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23044Re: Pythagorean Sets of Consecutive Primes

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  • djbroadhurst
    Sep 1, 2011
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      --- In primenumbers@yahoogroups.com,
      "Jens Kruse Andersen" <jens.k.a@...> wrote:

      > Puzzle:
      > If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?

      Yes.

      Proof: 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 and x is even.
      Hence both y^2 and -y^2 are congruent to 1 modulo 4,
      which is absurd.

      David Broadhurst
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