# A Diophantine Problem

• Mar 22nd 2008, 05:13 AM
enrique
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?
• Mar 22nd 2008, 08:14 PM
ThePerfectHacker
Case 1: x is even. Then we know $x^2 = 2st , y = s^2 - t^2, z=s^2+t^2$ where $\gcd(s,t)=1$ and the parity is different. It is safe to assume that $s$ is even (because the other way around everything is symettric except for $y$ but that not matter because we are squaring it, and thereby removing the negative sign). But $x=2x'$. Which means, $4(x')^2 = 2st \implies (x')^2 = (s/2)t$ (remember $s$ is even). Since $\gcd(s/2,t)=1$ it means $s/2=a^2\implies s = 2a^2$ and $t=b^2$. Thus, $x=2ab, y=4a^2 - b^2, z=4a^2+b^2$. For arbitrary relatively prime integers $a,b$.

Case 2: x is odd. Then we know $x^2 = s^2 - t^2, y=2st, z=s^2+t^2$. Thus, $x^2 + t^2 = s^2$. This forces $t$ to be even. And so, $x=a^2 - b^2, t=2ab, s=a^2+b^2$. Thus, $x=a^2-b^2, y=4ab(a^2+b^2), z=(a^2+b^2) + 4a^2b^2$.
• Mar 23rd 2008, 09:26 AM
enrique
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).
• Mar 23rd 2008, 10:44 AM
ThePerfectHacker
Quote:

Originally Posted by enrique
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?

I do not think so. Maybe if $n$ is a power of two.