p == 2),
                            > and rearrange slightly:
                            >
                            > ((B*2-1)^2+5)^2*5 == 4 (modulo p)
                            >
                            > In order for this to have solutions, 5 must be a quadratic
                            > residue modulo p, which eliminates primes p ending in 3 or 7.

                            Why?

                            ((2*B - 1)^2 + 5)^2*5 = 4 (mod p) ===>

                            ((2*B - 1)^2 + 5)^2 = 4*5^(-1) (mod p)

                            Then 4*5^(-1) must be a quadratic residue modulo p, no?

                            Best regards,

                            Ignacio Larrosa Cañestro
                            A Coruña (España)
                            ilarrosa@...
                          • Phil Carmody
                            ... Good question... ... To which you give the correct answer. See Riesel s appendices, or use QR. Phil () ASCII ribbon campaign () Hopeless ribbon
                            Message 13 of 23 , Mar 17, 2006
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                              --- Ignacio Larrosa Cañestro <ilarrosa@...> wrote:
                              > --- In primenumbers@yahoogroups.com, Jack Brennen <jb@...> wrote:
                              > > It can also be rewritten as:
                              > >
                              > > A = (((B*2-1)^2+5)^2*5-4)/16
                              > >
                              > >
                              > > Solving for A == 0 (modulo p), we can easily eliminate p == 2
                              > > and p == 5 directly from the original expression of A. Then
                              > > you get:
                              > >
                              > > (((B*2-1)^2+5)^2*5-4)/16 == 0 (modulo p)
                              > >
                              > > Multiply through by 16 (permissible because we eliminated p == 2),
                              > > and rearrange slightly:
                              > >
                              > > ((B*2-1)^2+5)^2*5 == 4 (modulo p)
                              > >
                              > > In order for this to have solutions, 5 must be a quadratic
                              > > residue modulo p, which eliminates primes p ending in 3 or 7.
                              >
                              > Why?

                              Good question...

                              > ((2*B - 1)^2 + 5)^2*5 = 4 (mod p) ===>
                              >
                              > ((2*B - 1)^2 + 5)^2 = 4*5^(-1) (mod p)
                              >
                              > Then 4*5^(-1) must be a quadratic residue modulo p, no?

                              To which you give the correct answer.

                              See Riesel's appendices, or use QR.

                              Phil

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                            • Mark Underwood
                              Talk about wonders never ceasing. Here s another 40 prime sequence I just stumbled across 10 minutes ago: 4x^2 -158x + 1601 is prime from x=0 to x=39, no
                              Message 14 of 23 , Mar 17, 2006
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                                Talk about wonders never ceasing. Here's another 40 prime sequence I
                                just stumbled across 10 minutes ago:

                                4x^2 -158x + 1601

                                is prime from x=0 to x=39, no repeats, all positive. This one has the
                                lovely feature of it's near symmetry: It starts at prime 1601, then
                                halfway at x=20 reaches a low at prime 41, then it swings back up
                                again to attain prime 1847 at x=39. It's four middle and lowest
                                primes are 53,43,41 and 47, four consecutive primes.

                                The sequence is bounded by
                                41*43 at x=-1
                                41*41 at x=40.
                                How close can you get!


                                Notice that the 40 prime polys thus far all have a starting
                                coefficient which is a square (1,4,9) :

                                1x^2 - x + 41
                                4x^2 - 158x + 1601
                                9x^2 - 231x + 1523

                                Have submitted to Mathworld. (None were in the Encyclopedia of
                                Integer Sequences.)

                                Mark




                                --- In primenumbers@yahoogroups.com, "Lelio sknet" <lelio@...> wrote:
                                >
                                > >BUT here's one I'd like to introduce with 40 consecutive primes,
                                all
                                > >positive and non repeating (!) :
                                > >
                                > >9x^2 -231x + 1523
                                > >
                                > >This sequence starts at prime 1523 and decreases to a low of 41
                                (at
                                > >x=13), then swings back up to finish off at prime 6203 at x=39.
                                > >
                                > >Wonders never cease!
                                > >
                                > >Mark
                                >
                                >
                                >
                                > So it is as good as Euler's
                                >
                                > Congratulations
                                >
                                > Next step you should submit it to Eric
                                >
                                > Lélio
                                >
                                >
                                >
                                >
                                >
                                >
                                >
                                >
                                > _____
                                >
                                >
                                >
                                > [Non-text portions of this message have been removed]
                                >
                              • Patrick Capelle
                                ... Two other little observations : 1. 41, 1601 and 1523 are prime numbers. 2. 1 - 1 + 41 = 41, which is prime. 4 - 158 + 1601 = 1447, which is prime. 9 - 231
                                Message 15 of 23 , Mar 18, 2006
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                                  --- In primenumbers@yahoogroups.com, "Mark Underwood"
                                  <mark.underwood@...> wrote:

                                  > Notice that the 40 prime polys thus far all have a starting
                                  > coefficient which is a square (1,4,9) :
                                  >
                                  > 1x^2 - x + 41
                                  > 4x^2 - 158x + 1601
                                  > 9x^2 - 231x + 1523
                                  -----------------------------------------------------------

                                  Two other little observations :
                                  1. 41, 1601 and 1523 are prime numbers.
                                  2. 1 - 1 + 41 = 41, which is prime.
                                  4 - 158 + 1601 = 1447, which is prime.
                                  9 - 231 + 1523 = 1301, which is prime.

                                  Regards,
                                  Patrick Capelle.
                                • Phil Carmody
                                  ... ? subst(4*x^2 -158*x + 1601,x, x/2+20) x^2 + x + 41 () ASCII ribbon campaign () Hopeless ribbon campaign / against HTML mail /
                                  Message 16 of 23 , Mar 18, 2006
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                                    --- Mark Underwood <mark.underwood@...> wrote:
                                    > Talk about wonders never ceasing. Here's another 40 prime sequence I
                                    > just stumbled across 10 minutes ago:
                                    >
                                    > 4x^2 -158x + 1601

                                    ? subst(4*x^2 -158*x + 1601,x, x/2+20)
                                    x^2 + x + 41



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                                  • Patrick Capelle
                                    ... These observations are in fact trivial. I suggested to search 40 prime polys ax^2 + bx + c such that : 1. a is a square 2. b is negative and non-prime 3. c
                                    Message 17 of 23 , Mar 18, 2006
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                                      --- In primenumbers@yahoogroups.com, "Mark Underwood"
                                      > <mark.underwood@> wrote:
                                      > Notice that the 40 prime polys thus far all have a starting
                                      > coefficient which is a square (1,4,9) :
                                      >
                                      > 1x^2 - x + 41
                                      > 4x^2 - 158x + 1601
                                      > 9x^2 - 231x + 1523

                                      --- In primenumbers@yahoogroups.com, "Patrick Capelle"
                                      <patrick.capelle@...> wrote:
                                      > Two other little observations :
                                      > 1. 41, 1601 and 1523 are prime numbers.
                                      > 2. 1 - 1 + 41 = 41, which is prime.
                                      > 4 - 158 + 1601 = 1447, which is prime.
                                      > 9 - 231 + 1523 = 1301, which is prime.
                                      >
                                      > Regards,
                                      > Patrick Capelle.
                                      ------------------------------------------------------------------

                                      These observations are in fact trivial.
                                      I suggested to search 40 prime polys ax^2 + bx + c such that :
                                      1. a is a square
                                      2. b is negative and non-prime
                                      3. c is prime
                                      4. a+b+c is prime

                                      But for x=0, c is always prime!
                                      And for x=1, the result a+b+c is also prime !
                                      Hence, my first observations are trivial.

                                      But what could be not trivial is to search 40 prime polys n^2x^2 +
                                      (q-p-n^2)x + p, with p,q primes and p+n^2 > q.

                                      Patrick Capelle.
                                    • Patrick Capelle
                                      ... wrote: But what could be not trivial is to search 40 prime polys n^2x^2 + (q-p-n^2)x + p, with p,q primes and p+n^2 q. ... n = 1 :
                                      Message 18 of 23 , Mar 18, 2006
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                                        --- In primenumbers@yahoogroups.com, "Patrick Capelle"
                                        <patrick.capelle@...> wrote:
                                        But what could be not trivial is to search 40 prime polys
                                        n^2x^2 + (q-p-n^2)x + p, with p,q primes and p+n^2 > q.
                                        ---------------------------------------------------------

                                        n = 1 : (p,q)=(41,41)
                                        n = 2 : (p,q)=(1601,1447)
                                        n = 3 : (p,q)=(1523,1301)

                                        Can someone propose a solution (p,q) for n = 4 ?

                                        Patrick Capelle.
                                      • Patrick Capelle
                                        ... wrote: Notice that the 40 prime polys thus far all have a starting coefficient which is a square (1,4,9) : 1x^2 - x + 41 4x^2 - 158x +
                                        Message 19 of 23 , Mar 18, 2006
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                                          --- In primenumbers@yahoogroups.com, "Mark Underwood"
                                          <mark.underwood@...> wrote:
                                          Notice that the 40 prime polys thus far all have a starting
                                          coefficient which is a square (1,4,9) :

                                          1x^2 - x + 41
                                          4x^2 - 158x + 1601
                                          9x^2 - 231x + 1523

                                          Have submitted to Mathworld. (None were in the Encyclopedia of
                                          Integer Sequences.)
                                          --------------------------------------------------------------

                                          The two last polynomials were already found in 2003.
                                          More informations here :
                                          http://www.primepuzzles.net/puzzles/puzz_232.htm

                                          Note that the substitutions are not forbidden, if the polynomials are
                                          able to give different sets of distinct primes when x = 0 to 39.
                                          Some of the prime numbers given by the three polynomials above are
                                          different ...
                                          Note also that unfortunately x^2 - x + 41 gives twice the number 41
                                          (it's not the case with x^2 + x + 41).

                                          Regards,
                                          Patrick Capelle.
                                        • Patrick Capelle
                                          We can keep Mark s idea about squares and modify the challenge. Let s start with the six polynomials of the Carlos Rivera web site (see
                                          Message 20 of 23 , Mar 18, 2006
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                                            We can keep Mark's idea about squares and modify the challenge.

                                            Let's start with the six polynomials of the Carlos Rivera web site
                                            (see http://www.primepuzzles.net/puzzles/puzz_232.htm ):
                                            x2 + x + 41
                                            x2 - 79x + 1601
                                            4x2 - 154x + 1523
                                            4x2 - 158x + 1601
                                            9x2 - 231x + 1523
                                            9x2 - 471x + 6203

                                            If we are interested by the 40 prime polynomials of the form
                                            n^2x^2 + (q-p-n^2)x + p, with p,q primes, we will find :
                                            For n = 1 : (p,q)=(41,43)
                                            (p,q)=(1601,1523)
                                            For n = 2 : (p,q)=(1523,1373)
                                            (p,q)=(1601,1447)
                                            For n = 3 : (p,q)=(1523,1301)
                                            (p,q)=(6203,5741)

                                            Can you propose two solutions for n = 4 ?

                                            Patrick Capelle.
                                          • Ignacio Larrosa Cañestro
                                            Wednesday, March 15, 2006 8:11 AM [GMT+1=CET], ... Thanks to Phil and Jose, I was slightly obtuse ... But, someone could tell us how to prove that it must be p
                                            Message 21 of 23 , Mar 19, 2006
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                                              Wednesday, March 15, 2006 8:11 AM [GMT+1=CET],
                                              Phil Carmody <thefatphil@...> escribió:

                                              > --- aldrich617 <aldrich617@...> wrote:
                                              >> I am searching for a way to prove the following Theorem:
                                              >>
                                              >> For any integer 'B', the value 'A' of the equation
                                              >> A = 5B^4 -10B^3 + 20B^2 -15B +11 will have as factors only
                                              >> integers that end in a one, excluding all others.
                                              >>
                                              >> This seems to be true at least up to 10^18. I think that proving
                                              >> it could give us new insights into primality testing,
                                              >> and factoring. Moreover there are similar equations, vast in
                                              >> number, apparently with the same property, that could then probably
                                              >> be verified to be similar threads of pure one. Together these
                                              >> would form an infinite interconnecting web.
                                              >
                                              > Can you tell us how you discovered that polynomial?
                                              > The discriminant is very smooth, being 5^3*11^2, and I suspect that
                                              > that's an essential ingredient to a proof.
                                              >
                                              > It seems that if p== +/-1 mod 10, then your polynomial splits at
                                              > least into 2 quadratics, and if p== +1 mod 10, then at least one of
                                              > those quadratics splits.
                                              >
                                              > This property should be easily explainable, but alas it's late and my
                                              > brain's on holiday. (I wrote this last night, but forgot to send).

                                              Thanks to Phil and Jose, I was slightly obtuse ...

                                              But, someone could tell us how to prove that it must be p = 1 (mod 10)?

                                              Thanks in advance,

                                              Ignacio Larrosa Cañestro
                                              A Coruña (España)
                                              ilarrosa@...
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