What is the general solution of x^4 + y^2 = z^2 ?
We know that primitive Pythagorean triples correspond to the triples (2st, s^2 - t^2, s^2 + t^2), where gcd (s,t)=1. So i separate the problem into two cases:
i) x is even : i found the general solution as
(sqrt(2st), s^2 - t^2, s^2 + t^2).
ii) y is even : i found the general solution as
(sqrt(s^2 - t^2), 2st, s^2 + t^2).
Is this solution true? Or in other words, do these triples produce every solution of this equation?