Solve the following system of equations:

$$
\begin{aligned}
& x_{1}+x_{1} x_{2}=1 \\
& x_{2}+x_{2} x_{3}=1 \\
& \vdots \\
& x_{99}+x_{99} x_{100}=1 \\
& x_{100}+x_{100} x_{1}=1
\end{aligned}
$$
Let $f(x)=\frac{1-x}{x}$; with this, our equations can be written as:

$$
x_{2}=f\left(x_{1}\right), \quad x_{3}=f\left(x_{2}\right), \ldots, x_{100}=f\left(x_{99}\right), \quad x_{1}=f\left(x_{100}\right)
$$

For any $i$, this implies

$$
x_{i}=f\left(x_{i-1}\right)=f(f\left(x_{i-2}\right)=\ldots=\underset{\underbrace{1}}{\qquad}\left(f\left(f\left(x_{i}\right)\right)\right) \ldots)
$$

For every positive integer $k$, let us introduce the notation

$$
f_{k}(x)=\underset{\underbrace{1}}{\underset{2}{2}}(f(\ldots . \underbrace{k}_{k}(x)) \ldots))
$$

Thus,

$$
x_{i}=f_{100}\left(x_{i}\right) \quad(i=1,2, \ldots, 100)
$$

By induction on $k$, we will show that

$$
f_{k}(x)=\frac{a_{k} x+b_{k}}{c_{k} x+d_{k}}
$$

where $a_{k}, b_{k}, c_{k}, d_{k}$ are suitable (depending only on $k$) real numbers. If $k=1$, then clearly $f_{1}(x)=f(x)=\frac{1-x}{x}$. Suppose that (3) holds for some $n$; then

$$
\begin{aligned}
& f_{n+1}(x)=f\left(f_{n}(x)\right)=f\left(\frac{a_{n} x+b_{n}}{c_{n} x+d_{n}}\right)= \\
= & \frac{1-\frac{a_{n} x+b_{n}}{c_{n} x+d_{n}}}{\frac{a_{n} x+b_{n}}{c_{n} x+d_{n}}}=\frac{\left(a_{n}-c_{n}\right) x+\left(b_{n}-d_{n}\right)}{-a_{n} x-b_{n}}
\end{aligned}
$$

Thus, (3) also holds for $n+1$, so it is true for all $k$; in particular,

$f_{100}(x)=\frac{a x+b}{c x+d}$

(The domain of $f_{100}(x)$ is, of course, narrower than that of $\frac{a x+b}{c x+d}$, but this has no significant importance here.) According to (2), each $x_{1}$ is a root of the equation $x=f_{100}(x)$, which, after multiplying by $(c x+d)$, leads to a quadratic equation at most. This equation is certainly not an identity, since for $x=1$, $f(x)<0$, so for all $x>1$, $f_{100}(x)$ is defined and negative. Therefore, the equation $x=f_{100}(x)$ has at most two roots, which we denote by $\alpha$ and $\beta$; the unknown values of $x_{i}$ can only be chosen from these $\alpha$ and $\beta$.

Suppose that for some $k$, $x_{k+1}=x_{k-1}=\alpha \neq \beta=x_{k}$; then

$$
\beta=x_{k}=\frac{1-x_{k-1}}{x_{k-1}}=\frac{1-\alpha}{\alpha}
$$

and

$$
\alpha=x_{k+1}=\frac{1-x_{k}}{x_{k}}=\frac{1-\beta}{\beta}
$$

implying $1-\alpha=\alpha \beta=1-\beta$, hence $\alpha=\beta$, which is a contradiction. Therefore, for some $k$, $x_{k}=x_{k+1}$, so $x$ satisfies the equation $x=f(x)$. Thus, by (1), $x_{1}=x_{2}=\ldots=x_{99}=x_{100}$, and this common value is the solution of the equation $x=\frac{1-x}{x}$, which is $\frac{-1 \pm \sqrt{5}}{2}$.

Remark. In the form of the rational function $f_{k}(x)$, after some experimentation, we can conjecture that

$$
f_{k}(x)=\frac{F_{k+2} x+F_{k+1}}{F_{k+1} x-F_{k}}
$$

where $F_{n}$ denotes the $n$-th Fibonacci number, i.e., $F_{1}=0, F_{2}=1, \ldots, F_{n+2}=F_{n+1}+F_{n}$. (Our conjecture can be easily proven by induction based on (*).) Using this, we can directly show that the equation $x=f_{100}(x)$ leads to the equation $x^{2}+x-1=0$.