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Re: Diophantine equation and twin primes

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  • Mark
    ... Based on Warren s (corrected) simplification to your equation -3x^2 + y^2 -2xy - 4cx + 4cy + 4 = 0 namely r^2 - x^2 - c^2 + 1 = 0 Then (r-x)(r+x) =
    Message 1 of 2 , May 27, 2013
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      --- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
      >
      > ________________________________
      >
      > Prove: 
      >
      > Theorem:
      >
      > Let c a positive even number >2
      >
      > If (x,y)= (1,1) and (c^2/2-1,3c^2/2-2c-1) are the only positive integer solutions
      >
      >  of the polynomial 
      >
      > -3x^2+y^2-2xy-4cx+4cy+4=0
      >
      >  then c +1 and c-1 are twin primes
      >
      >
      > Sincerely
      >
      > Sebastián Martín Ruiz
      >

      Based on Warren's (corrected) simplification to your equation

      -3x^2 + y^2 -2xy - 4cx + 4cy + 4 = 0

      namely

      r^2 - x^2 - c^2 + 1 = 0

      Then

      (r-x)(r+x) = (c-1)(c+1)

      If both c-1 and c+1 are prime, then clearly there are only two options:

      r-x = c-1 and r+x = c+1

      OR

      r-x = 1 and r+x = (c-1)(c+1)

      Follow each of those options and you easily arrive at the two results.

      (x,y) = (1,1) and ( (c^2)/2 - 1,3(c^2)/2 - 2c - 1)

      Mark
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