3-ча 1. Solve the system of fifteen equations with fifteen unknowns:

\[
\left\{\begin{aligned}
1-x_{1} x_{2} & =0 \\
1-x_{2} x_{3} & =0 \\
1-x_{3} x_{4} & =0 \\
\cdots & \cdots \\
1-x_{14} x_{15} & =0 \\
1-x_{15} x_{1} & =0
\end{aligned}\right.
\]
Solve 1. Answer: \(x_{1}=x_{2}=\cdots=x_{15}= \pm 1\). Clearly, \(x_{2} \neq 0\), 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. Moreover, from the first and last equations we get \(x_{15}=x_{2}\). In the end, we get \(x_{1}=x_{3}=\cdots=x_{15}=x_{2}=x_{4}=\cdots=x_{14}\). Therefore, from the first equation we get \(x_{1}^{2}=1\), i.e., \(x_{1}= \pm 1\). Obviously, both sets of unknowns specified in the answer are indeed solutions to the system.