- --- In primenumbers@yahoogroups.com, "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 :-) - --- In primenumbers@yahoogroups.com,

"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

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) - David wrote:
> See

It was apparently harder than I thought.

> 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)

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