(given to Ambroise Marigot). Solve $5 x^{3}+11 y^{3}+13 z^{3}=0$ in relative integers.
. We consider this equation modulo 13. An exhaustive study shows that a cube modulo 13 can only take the following values: $0,1,5,8$ and 12. Thus the quantity $5 x^{3}$ can only take the values $0,1,5,8$ and 12, while $11 y^{3}$ can only take the values $0,2,3,10$ and 11. The term $13 z^{2}$, on the other hand, obviously vanishes modulo 13 regardless of the value of $z$. From the above, we deduce that the only way to have a zero sum is $5 x^{3} \equiv 11 y^{3}(\bmod 1) 3$. It follows, since 13 is prime, that $x$ and $y$ are both multiples of 13. By writing $x=13 x^{\prime}$ and $y=13 y^{\prime}$ and substituting into the original equation, we realize that $z$ is also necessarily a multiple of 13, i.e., can be written as $z=13 z^{\prime}$. However, the triplet $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ is again a solution to the same equation, and an application of the principle of infinite descent ensures that the only solution is the triplet $(0,0,0)$.