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Re: Pythagorean Sets of Consecutive Primes

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  • djbroadhurst
    ... [Then I screwed up, sorry: I *should* have said:] and x is odd. I am pondering the unfortunate consequences.... David (feeling more odd then even :-)
    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: I *should* have said:]

      and x is odd.

      I am pondering the unfortunate consequences....

      David (feeling more odd then even :-)
    • 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 2 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 3 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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