\[
\left.\begin{array}{l}
\text { Systems of algebraic nonlinear equations } \\
\text { [Symmetric systems. Involutory transformations ]}
\end{array}\right]
\]

Solve the system of equations: \( x_{1} x_{2}=x_{2} x_{3}=\ldots=x_{n-1} x_{n}=x_{n} x_{1}=1 \).
Clearly, $x_{2} \neq 0$, so from the first equation we get $x_{1}=x_{3}$. Similarly, $x_{2}=x_{4}$, and so on.

Moreover, for odd $n$, $x_{n}=x_{2}$, meaning all unknowns are equal to each other. Hence, $x_{1}^{2}=1$, which means $x_{1}= \pm 1$.

For even $n$, $x_{1}=x_{3}=\ldots=x_{n-1}, x_{2}=x_{4}=\ldots=x_{n-2}=x_{n}$.

## Answer

$x_{1}=x_{2}=\ldots=x_{n}= \pm 1$ for odd $n$, and $x_{1}=x_{3}=\ldots=x_{n-1}=a$ and $x_{2}=x_{4}=\ldots=x_{n}=1 / a(a \neq 0)$ for even $n$.