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Re: [PrimeNumbers] Re: Is there a simple form for prmes of X^2+11Y^2?

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  • mikeoakes2@aol.com
    In a message dated 06/07/03 06:57:13 GMT Daylight Time, ... In fact, the fraction of primes p congruent to 1 or 3 or 5 or 9 or 15 mod 22 which can be so
    Message 1 of 12 , Jul 6, 2003
      In a message dated 06/07/03 06:57:13 GMT Daylight Time,
      mark.underwood@... writes:


      > For p=11:
      >
      > All primes that can be expressed as x^2 + 11y^2 can also be
      > represented as 22n+1, 22n+3, 22n+5, 22n+9 or 22n + 15. But the
      > converse is not true; not all primes of the form 22n+1, 22n+3, 22n+9
      > or 22n+15 can be written as x^2 + 11y^2.
      >

      In fact, the fraction of primes p congruent to 1 or 3 or 5 or 9 or 15 mod 22
      which can be so expressed is certainly 1/3.

      Experimental evidence:-
      p_max fraction
      1000 0.29762
      10000 0.3271
      100000 0.32840
      1000000 0.33270
      10000000 0.33283
      100000000 0.33322
      (execution time 35 GHz-mins.)

      Theoretical evidence:-
      David Cox's "Primes of the Form x^2 + ny^2" (Wiley, 1989) proves the following
      Theorem 9.2: Let n > 0 be an integer. Then there is a monic irreducible
      polynomial f(x) of degree h(-4n) such that if an odd prime p divides neither n nor
      the discriminant of f(x) then
      p = x^2 + ny^2 iff
      (-n|p) = 1 and f(x) = 0 mod p has an integer solution.

      [In the above, (-n|p) = is the Legendre symbol, and h(-4n) is the class
      number of the quadratic field Q(sqrt(-4n)).]

      Now (-11|p) = 1 precisely for the above set of 5 residues.

      And h(-44) = 3 - see e.g. H.Cohen "A Course in Computational Algebraic Number
      Theory" (Springer, 1996), Appendix B.
      So the extra condition is that a cubic polynomial has a root mod p; I haven't
      been able to determine the polynomial explicitly (this stuff is /hard/), but
      it's reasonable to assume that it has a root mod p for 1 in 3 values of p, by
      analogy with cubic residuocity.

      > All primes that can be expressed as x^2 - 11y^2 can be also
      > represented as 22n+1 or 22n+3 or 22n+9 or 22n+15. But again, I don't
      > think the converse is true.
      >

      This case is much easier:-
      p = x^2 - 11y^2 iff p = 1 or 5 or 9 or 25 or 37 mod 44.

      Mike Oakes



      [Non-text portions of this message have been removed]
    • mikeoakes2@aol.com
      ... haven t been able to determine the polynomial explicitly By searching a subspace of the space of all cubics, I have found a suitable one (it is not
      Message 2 of 12 , Jul 6, 2003
        > So the extra condition is that a cubic polynomial has a root mod p; I
        haven't been > able to determine the polynomial explicitly

        By searching a subspace of the space of all cubics, I have found a suitable
        one (it is not unique), so can state, finally, the following Theorem:-

        If p is not 2 or 11,
        then p = x^2 + 11*y^2 iff
        (i) p = 1 or 3 or 5 or 9 or 15 mod 22, and
        (ii) there is an integer solution x to the equation x^3 - 4*x + 4 = 0 mod p.

        The exclusion of 2 and 11 is because the cubic has discriminant -176 =
        -11*2^4.

        It's interesting that all the candidate cubics had discriminants = 0 mod 44.

        Mike Oakes


        [Non-text portions of this message have been removed]
      • Mark Underwood
        Thanks Mike. Your findings are starting to go over my head but are very appreciated. I shall have to look into this further. Based on your conclusion that p
        Message 3 of 12 , Jul 6, 2003
          Thanks Mike. Your findings are starting to go over my head but are
          very appreciated. I shall have to look into this further.

          Based on your conclusion that "p = x^2 - 11y^2 iff p = 1 or 5 or 9 or
          25 or 37 mod 44", that would mean it covers 1/4 of the primes, since
          these are 5 of the possible twenty residues mod 44. It turns out then
          that the table at Wolfram's site was never meant to be complete but
          just a small sample.

          I found your data that x^2 + 11y^2 covers 1/3 of the primes very
          interesting. And your finding that

          If p is not 2 or 11, then p = x^2 + 11*y^2 iff
          (i) p = 1 or 3 or 5 or 9 or 15 mod 22, and
          (ii) there is an integer solution x to the equation x^3 - 4*x + 4 = 0
          mod p.

          is simply amazing (and beyond my reach right now!).

          Incidentally, a clever and seasoned veteran in these matters informed
          me via email that the three quadratic forms x^2+y^2, x^2+2y^2 and x^2-
          2y^2 are not unique in their ability to cover all the primes as I had
          supposed. As a hint, he showed that x^2+y^2 and x^2+3y^2 cover the
          12n+1, 12n+5, and 12n+7 residues, leaving only the 12n+11 residue
          uncovered. Then he said there is indeed a quadratic form with a small
          b that does cover all the primes of 12n+11 but left it to me to find
          it (!) I wonder if the b is positive. Another hint was to do a seach
          on Bernstein and Atkin's sieve, which I shall do as time allows.

          Mark



          --- In primenumbers@yahoogroups.com, mikeoakes2@a... wrote:
          > > So the extra condition is that a cubic polynomial has a root mod
          p; I
          > haven't been > able to determine the polynomial explicitly
          >
          > By searching a subspace of the space of all cubics, I have found a
          suitable
          > one (it is not unique), so can state, finally, the following
          Theorem:-
          >
          > If p is not 2 or 11,
          > then p = x^2 + 11*y^2 iff
          > (i) p = 1 or 3 or 5 or 9 or 15 mod 22, and
          > (ii) there is an integer solution x to the equation x^3 - 4*x + 4 =
          0 mod p.
          >
          > The exclusion of 2 and 11 is because the cubic has discriminant -
          176 =
          > -11*2^4.
          >
          > It's interesting that all the candidate cubics had discriminants =
          0 mod 44.
          >
          > Mike Oakes
          >
          >
          > [Non-text portions of this message have been removed]
        • mikeoakes2@aol.com
          In a message dated 05/07/03 15:52:25 GMT Daylight Time, sleephound@yahoo.com ... For there to be a characterisation of the form p = x^2 + n*y2 iff p = r_1 or
          Message 4 of 12 , Jul 6, 2003
            In a message dated 05/07/03 15:52:25 GMT Daylight Time, sleephound@...
            writes:


            > ? Is there a simple rule for when a
            > linear characterization exists?
            >

            For there to be a characterisation of the form
            "p = x^2 + n*y2 iff p = r_1 or r_2 or .. or r_k mod q"
            it is necessary that the class number h(-4*n) = 1 or 2, since that is the
            degree of the polynomial in the subsidiary condition (see my earlier post).

            [Note: h(-4*n) = 2 is ok because it will give a subsidiary equation of the
            form x^2 - b*x + c = 0 mod p, i.e. (b^2 - 4*c)/4 is to be a quadratic residue
            mod p, which, by Gauss's law of quadratic reciprocity, will be true iff p is a
            quadratic residue or nonresidue mod some integer s, which in turn will give a
            condition of the form p = r_1 or r_2 or .. or r_k mod s.]

            The /only/ values of (square-free) n for which this class number condition
            holds are (again from Cohen, Appendix B):-
            h(-4*n) = 1 for n=1,2,3,7.
            h(-4*n) = 2 for n=5,6,10,13,15,22,37,58.

            This explains why tables like that at
            http://mathworld.wolfram.com/PrimeRepresentation.html
            only have entries which are a subset of these particular values of n.

            Mike Oakes


            [Non-text portions of this message have been removed]
          • Satoshi Tomabechi
            On Sun, 6 Jul 2003 08:19:29 EDT ... Hilbert polynomial of the quadratic field with discriminant -44 is x^3-1122662608*x^2+270413882112*x-653249011576832
            Message 5 of 12 , Jul 6, 2003
              On Sun, 6 Jul 2003 08:19:29 EDT
              mikeoakes2@... wrote:
              > And h(-44) = 3 - see e.g. H.Cohen "A Course in Computational Algebraic Number
              > Theory" (Springer, 1996), Appendix B.
              > So the extra condition is that a cubic polynomial has a root mod p; I haven't
              > been able to determine the polynomial explicitly (this stuff is /hard/), but

              Hilbert polynomial of the quadratic field with discriminant -44 is
              x^3-1122662608*x^2+270413882112*x-653249011576832

              Satoshi Tomabechi
            • sleephound
              ... I ve found two places on the web where I think h(-44)=3 ought to show up, and it isn t in either place: http://mathworld.wolfram.com/ClassNumber.html
              Message 6 of 12 , Jul 6, 2003
                --- In primenumbers@yahoogroups.com, mikeoakes2@a... wrote:

                > And h(-44) = 3 - see e.g. H.Cohen "A Course in Computational
                >Algebraic Number Theory" (Springer, 1996), Appendix B.

                I've found two places on the web where I think h(-44)=3 ought to show
                up, and it isn't in either place:

                http://mathworld.wolfram.com/ClassNumber.html

                http://www.research.att.com/cgi-bin/access.cgi/as/
                njas/sequences/eisA.cgi?Anum=A006203

                I also see some differences from your list here:

                > h(-4*n) = 1 for n=1,2,3,7.
                > h(-4*n) = 2 for n=5,6,10,13,15,22,37,58.

                12 and 28 don't show up for h(-d)=1.
                60 doesn't show up for h(-d)=2.

                Am I doing something wrong, or so these sources disagree with your
                source?
              • mikeoakes2@aol.com
                In a message dated 07/07/03 04:23:08 GMT Daylight Time, sleephound@yahoo.com ... h(-12) = h(-28) = 1 give n = 3 resp. 7. h(-60) = 2 gives n = 15. Ok? Notice
                Message 7 of 12 , Jul 7, 2003
                  In a message dated 07/07/03 04:23:08 GMT Daylight Time, sleephound@...
                  writes:


                  > I also see some differences from your list here:
                  >
                  > > h(-4*n) = 1 for n=1,2,3,7.
                  > > h(-4*n) = 2 for n=5,6,10,13,15,22,37,58.
                  >
                  > 12 and 28 don't show up for h(-d)=1.
                  > 60 doesn't show up for h(-d)=2.
                  >
                  > Am I doing something wrong, or so these sources disagree with your
                  > source?
                  >

                  h(-12) = h(-28) = 1 give n = 3 resp. 7.
                  h(-60) = 2 gives n = 15.
                  Ok?

                  Notice that I said "..(squarefree) n...". The list for /all/ n is a bit
                  longer:-
                  h(-4*n) = 1 for n=1,2,3,4,7.
                  h(-4*n) = 2 for n=5,6,8,9,10,12,13,15,16,18,22,25,28,37,58.

                  In fact, there's quite a serious error in my earlier claim that
                  > it is necessary that the class number h(-4*n) = 1 or 2, since that is the
                  > degree of the polynomial in the subsidiary condition

                  Further reading has made it clear that not only the great Gauss (1801) but
                  even before him the mighty Euler (c. 1750) were in posession of /65/ values of n
                  which have a "linear" characterisation of the type we are discussing.
                  To the above must be added (using our modern notation of class numbers of
                  fields):-
                  h(-4*n) = 4 for n=21,24,30,33,40,42,45,48,57,60,70,72,
                  78,85,88,93,102,112,130,133,177,190,232,253
                  h(-4*n) = 8 for n=105,120,165,168,210,240,273,280,312,330,
                  345,357,385,408,462,520,760.
                  h(-4*n) = 16 for n=840,1320,1365,1848.
                  [These values come from David Cox's superb book.]

                  Apparently it has since been proved that there is at most one other value of
                  n, but no-one has been able to show whether or not it exists!
                  This is one of the deepest and richest areas of number theory, still not
                  exhausted by centuries of mining.

                  Mike Oakes


                  [Non-text portions of this message have been removed]
                • mikeoakes2@aol.com
                  In a message dated 07/07/03 02:08:44 GMT Daylight Time, ... Thanks! Can you reduce it? (I m no pari expert) - it ought to be equivalent to my cubic. Mike Oakes
                  Message 8 of 12 , Jul 7, 2003
                    In a message dated 07/07/03 02:08:44 GMT Daylight Time,
                    tomabeti@... writes:


                    > Hilbert polynomial of the quadratic field with discriminant -44 is
                    > x^3-1122662608*x^2+270413882112*x-653249011576832
                    >

                    Thanks!
                    Can you reduce it? (I'm no pari expert) - it ought to be equivalent to my
                    cubic.

                    Mike Oakes



                    [Non-text portions of this message have been removed]
                  • Satoshi Tomabechi
                    On Mon, 7 Jul 2003 04:31:45 EDT ... x^3 - x^2 + x + 1 ... It says that two polynomials define the same (class) filed. You can use another Weber polynomial x^3
                    Message 9 of 12 , Jul 7, 2003
                      On Mon, 7 Jul 2003 04:31:45 EDT
                      mikeoakes2@... wrote:

                      > > Hilbert polynomial of the quadratic field with discriminant -44 is
                      > > x^3-1122662608*x^2+270413882112*x-653249011576832
                      > >
                      >
                      > Thanks!
                      > Can you reduce it? (I'm no pari expert) - it ought to be equivalent to my
                      > cubic.


                      x^3 - x^2 + x + 1

                      David Broadhurst told me:
                      >? polred(x^3-1122662608*x^2+270413882112*x-653249011576832)
                      >%1 = [x - 1, x^3 - x^2 - x - 1, x^3 - x^2 + x + 1]
                      >so x^3-x^2-x+1 seems to do the same job?
                      It says that two polynomials define the same (class) filed.

                      You can use another Weber polynomial x^3 - 2*x^2 + 2*x - 2,
                      which also defines the same field.

                      Satoshi Tomabechi
                    • mikeoakes2@aol.com
                      In a message dated 08/07/03 02:14:47 GMT Daylight Time, ... Thank you. Meanwhile, (pari Grand Master) David Broadhursthas emailed me with ... So that s good !
                      Message 10 of 12 , Jul 8, 2003
                        In a message dated 08/07/03 02:14:47 GMT Daylight Time,
                        tomabeti@... writes:

                        >> Can you reduce it? (I'm no pari expert) - it ought to be equivalent to my
                        >> cubic.
                        >
                        > x^3 - x^2 + x + 1

                        Thank you.
                        Meanwhile, (pari Grand Master) David Broadhursthas emailed me with
                        > henri cohen's "minimal height" form for your d=-44 cubic is
                        > x^3-x^2+x+1

                        So that's good ! - there's complete agreement between your theoretical
                        Hilbert polynomial and my experimentally discovered cubic.

                        Mike Oakes



                        [Non-text portions of this message have been removed]
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