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25107Re: Diophantine equation

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  • WarrenS
    May 26, 2013
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      On 5/26/13, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
      > you made a sign error (+-) in(*****)

      --you are correct... trying again below. Hopefully no stupid errors this time, but
      I again am not checking carefully.

      >> 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 (****have corrected sign error here***)
      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.
      or equivalently
      (r-x)*(r+x) = c^2 - 1 = (c-1)*(c+1)
      Does a solution of this always exist if c is large enough?

      If c is even, then r=x+1 where 2r+2=c^2 will work.

      If c is odd (then my old erroneous disproof with the sign error
      fails to work) then since r>c and x>1 are demanded, the
      obvious solution r=c, x=1 is not eligible.

      But since the right hand side is divisible by 4 since c-1 and c+1 both are even,
      r=x+2, 4x+4=(c-1)*(c+1) which implies x=(c^2-5)/4 works.

      So, now your conjecture is PROVEN -- solutions always exist for c large enough.

      Warren D. Smith
      http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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