--- 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