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

I realised that this looks a lot like the Pythagorean theorem -- it could be written as $𝑥^2 + (\sqrt{2}y)^2 = (2z)^2$ 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.

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    $\begingroup$ Your rewrite isn't so helpful because is doesn't lead to factorization and introduces an irrational. $\endgroup$ – Yves Daoust Sep 4 '17 at 17:22
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$ 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.

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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.

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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 $4$.

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  • $\begingroup$ @samjoe: use my hint :-) $\endgroup$ – Yves Daoust Sep 4 '17 at 17:28
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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.

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