Tutescu L.

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

Author: Galnier [.

At each vertex of a cube, there is a number +1 or -1. In the center of each face of the cube, a number is placed that is equal to the product of the numbers at the vertices of that face.

Can the sum of the resulting 14 numbers be equal to 0?

## Solution

The product of all 14 numbers is equal to the fourth power of the product of the numbers at the vertices and is therefore positive. Consequently, there is an even number of minus ones among the 14 numbers, and their sum is 0 only when there are exactly 7 minus ones.

## Answer

It cannot.