<s>[INST] All solutions $(x, y, z) \in (\mathbb{N},\mathbb{N},\mathbb{N})$ for $3x^2 + 6y^2 = z^2$ 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.
 [/INST] There is a systematic way to solve such equation:  the idea is to parametrize the ellipse $C: 3X^2+6Y^2=1$ by a family of lines through a fixed rational point. (in other words, an ellipse is a rational curve).
Here $A=({1\over 3}, {1\over 3})$ is a rational point on this ellipse.  Note that a line through $A$ has equation $L_t:(X-1/3)=t(Y-1/3)$ must cut the ellipse in exactly one otherpoint which  is a rational fonction of $t.
It is preferable to write $C= (3.(X-1/3)^2+6.(Y-1/3)^2+6(X-1/3)+12(Y-1/3)=0$
Let $U=X-1/3, V=Y-1/3$
The intersection of $L_t\cap C$ is $U=tV, 3U^2+6V^2+6U+12V=0$
Or $3t^2V^2+6V^2+6tV+12V=0$
If we exclude the point A $(V=0)$ we get; $V(3t^2+6)=-6T-12$, $V=-{t+4\over t^2+3}, U=t V$
Now $t={a\over b}\in \bf Q$, ${x\over z}=U+1/3=-t{t+4\over t^2+3}+1/3$ and ${y\over z}=V+1/3=-{t+4\over t^2+3}$
${x\over z}=-{a^2+4ab\over a^2+3b^2}+1/3={-2a^2+3b^2-12ab\over a^2+3b^2}$
${y\over z}=-{a/b+4\over (a/b)^2+3}+1/3=$
$-{ab+4b^2\over a^2+3b^2}+1/3=$ $a^2-3ab-9b^2\over 3(a^2+b^2)$
So $x=-2a^2+3b^2-12ab$, $y=a^2-3ab-9b^2$, $z= 3(a^2+b^2)$ is a solution if $(a,b,c)\in \bf Z$.
Conversely, for every solution, the point  $ P=(x/z, y/z)$ is on the ellipse, so $(x,y,z)$ is proportional to the triple $(-2a^2+3b^2-12ab , a^2-3ab-9b^2,3(a^2+b^2))$, for $(a,b)$ the slope of the line $(A,P)$.
Note that a different choice for $A$   yields a different parametrization of the set of solution...
</s>