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25185Pell Equation

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  • Kermit Rose
    Jun 22, 2013
      > 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
      a2^2 - d b2^2 = 1


      (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

      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]


      [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]


      [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.

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