Prove that there are no integer solutions for $x^3+2y^3=4z^3$
My attempt:
Assume by contradiction that the equation $x^3+2y^3=4z^3$ has an integer solution. We'll choose the $(x,y,z)$ solution with the minimal z. We will also mark $d=\gcd(x,y)$. Therefore $d \mid x$ and $d \mid y$, so $d^3 \mid x^3$ and $d^3 \mid 2y^3$, and thus $d^3 \mid 4z^3=x^3+2y^3$.
I'm trying to find another triple that looks something like $(\frac{x}{d}, \frac{y}{d}, ...)$ that also satisfy the equation, but I don't know how to continue from here.
