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. 
 A: 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.
A: Hint:
Write
$$(2z-x)(2z+x)=2y^2$$ and discuss how this identity can be decomposed in integer factors.

Alternatively:
Rewrite the equation modulo $2$ and modulo $4$.
A: $ x^2+2y^2=4z^2 \implies (kx)^2 + 2(ky)^2 = 4(kz)^2 $, therefore if at least one solution exists, infinitely many solutions exist.
$ 2^2+2\cdot4^2=4\cdot3^2$, therefore at least one solution exists, therefore infinitely many solutions exist.
A: I f nothing else, you can take positive integers $v < u \sqrt 2$
with
$$  x = 4 u^2 - 2 v^2, \; \;  y = 4 u v, \; \; z = 2 u^2 + v^2  $$
I will check more carefully in a minute, but if $v$ is odd and $\gcd(u,v) = 1,$ it should be true that $\gcd(x,y,z) = 1$ as well.
