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25101Re: [PrimeNumbers] Re: Diophantine equation

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  • Sebastian Martin Ruiz
    May 26, 2013
      you made a sign error (+-) in(*****)

      De: WarrenS <warren.wds@...>
      Para: primenumbers@yahoogroups.com
      Enviado: Domingo 26 de Mayo de 2013 6:15
      Asunto: [PrimeNumbers] Re: Diophantine equation


      > Hello all:
      > Let c an even positive integer number fixed.
      > Consider the quadratic equation in two variables:
      > -3x^2+y^2-2xy-4cx+4cy+4=0
      > Can anyone prove that for every c large enough there is at least one other than the trivial solution (1,1)  of the equation with x and y both positive integers? i.e. (y>x>1 integers)
      > Sincerely
      > Sebastián Martín Ruiz

      --Let z=y-x. Then your equation is equivalent to
      4x^2 - z^2 - 4cz = 4
      with x>1 and z>0 integer unknowns.
      Now make the further change of variable t=z+2c.
      Then the equation is
      (2x)^2 - t^2 = 4+4c^2 (*******)
      with x>0 and t>2c integer unknowns,
      which forces t to be even. Letting t=2r
      we finally transform the equation to the form
      x^2 - r^2 = 1+c^2
      in integer unknowns x>1 and r>c.

      This in turn may be written
      (x-r)*(x+r) = 1+c^2.

      Does a solution of this always exist if c is large enough?
      If c is even, then x=r+1 where 2r=c^2 will work.
      If c is odd, then there is no solution because
      right hand side is 2 mod 4 and since x and r must have same parity,
      the left hand side is divisible by 4.

      So, in conclusion, your conjecture is DISPROVEN.
      Assuming I did not make any stupid errors.

      --Warren D. Smith.

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