How many positive integer solutions are there to the equation $^2 + 2^2 = 4^2$? 
How many positive integer solutions are there to the equation $^2 + 2^2 = 4^2$? 

I realised that this looks a lot like the Pythagorean theorem -- it could be written as $^2 + (\sqrt{2}y)^2 = (2z)^2$ as well. Then wouldn't there be an infinite number of solutions that are positive integers? For some reason they don't have the answer to this one, so I just wanted to check that I got this right. 

Observe $x^2$ divisible by $2$, thus $x$ is divisible by $2$. Write 
$$ \begin{align*} x = 2n &\implies 4n^2 + 2y^2 = 4z^2 \\
&\implies y^2+2n^2=2z^2 \\
&\implies y = 2m \\ 
&\implies 4m^2+2n^2=2z^2 \\
&\implies n^2+2m^2=z^2 \\
&\implies n = k(a^2-2b^2), \end{align*}$$
where $m = 2kab, z = k(a^2+2b^2), a,b,k \in \mathbb{Z}^{+}$ (see a solution of this equation by member Ivan Loh also at MSE in 2013 post). So $x = 2n = 2k(a^2-2b^2), y = 2m = 4kab, z = k(a^2+2b^2)$. I hope this helps.
