23046Re: Pythagorean Sets of Consecutive Primes
- Sep 1, 2011--- In firstname.lastname@example.org, "djbroadhurst" <d.broadhurst@...> wrote:
> > Puzzle:[Then I screwed up, sorry: I *should* have said:]
> > 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
and x is odd.
I am pondering the unfortunate consequences....
David (feeling more odd then even :-)
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