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25177some questions about algebraic factoring in the field of adjoined square

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  • bhelmes_1
    Jun 12, 2013
      A beautiful day to all,

      i have tried to express the following ideas as good as possible.
      Nevertheless it is not perfect :-(

      f:=p*q the number which should be factorized, p and q primes
      d:=4n+2 or d=4n+3 with jacobi (d, f)=-1
      D:=sqrt (d)
      F: a+bD with a, b element Z
      and the equation of Pell a²-db²=1

      As i know there are always infinite solutions for this equation

      1. Question
      What is the fastest way to solve the equation of Pell.
      Is continued fraction preferable
      or is it better to use Hensel lifting
      for the polynomial f(b)=db²+1 and to look for a solution
      with a²=f(b)

      2. Question
      Are the solutions of the equation of Pell always useful
      for the factoring of f
      if you reduce the solutions to a²-db²=1 mod f

      you get a1²-db1² = a2²-db2² mod f
      if a1=a2 or b1=b2 you get the equation of the form
      x^2=y^2 mod f

      3. Question
      Does it make a difference if you look for a factoring solution in N
      or in F(D)

      5. f=35
      solutions of Pell a²-db²=1 mod 35 :
      1, 6, 29, 34, 21+5D, 14+5D, 21+30D, 14+30D
      gcd (6-1, 35) = 5
      gcd (34-29, 35)= 5
      gcd (21, 35) = 7
      gcd (14, 35) = 7 and so on

      6. f=55
      solutions of Pell a²-db²=1 mod 55 :
      1, 21, 34, 54, 44+20D, 44+35D, 11+20D, 11+35D
      gcd (44, 55)=11
      gcd (20, 55)=5
      gcd (11, 55)=11

      Perhaps this application helps more for small values f<100
      I. Primes
      4. cycle structur of primes
      c) mit adjungierter Wurzel / with adjoined square
      Choose p=35 Adjoined square root=2
      You find the solutions of Pell by clicking on the |1+A|=1

      Nice greetings from the primes
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