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23046Re: Pythagorean Sets of Consecutive Primes

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  • djbroadhurst
    Sep 1, 2011
      --- 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 :-)
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