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Re: Pythagorean Sets of Consecutive Primes

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  • djbroadhurst
    ... See http://www.mathmeth.com/tom/files/pyth.pdf which claims a proof by infinite descent that x^2+y^2 and x^2-y^2 cannot both be odd squares. David (who
    Message 1 of 7 , Sep 1, 2011
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      --- In primenumbers@yahoogroups.com,
      "djbroadhurst" <d.broadhurst@...> wrote:

      > > Puzzle:
      > > If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?
      > >
      > > Suppose the converse. Then we have 4 possibilities:
      > > 1) a^2 = b^2 + c^2, d^2 = b^2 - c^2
      > > 2) a^2 = b^2 + c^2, d^2 = c^2 - b^2
      > > 3) a^2 = b^2 - c^2, d^2 = b^2 + c^2
      > > 4) a^2 = c^2 - b^2, d^2 = b^2 + c^2
      > > Hence there exists a pair of coprime integers (x,y)
      > > such that x^2+y^2 and x^2-y^2 are both odd squares.
      > > But each odd square is congruent to 1 modulo 8.
      > > Hence 2*x^2 = 2 mod 8
      >
      > [Then I screwed up, sorry

      See
      http://www.mathmeth.com/tom/files/pyth.pdf
      which claims a proof by infinite descent that
      x^2+y^2 and x^2-y^2 cannot both be odd squares.

      David (who miserably failed to find a simpler proof)
    • Jens Kruse Andersen
      ... It was apparently harder than I thought. I didn t have a proof but guessed it would be relatively easy when there is a known parametrization of all
      Message 2 of 7 , Sep 1, 2011
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        David wrote:
        > See
        > http://www.mathmeth.com/tom/files/pyth.pdf
        > which claims a proof by infinite descent that
        > x^2+y^2 and x^2-y^2 cannot both be odd squares.
        >
        > David (who miserably failed to find a simpler proof)

        It was apparently harder than I thought.
        I didn't have a proof but guessed it would be relatively easy when
        there is a known parametrization of all Pythagorean triples.
        I only did a brute force search to see there were no small solutions.

        --
        Jens Kruse Andersen
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