Find all solutions to the equation $x^2 + 3y^2 = z^2$ Find all positive integer solutions to the equation $x^2 + 3y^2 = z^2$
So here's what I've done thus far:
I know that if a solution exists, then there's a solution where (x,y,z) = 1, because if there is one where $(x,y,z) = d$, then $\frac{x}{d}, \frac{y}{d}, \frac{z}{d}$. is also a solution.
I'm trying to mimic the pythagorean triple proof where they have that $x = u^2 - v^2$ and $y = 2uv$ and $z = u^2 + v^2$
So, looking at the original equation mod 4, I can see that it's in the form:
$(0 or 1) - (0 or 1) = 3(0 or 1)$. Thus we for sure know that $3y^2$ is congruent to 0 mod 4 and that $y$ is even. We also learned that $z$ and $x$ are either both even or both odd.
From there I'm guessing I need to handle each case, but I'm stuck as to where to go from here.

I don't think y is necessarily even, $13^2 + 3*3^2 = 14^2.$
There seem to be two sets of solutions
$2x = p^2 - 3q^2, y = 2pq, 2z = p^2 + 3q^2$ and $pq$ odd
$x = p^2 - 3q^2, y =  pq,  z = p^2 + 3q^2$ and $pq$ even
$gcd(p,q) = 1 = gcd(p,3)$
