25. (Vietnam) Let $a$ be a known number, $|a|>1$. Solve the system of equations
$$
\left\{\begin{array}{l}
x_{1}^{2}=a x_{2}+1, \\
x_{2}^{2}=a x_{3}+1, \\
\cdots \cdots \\
x_{090}^{2}=a x_{1000}+1, \\
x_{1000}^{2}=a x_{1}+1 .
\end{array}\right.
$$
Solve for the case when $a1$, so we only need to solve for the case $a>1$.

Since the left side of the equation is non-negative, we have $x_{i} \geqslant-\frac{1}{a}>-1$, $i=1,2, \cdots, 1000$. Also, because the system of equations is cyclically symmetric, we can assume
$$
x_{1}=\max \left\{x_{1}, x_{2}, \cdots, x_{1000}\right\} \text {. }
$$

If $x_{1} \geqslant 0$, then from the last equation in the system, we derive $x_{1000}^{2} \geqslant 1$, i.e., $\left|x_{1000}\right| \geqslant 1$. But since $x_{1000}>-1$, it must be that $x_{1000} \geqslant 1$. This similarly leads to $x_{900} \geqslant 1$, and so on. Therefore, the solutions to the system of equations have only two possible scenarios: $x_{1}, x_{2}, \cdots, x_{1000}$ are all not less than 1 or all less than 0.
(i) Suppose $x_{i} \geqslant 1, i=1,2, \cdots, 1000$. Condition (1) implies
$$
x_{1} \geqslant x_{2} \Rightarrow x_{1}^{2} \geqslant x_{2}^{2} \Rightarrow a x_{2}+1 \geqslant a x_{3}+1 \Rightarrow x_{2} \geqslant x_{3} \text {. }
$$

Similarly, we have
$$
x_{1} \geqslant x_{2} \geqslant x_{3} \geqslant x_{4} \geqslant \cdots \geqslant x_{1001} \geqslant x_{1} .
$$

Thus, $x_{1}=x_{2}=\cdots=x_{1000}=t_{1}$, where $t_{1}$ is the root of the equation
$$
t^{2}-a t-1=0
$$

that is not less than 1, i.e., $t_{1}=\frac{1}{2}\left(a+\sqrt{a^{2}+4}\right)$.
(ii) Suppose $x_{i}<0, i=1,2, \cdots, 1000$. Condition (1) implies
$$
x_{1} \geqslant x_{2} \Rightarrow x_{1}^{2} \leqslant x_{2}^{2} \Rightarrow a x_{2}+1 \leqslant a x_{3}+1 \Rightarrow x_{2} \leqslant x_{3} \text {. }
$$

Similarly, we have
$$
x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant \cdots \leqslant x_{1001} \leqslant x_{1} .
$$

Thus, $x_{1}=x_{2}=\cdots=x_{1000}=t_{2}$, where $t_{2}$ is the root of the equation
$$
t^{2}-a t-1=0
$$

that is less than 0, i.e., $t_{2}=\frac{1}{2}\left(a-\sqrt{a^{2}+4}\right)$.
(iii) When $a > \frac{2}{\sqrt{3}}$, equation (3) has no solution; when $1 < |a| \leq \frac{2}{\sqrt{3}}$, the system of equations has two additional solutions:
$$
\begin{array}{c}
\left\{\begin{array}{l}
x_{1}=x_{3}=\cdots=x_{999}=\frac{1}{2}\left(-a+\sqrt{4-3 a^{2}}\right), \\
x_{2}=x_{4}=\cdots=x_{1000}=\frac{1}{2}\left(-a-\sqrt{4-3 a^{2}}\right) ;
\end{array}\right. \\
\left\{\begin{array}{l}
x_{1}=x_{3}=\cdots=x_{999}=\frac{1}{2}\left(-a-\sqrt{4-3 a^{2}}\right), \\
x_{2}=x_{4}=\cdots=x_{1000}=\frac{1}{2}\left(-a+\sqrt{4-3 a^{2}}\right) .
\end{array}\right.
\end{array}
$$

When $|a|=\frac{2}{\sqrt{3}}$, the last two solutions are the same.