25. (VIE 1) Solve the following system of equations, in which $a$ is a given number satisfying $|a|>1$ :
$$
\begin{aligned}
x_{1}^{2}= & a x_{2}+1, \\
x_{2}^{2}= & a x_{3}+1, \\
\cdots & \cdots \\
x_{999}^{2}= & a x_{1000}+1, \\
x_{1000}^{2}= & a x_{1}+1 .
\end{aligned}
$$
25. We need only consider the case $a>1$ (since the case $a1$ by taking $a^{\prime}=-a, x_{i}^{\prime}=-x_{i}$ ). Since the left sides of the equations are nonnegative, we have $x_{i} \geq-\frac{1}{a}>-1, i=1, \ldots, 1000$. Suppose w.lo.g. that $x_{1}=\max \left\{x_{i}\right\}$. In particular, $x_{1} \geq x_{2}, x_{3}$. If $x_{1} \geq 0$, then we deduce that $x_{1000}^{2} \geq 1 \Rightarrow x_{1000} \geq 1$; further, from this we deduce that $x_{900}>1$ etc., so either $x_{i}>1$ for all $i$ or $x_{i}1$ for every $i$. Then $x_{1} \geq x_{2}$ implies $x_{1}^{2} \geq x_{2}^{2}$, so $x_{2} \geq x_{3}$. Thus $x_{1} \geq x_{2} \geq \cdots \geq x_{1000} \geq x_{1}$, and consequently $x_{1}=\cdots=x_{1000}$. In this case the only solution is $x_{i}=\frac{1}{2}\left(a+\sqrt{a^{2}+4}\right)$ for all $i$.
(ii) $x_{i}\frac{2}{\sqrt{3}}$, this case yields no solutions, while if $a \leq \frac{2}{\sqrt{3}}$, we obtain $x_{1}=$ $\frac{1}{2}\left(-\alpha-\sqrt{4-3 \alpha^{2}}\right), x_{2}=\frac{1}{2}\left(-a+\sqrt{4-3 \alpha^{2}}\right)$, or vice versa.