Re: Pythagorean Sets of Consecutive Primes
- --- In firstname.lastname@example.org,
"djbroadhurst" <d.broadhurst@...> wrote:
> > Puzzle:See
> > 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
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)
- David wrote:
> SeeIt was apparently harder than I thought.
> 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)
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