(given to Ambroise Marigot). Solve $x^{3}-3 y^{3}-9 z^{3}=0$ in relative integers.
. We reason by infinite descent. The initial equation implies that $x^{3}$ is a multiple of 3, and therefore the same is true for $x$. We write $x=3 x^{\prime}$ and the equation becomes $9 x^{\prime 3}-y^{3}-$ $3 z^{3}=0$. By repeating the same reasoning, we show that $y$ is a multiple of 3, and then by writing $y=3 y^{\prime}$, the equation transforms again to give $3 x^{\prime 3}-9 y^{\prime 3}-z^{3}=0$. Once again, it follows that $z=$ $3 z^{\prime}$ for some integer $z^{\prime}$, and the triplet $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ is then again a solution to the initial equation. By the principle of infinite descent, it follows that the only solution is $(0,0,0)$.