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

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  • Mark Underwood
    I find this all very interesting, it is new to me. What specifically caught my attention was the linear representation of primes generated by the equation x^2
    Message 1 of 12 , Jul 5, 2003
      I find this all very interesting, it is new to me. What specifically
      caught my attention was the linear representation of primes generated
      by the equation x^2 + p*y^2, where p is prime. The nature of the
      linear representation is not always the same from prime to prime. Is
      there are rule as Sleephound queried, I'm not sure.

      Below are the representations for p=2, p=3, p=5, p=7 and p=11. The
      p=11 result is courtesy of Sleephound's data, not Wolfram's site.


      From the Wolfram site,

      For p=2 :

      All primes that can be expressed as x^2 + 2y^2 can also be
      represented as 8n+1 or 8n+3. Conversely, if I understand it
      correctly, all primes of the form 8n+1 or 8n+3 can be expressed as
      x^2 + 2y^2.

      All primes that can be expressed as x^2 - 2y^2 can also be
      represented as 8n+1 or 8n+7.

      Observation: There is partial intersection between the two groups.
      One quarter of the primes are not represented at all, that is primes
      of the form 8n + 5.

      For p=3:

      All primes that can be expressed as x^2 + 3y^2 can also be
      represented as 6n+1.

      All primes that can be expressed as x^2 - 3y^2 can also be
      represented as 12n+1.

      Observation: One group is entirely a subset of the other. One half of
      the primes are not represented, that is primes of the form 6n-1.


      For p=5:

      All primes that can be expressed as x^2 + 5y^2 can be also be
      represented as 20n+1 or 20n+9.

      All primes that can be expressed as x^2 - 5y^2 can also be
      represented as 10n+1 or 10n+9.

      Observation: One group is entirely a subset of the other. One half of
      the primes are not represented, that is primes of the form 10n+3 and
      10n+7.

      For p=7:

      All primes that can be expressed as x^2 + 7y^2 can also be
      represented as 14n+1 or 14n+9 or 14n+25.

      All primes that can be expressed as x^2 - 7y^2 can also be
      represented as 28n+1 or 28n+9 or 28n+25.

      Observation: One group is entirely a subset of the other. One half of
      the primes are not represented, that is primes of the form 14n+3 and
      14n+5 and 14n+9.

      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.

      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.

      Observation: I'm not sure if the two groups overlap even though they
      have the same linear representation. And the fact that the converse
      is not true in this cases sets it apart from the earlier primes. I
      wonder if all the rest of the primes after 11 (13,17,19 ...) are like
      11 in this regard?


      Final Note:

      Also, I noticed from Wolfram's table that all primes of the form x^2
      + y^2 can be written as 4n+1, and (I am supposing) the converse is
      true, all primes of the form 4n+1 can be written as x^2 + y^2. That
      is half the primes. And as we know all of the primes can be written
      as x^2 - y^2.

      It seems that no expression x^2 - by^2 combined with x^2 + by^2 will
      yield all the primes, except of course when b = 1. But it appears
      that the combined *three* expressions, x^2 + y^2, x^2 + 2y^2 and x^2 -
      2y^2 will yield all the primes since they cover all the
      possibilities: 8n+1, 8n+3, 8n+5 and 8n+7. I doubt that any other
      combination of three will yield all the primes. Wait, could it be
      that no combination of four will yield all the primes, even if up to
      two of the above three expressions are used? Further, no combination
      of 5, or 6, or 7 or ....?

      Mark





      --- In primenumbers@yahoogroups.com, "sleephound" <sleephound@y...>
      wrote:
      > Primes of the form X^2+Y^2 are of the form 4n+1, Primes of the form
      > X^2+2Y^2 are of the form 8n+1 or 8n+1. Some other examples of
      > quadratic forms yielding primes of linear form are listed here:
      >
      > http://mathworld.wolfram.com/PrimeRepresentation.html
      >
      > Primes of the form X^2+11Y^2 begin 47, 53, 103, 163, 199. There
      > doesn't appear to be a linear form that separates these primes into
      > two groups - I've checked through 4000. Is there some simple rule
      > that characterizes these primes? Is there a simple rule for when a
      > linear characterization exists?
    • 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 2 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 3 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 4 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 5 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 6 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 7 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 8 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 9 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 10 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 11 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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