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Pell Equation

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  • Kermit Rose
    ... Hello Bernhard, Hello David. More generally, for any given integer d, if a1^2 - d b1^2 = 1 and a2^2 - d b2^2 = 1 then (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1
    Message 1 of 5 , Jun 22, 2013
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      > 1a. Re: some questions about algebraic factoring in the field of adjoine


      > --- In primenumbers@yahoogroups.com,
      > "bhelmes_1" <bhelmes@...> asked:

      > > d=4n+2 or d=4n+3
      > > and the equation of Pell a^2-d*b^2=1
      > > What is the fastest way to solve the equation of Pell

      > Posted by: "djbroadhurst" d.broadhurst@... djbroadhurst
      > Date: Fri Jun 21, 2013 5:10 pm ((PDT))


      > In the case of positive d = 3 mod 4,
      > used quadunit(4*d), in Pari-GP,
      > and then take powers of this unit,
      > which has norm = +1.


      Hello Bernhard, Hello David.

      More generally, for any given integer d,

      if a1^2 - d b1^2 = 1
      and
      a2^2 - d b2^2 = 1

      then

      (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1

      and

      (a1 a2 + d b1 b2)^2 - d (a1 b2 + b1 a2)^2 = 1

      Thus if

      [a1,b1,d] and [a2,b2,d]

      are solutions to Pells equation,

      then so are:

      [a1^2 + d b1^2,2 d a1 b1,d]

      and

      [a1 a2 + d b1 b2, a1 b2 + b1 a2,d]

      Then by finding, for a given integer d,
      the smallest solution in a and b,
      such that

      a^2 - d b^2 = 1,

      it is a simple matter to

      repeated use the recursively generated solutions

      [a1^2 + d b1^2,2 d a1 b1,d]

      and

      [a1 a2 + d b1 b2, a1 b2 + b1 a2,d]

      to find all the solutions, within some bound on a and b, of the Pell
      equation for a given integer d.


      Kermit
    • djbroadhurst
      ... Indeed. Yet it is both clearer and more efficient to take powers of a unit. I illustrated this for positive d=4 mod 3, where one takes uses quadunit(4*d),
      Message 2 of 5 , Jun 23, 2013
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        --- In primenumbers@yahoogroups.com,
        Kermit Rose <kermit@...> wrote:

        > if
        > a1^2 - d b1^2 = 1
        > a2^2 - d b2^2 = 1
        > then
        > (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1
        > (a1 a2 + d b1 b2)^2 - d (a1 b2 + b1 a2)^2 = 1

        Indeed. Yet it is both clearer and more efficient to
        take powers of a unit. I illustrated this for positive
        d=4 mod 3, where one takes uses quadunit(4*d),
        with positive norm. The extension to other
        discriminants is straightforward. Moreover
        "quadunit" yields the building block that Kermit
        assumed to have been found, by undisclosed methods :-)

        David
      • djbroadhurst
        typo: I meant d=3 mod 4
        Message 3 of 5 , Jun 23, 2013
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          typo: I meant d=3 mod 4

          --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
          >
          >
          >
          > --- In primenumbers@yahoogroups.com,
          > Kermit Rose <kermit@> wrote:
          >
          > > if
          > > a1^2 - d b1^2 = 1
          > > a2^2 - d b2^2 = 1
          > > then
          > > (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1
          > > (a1 a2 + d b1 b2)^2 - d (a1 b2 + b1 a2)^2 = 1
          >
          > Indeed. Yet it is both clearer and more efficient to
          > take powers of a unit. I illustrated this for positive
          > d=4 mod 3, where one takes uses quadunit(4*d),
          > with positive norm. The extension to other
          > discriminants is straightforward. Moreover
          > "quadunit" yields the building block that Kermit
          > assumed to have been found, by undisclosed methods :-)
          >
          > David
          >
        • Maximilian Hasler
          ... no : the 3rd equation is wrong. M. [Non-text portions of this message have been removed]
          Message 4 of 5 , Jun 23, 2013
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            On Sun, Jun 23, 2013 at 7:34 AM, djbroadhurst <d.broadhurst@...>wrote:

            > **
            > Kermit Rose <kermit@...> wrote:
            > > if
            > > a1^2 - d b1^2 = 1
            > > a2^2 - d b2^2 = 1
            > > then
            > > (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1
            > > (a1 a2 + d b1 b2)^2 - d (a1 b2 + b1 a2)^2 = 1
            >
            > Indeed.
            >

            no : the 3rd equation is wrong.

            M.


            [Non-text portions of this message have been removed]
          • Kermit Rose
            On 6/24/2013 12:07 PM, Maximilian Hasler ... (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1 should have been (a1^2 + d b1^2)^2 - d (2 a1 b1)^2 = 1 Kermit
            Message 5 of 5 , Jun 24, 2013
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              On 6/24/2013 12:07 PM,

              "Maximilian Hasler"

              wrote:
              > ________________________________________________________________________
              > 1a. Re: Pell Equation
              > Posted by: "Maximilian Hasler" maximilian.hasler@... maximilian_hasler
              > Date: Sun Jun 23, 2013 12:37 pm ((PDT))
              >
              >
              >> **
              >> Kermit Rose <kermit@...> wrote:
              >>> if
              >>> a1^2 - d b1^2 = 1
              >>> a2^2 - d b2^2 = 1
              >>> then
              >>> (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1
              >>> (a1 a2 + d b1 b2)^2 - d (a1 b2 + b1 a2)^2 = 1
              >>
              > no : the 3rd equation is wrong.
              >
              > M.
              >

              (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1

              should have been

              (a1^2 + d b1^2)^2 - d (2 a1 b1)^2 = 1


              Kermit
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