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

Since the equation is homogeneous, we may WLOG assume $\gcd(x, y, z)=1$. (So that all solutions will be given by multiplying all the primitive solutions by any positive integer $k$) 
Now if $x$ is even, then $z$ is even, so $y$ is also even, a contradiction. Thus $x$ is odd, so $z$ is odd, and so $x^2 \equiv z^2 \equiv 1 \pmod{4}$. Thus $4 \mid 2y^2$, so $y$ is even. Note that if $p \mid x, z$ for some prime $p$, then $p$ is odd and $p \mid 2y^2$, so $p \mid y$, a contradiction, so $\gcd(x, z)=1$.
Let $y=2y'$, so that $2y'^2=\frac{z^2-x^2}{4}=(\frac{z-x}{2})(\frac{z+x}{2})$. Now $\gcd((\frac{z-x}{2}),(\frac{z+x}{2}))=\gcd(x, z)=1$, so we have 2 cases:
Case 1: $4 \mid z-x$. Then we have $\frac{z-x}{2}=2a^2, \frac{z+x}{2}=b^2, y'=ab$ for some $a, b \in \mathbb{Z}^+$, so $z=b^2+2a^2, x=b^2-2a^2, y=2ab, \, b>a\sqrt{2}>0$. Checking, these are indeed solutions.
Case 2: $4 \mid z+x$. Then we have $\frac{z-x}{2}=b^2, \frac{z+x}{2}=2a^2, y'=ab$ for some $a, b \in \mathbb{Z}^+$, so $z=b^2+2a^2, x=2a^2-b^2, y=2ab, \, a\sqrt{2}>b>0$. Checking, these are indeed solutions.
Therefore all primitive solutions are given by $(x, y, z)=(|b^2-2a^2|, 2ab, b^2+2a^2), a, b \in \mathbb{Z}^+$.
Therefore all positive integer solutions are given by $$(x, y, z)=(k|b^2-2a^2|, k(2ab), k(b^2+2a^2)), a, b, k \in \mathbb{Z}^+$$
