4・141 Solve the system of equations
$$\left\{\begin{array}{l}
x_{5} + x_{2} = y x_{1}, \\
x_{1} + x_{3} = y x_{2}, \\
x_{2} + x_{4} = y x_{3}, \\
x_{3} + x_{5} = y x_{4}, \\
x_{4} + x_{1} = y x_{5}.
\end{array}\right.$$

where $y$ is a parameter.
[Solution]
$$\left\{\begin{array}{l}
x_{5}+x_{2}=y x_{1} \\
x_{1}+x_{3}=y x_{2} \\
x_{2}+x_{4}=y x_{3} \\
x_{3}+x_{5}=y x_{4} \\
x_{4}+x_{1}=y x_{5}
\end{array}\right.$$

Obviously, for any value of $y$,
$$x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$$

is a solution to the system of equations.
Next, we seek the "non-trivial solutions" (i.e., at least one $x_{i} \neq 0, i=1,2,3,4,5$).

From (1) and (5), we get
$$\begin{array}{l}
x_{2}=y x_{1}-x_{5} \\
x_{4}=y x_{5}-x_{1}
\end{array}$$

From (2) and (4), we get
$$\begin{array}{l}
x_{3}=y x_{2}-x_{1} \\
x_{3}=y x_{4}-x_{5}
\end{array}$$

Substituting (6) and (7) into (8) and (9) respectively, we get
$$\begin{array}{l}
x_{3}=\left(y^{2}-1\right) x_{1}-y x_{5} \\
x_{3}=\left(y^{2}-1\right) x_{5}-y x_{1}
\end{array}$$

From (10) and (11), we get
$$\left(y^{2}-1\right) x_{1}-y x_{5}=\left(y^{2}-1\right) x_{5}-y x_{1}$$

which simplifies to $\left(y^{2}+y-1\right)\left(x_{1}-x_{5}\right)=0$.
When $y^{2}+y-1 \neq 0$, it must be that $x_{1}=x_{5}$.
Similarly, we can get $x_{1}=x_{2}, x_{2}=x_{3}, x_{3}=x_{4}$, and so on. Thus, when $y^{2}+y-1 \neq 0$, we have $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$.

Substituting this result into any of equations (1) or (5), we get $y=2$. Clearly, $y^{2}+y-1 \neq 0$ in this case.

Thus, when $y=2$, $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$ is a non-trivial solution to the system of equations. When $y^{2}+y-1 \neq 0$ and $y \neq 2$, the system has only the trivial solution.
When $y^{2}+y-1=0$, from $y^{2}-1=-y$, equation (10) gives
$$x_{3}=-y\left(x_{1}+x_{5}\right)$$

Also, from equations (6) and (7), we get $x_{2}=y x_{1}-x_{5}, x_{4}=y x_{5}-x_{1}$. Combining these, the solutions to the original system of equations are:
(i) When $y^{2}+y-1 \neq 0$ and $y \neq 2$, the system has only the trivial solution
$$x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$$
(ii) When $y=2$, the system has the solution $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=t$, where $t$ is any real number.
(iii) When $y^{2}+y-1=0$, the solutions to the system are
$$\left\{\begin{array}{l}
x_{1}=u \\
x_{5}=v \\
x_{2}=y u-v \\
x_{3}=-y(u+v) \\
x_{4}=y v-u
\end{array}\right.$$

where $u, v$ are any numbers, and $y=\frac{-1 \pm \sqrt{5}}{2}$.