Let $n$ be a given positive integer. Give all solutions of the following system of equations:

$$
x_{1}^{2 n}+\left(x_{1} x_{2}\right)^{2 n}+\ldots+\left(x_{1} x_{2} \ldots x_{n}\right)^{2 n}=1
$$

$$
x_{1}^{2 n+1}+\left(x_{1} x_{2}\right)^{2 n+1}+\ldots+\left(x_{1} x_{2} \ldots x_{n}\right)^{2 n+1}=1
$$
We denote the product $x_{1} x_{2} \ldots x_{k}$ by $s_{k}$ $(k=1,2, \ldots, n)$. Since in equation (1) all exponents are even, these powers are non-negative, and thus none of them can be greater than 1. Therefore,

$$
\left|s_{k}\right| \leqq 1, \quad k=1,2, \ldots, n
$$

On the interval $-1 \leqq s \leqq 1$, $s^{2 n} \geqq s^{2 n+1}$, since the difference between the two sides is $s^{2 n}(1-s) \geqq 0$. The difference can only be 0 when $s=0$ or $s=1$. Therefore, (1) and (2) can only be satisfied simultaneously if each of the products $s_{1}, s_{2} \ldots, s_{n}$ is either 0 or 1. However, due to (1), only one of the products can be equal to 1, and it must be the first one:

$$
s_{1}=1, s_{2}=\ldots=s_{n}=0
$$

Returning to the original variables, we find that the solution to the system of equations (1), (2) is:

$$
x_{1}=1, x_{2}=1, x_{3}, \ldots x_{n} \text { arbitrary. }
$$

(If $n=1$ or 2, then of course only $x_{1}=1$, or $x_{1}=1, x_{2}=0$ is the solution.)

Peter Fischer (Budapest, Istvan I. Gymnasium, 4th grade)