Find all solutions: $x^2 + 2y^2 = z^2$ I'm use to finding the solutions of linear Diophantine equations, but what are you suppose to do when you have quadratic terms? For example consider the following problem:

Find all solutions in positive integers to the following Diophantine equation
$x^2 + 2y^2 = z^2$

I'd usually start by finding the gcd and use some other tricks, but I'm not sure how to approach this type of problem

We can approach this in the same manner as Pythagorean Triples.
Let's only look for primitive solutions, $\gcd(x,y,z)=1$. Since
$$
z^2-x^2=2y^2
$$
$z$ and $x$ must have the same parity. That means both $z-x$ and $z+x$ are even so $y$ must be even. Therefore, for the triple to be primitive, $x$ and $z$ must be odd. Let $y=2ab$ where $(a,b)=1$ and $b$ is odd. Then, since one of $z+x$ or $z-x$ must be $2\bmod4$ and the other must be $0\bmod4$,
$$
\overbrace{2y^2}^{8a^2b^2}=\overbrace{(z+x)}^{4a^2}\overbrace{(z-x)}^{2b^2}\qquad\text{smaller factor is }2\bmod4
$$
or
$$
\overbrace{2y^2}^{8a^2b^2}=\overbrace{(z+x)}^{2b^2}\overbrace{(z-x)}^{4a^2}\qquad\text{larger factor is }2\bmod4
$$
Solving for $x$ and $z$ gives $x=2a^2-b^2$ if the smaller factor is $2\bmod4$, or $x=b^2-2a^2$ if the larger factor is $2\bmod4$. That is,
$$
x=\left|\,2a^2-b^2\right|,y=2ab,z=2a^2+b^2
$$
where $(a,b)=1$ and $b$ is odd.
For example,
$$
\begin{array}{c|cc}
a\backslash b\!\!&1&3&5&7\\\hline
1&(1,2,3)&(7,6,11)&(23,10,27)&(47,14,51)\\
2&(7,4,9)&(1,12,17)&(17,20,33)&(41,28,57)\\
3&(17,6,19)&\text{n/a}&(7,30,43)&(31,42,67)\\
4&(31,8,33)&(23,24,41)&(7,40,57)&(17,56,81)\\
5&(49,10,51)&(41,30,59)&\text{n/a}&(1,70,99)\\
6&(71,12,73)&\text{n/a}&(47,60,97)&(23,84,121)\\
7&(97,14,99)&(89,42,107)&(73,70,123)&\text{n/a}
\end{array}
$$
