Example 5 If real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{1} x_{5}=-1, \\
x_{2} x_{1}+x_{2} x_{3}+x_{2} x_{4}+x_{2} x_{5}=-1, \\
x_{3} x_{1}+x_{3} x_{2}+x_{3} x_{4}+x_{3} x_{5}=-1, \\
x_{4} x_{1}+x_{4} x_{2}+x_{4} x_{3}+x_{4} x_{5}=-1, \\
x_{5} x_{1}+x_{5} x_{2}+x_{5} x_{3}+x_{5} x_{4}=-1,
\end{array}\right.
$$

Find all possible values of $x_{1}$.
(2009, Japan Mathematical Olympiad (Preliminary))
Let $S=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}$.
From the given, $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ all satisfy the equation $x^{2}-S x-1=0$.
Thus, $x_{i} \in\left\{\frac{S+\sqrt{S^{2}+4}}{2}, \frac{S-\sqrt{S^{2}+4}}{2}\right\}(i=1,2, \cdots, 5)$.
Therefore, $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}$
$$
=\frac{5 S+\varepsilon \sqrt{S^{2}+4}}{2}(\varepsilon \in\{ \pm 1, \pm 3, \pm 5\}) \text {. }
$$

Thus, $S=\frac{5 S+\varepsilon \sqrt{S^{2}+4}}{2}$, which means
$$
3 S=-\varepsilon \sqrt{S^{2}+4} .
$$

Further, from $|3 S|=1-\varepsilon \sqrt{S^{2}+4}|>| \varepsilon S \mid$, we know that $\varepsilon$ can only be $\pm 1$, hence $S= \pm \frac{\sqrt{2}}{2}$.
When $S=\frac{\sqrt{2}}{2}$,
$x_{1}^{2}-\frac{\sqrt{2}}{2} x_{1}-1=0 \Rightarrow x_{1}=\sqrt{2}$ or $-\frac{\sqrt{2}}{2}$.
Similarly, when $S=-\frac{\sqrt{2}}{2}$, $x_{1}=-\sqrt{2}$ or $\frac{\sqrt{2}}{2}$.
In summary, all possible values of $x_{1}$ are $\pm \sqrt{2}, \pm \frac{\sqrt{2}}{2}$.
[Note] The five equations in the condition have a clear cyclic characteristic, which can be considered as a whole, leading to the introduction of the auxiliary quantity $S=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}$, transforming the original five-variable equation system problem into considering "how many times $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ can take the two roots of $x^{2}-S x-1=0$".