3.5. $\left\{\begin{array}{l}1-x_{1} x_{2}=0, \\ 1-x_{2} x_{3}=0, \\ 1-x_{3} x_{4}=0, \\ \ldots \ldots \ldots \ldots \ldots \ldots \\ 1-x_{n-1} x_{n}=0, \\ 1-x_{n} x_{1}=0 .\end{array}\right.$
3.5. Answer: $x_{1}=x_{2}=\ldots=x_{n}= \pm 1$ for odd $n$, $x_{1}=x_{3}=\ldots$ $\ldots=x_{n-1}=a$ and $x_{2}=x_{4}=\ldots=x_{n}=1 / a(a \neq 0)$ for even $n$.

Let $n$ be odd. It is clear that $x_{2} \neq 0$, so from the first and second equations we get $x_{1}=x_{3}$. From the second and third equations we get $x_{2}=x_{4}$, and so on. In addition, from the first and last equations we get $x_{n}=x_{2}$. In the end, we get $x_{1}=x_{3}=\ldots=x_{n}=x_{2}=$ $=x_{4}=\ldots=x_{n-1}$. Therefore, from the first equation we get $x_{1}^{2}=1$, i.e., $x_{1}= \pm 1$. It is obvious that both sets of values of the unknowns specified in the answer are indeed solutions to the system.

For even $n$, we similarly get $x_{1}=x_{3}=\ldots=x_{n-1}, x_{2}=$ $=x_{4}=\ldots=x_{n-2}$ and $x_{2}=x_{n}$. It is obvious that such sets of numbers are solutions to the given system of equations.