A Diophantine Problem
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?
Case 1: x is even. Then we know where and the parity is different. It is safe to assume that is even (because the other way around everything is symettric except for but that not matter because we are squaring it, and thereby removing the negative sign). But . Which means, (remember is even). Since it means and . Thus, . For arbitrary relatively prime integers .
Case 2: x is odd. Then we know . Thus, . This forces to be even. And so, . Thus, .
Thanks for the help ThePerfectHacker. Can we use the same strategy for the equations x^n + y^2 = z^2, where n>2? Or for which n we can apply the same procedure?
Note: I have noticed a small typo in the solution:z must be (a^2 + b^2)^2 + 4*((ab)^2).
I do not think so. Maybe if is a power of two.
Originally Posted by enrique