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 developed an alternative to Euclid's formula that may help here.
$$A=(2n-1)^2+2(2n-1)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2$$
Whenever $(2n-1)$ and $k$ are both multiples of $3$, then $9$ will be a factor of $GCD(A,B,C)$.
For example:
$F(2,3)=(27,36,45)\qquad F(2,6)=(45,108,117)\qquad F(2,9)=(63,216,225)\quad$
$F(5,3)=(135,72,153)\qquad F(5,6)=(189,180,261)\qquad F(5,9)=(243,324,405)\quad$
If you divide the $B$ term by $3$ in any of these triples, it will meet your criteria. Just remember let $$n\in\{2,5,8,11,...\}\quad\land\quad k\in\{3,6,9,12,...\}$$
and you will find an infinite number of these triples for any given $n$ or $k$.
