All solutions $(x, y, z) \in (\mathbb{N},\mathbb{N},\mathbb{N})$ for $3x^2 + 6y^2 = z^2$ https://ift.tt/eA8V8J

My attempt:

$z^2 = 3(x^2 + 2y^2)$

so $3|z^2$ and thus $3|z$

letting $z = 3k$, we reduce the equation to $x^2+2y^2=3k^2$

Obviously $(n,n,n) $ represents infinitely many solutions, but not all since $(5,1,3)$ is a solution for example.

Also, if $x$ and $k$ are even then $4|2y^2$, then $y$ is even

If $x$ and $k$ are odd, then $x^2$ and $k^2$ are congruent to 1 mod 8 so $2y^2$ is congruent to 2 mod 8 so y is also odd.Therefore, it's sufficient to find the odd solutions.

But I don't know how to proceed from here. Any hints on how to parametrize the odd solutions or a suitable mod $n$ to consider to simplify it further would be appreciated.

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