[math-fun]

The solutions of x^2 + y^2 = z^2 the well known "Pythagorean triples" goes back to Euclid. I'd like to know if anyone in the history of Diophantine equations has looked at the same problem with 2 replaced by -2. That is, find all solutions in integers of 1/x^2 + 1/y^2 = 1/z^-2. The solution is easy and rather nice. David