Tutescu,

Solve the system of equations:

$$
\begin{aligned}
& \left(x_{3}+x_{4}+x_{5}\right)^{5}=3 x_{1}, \\
& \left(x_{4}+x_{5}+x_{1}\right)^{5}=3 x_{2}, \\
& \left(x_{5}+x_{1}+x_{2}\right)^{5}=3 x_{3}, \\
& \left(x_{1}+x_{2}+x_{3}\right)^{5}=3 x_{4}, \\
& \left(x_{2}+x_{3}+x_{4}\right)^{5}=3 x_{5} .
\end{aligned}
$$

#
The system is symmetric with respect to cyclic permutations, so we can assume that $x_{1}$ is the largest of the five numbers. Then (since the function $f(x)=x^{5}$ is increasing), $3 x_{2}=\left(x_{4}+x_{5}+x_{1}\right)^{5} \geq\left(x_{3}+x_{4}+x_{5}\right)^{5}=3 x_{1} \geq 3 x_{2}$, which means $x_{2}=x_{1}$. Similarly, it can be shown that all numbers are equal. It remains to solve the equation $(3 x)^{5}=3 x$.

## Answer

Three solutions: $(0,0,0,0,0),( \pm 1 / 3, \pm 1 / 3, \pm 1 / 3, \pm 1 / 3, \pm 1 / 3)$.