Solve the equation $5 x^{3}+11 y^{3}+13 z^{3}=0$ in integers.

#
As in problem 31291, the equation is homogeneous, so it is sufficient to find non-zero solutions in which the variables are coprime.

Let such a solution exist. From the condition, it follows that $5 x^{3} \equiv 2 y^{3}(\bmod 13)$. It is easy to verify that a cube, when divided by 13, can give a remainder of $0, \pm 1$ or $\pm 5$. From this, it follows that $x$ and $y$ are divisible by 13. But then $z$ is also divisible by 13.

Contradiction.

## Answer

$(0,0,0)$