What are the $(x, y, z)$ integers such that $x^{3}+2 y^{3}=4 z^{3}$?
If we take a solution triplet, we realize that $x^{3}$ must be even, and therefore $x$ must be even. We set $x=2 x^{\prime}$, which gives $4 x^{\prime 3}+y^{3}=2 z^{3}$. We deduce that $y$ is even, and we show similarly that $z$ is even. We thus arrive at an equation $x^{\prime 3}+2 y^{\prime 3}=4 z^{\prime 3}$ as in the previous exercise, and we conclude in the same way by infinite descent: the only solution is $(0,0,0)$