<s>[INST] Equation in integers $x y^2 + 2 y z^2 + z x^2 = 0$ The equation in integers
$$x y^2 + 2 y z^2 + z x^2 = 0$$
has some integral solution $[1\colon 0\colon 0]$, $[0\colon 1\colon 0]$, $[0\colon 0\colon 1]$, $[1\colon -1 \colon 1]$. I wonder if there are other  solutions ( non-proportional to the one listed).
Notes: If we expand $(x + y \sqrt[3]{2} + z \sqrt[3]{4})^3$ the coefficient of $\sqrt[3]{4}$ is $3(x y^2 + 2 y z^2 + z x^2)$. I was looking for an $x + y \sqrt[3]{2}+ z\sqrt[3]{4}$ such that
$$(x + y \sqrt[3]{2}+ z\sqrt[3]{4})^3= a+b \sqrt[3]{2}$$
We have the example
$$(1 - \sqrt[3]{2}+\sqrt[3]{4})^3= 9(-1+\sqrt[3]{2})$$
and was wondering if we could get other examples.
What I've tried: the curve above is a smooth cubic, so an elliptic curve. However, no extra rational points using secants and tangents can be gotten out of the given points.
It is easy to show that if $p\mid y$ then $p$ also divides one of the $x$, $z$, but chasing prime divisors does not seem to get a contradiction that quickly.
Thank you for your attention!
 [/INST] The case where $xyz=0$ is easy, so I will skip it. Assume then that $xyz\neq 0$.
Your equation is then $(\frac{x}{z})(\frac{y}{z})^2+2(\frac{y}{z})+(\frac{x}{z})^2=0,$ that is $(\frac{xy}{z^2})^2+2(\frac{xy}{z})+(\frac{x}{z})^3=0$.
Set $u=-\frac{x}{z}$ and $v=\frac{xy}{z}+1$. Then we get $v^2-1-u^3=0$.
According to LMFDB database , the elliptic curve $v^2=u^3+1$ has rank $0$, and the torsion points are: $(-1,0),(0,-1),(0,1),(2,-3),(2,3)$. This should be enough to solve your question.
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