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 \(a < -1\) can be reduced to \(a > 1\) by taking \(a' = -a, x_i' = -x_i\)). Since the left sides of the equations are nonnegative, we have \(x_i \geq -\frac{1}{a} > -1, i = 1, \ldots, 1000\). Suppose w.l.o.g. that \(x_1 = \max \{x_i\}\). 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 \(x_{999} > 1\) 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) \(x_i < 0\). If \(a > \frac{2}{\sqrt{3}}\), this case yields no solutions, while if \(a \leq \frac{2}{\sqrt{3}}\), we obtain \(x_1 = \frac{1}{2}\left(-a - \sqrt{4 - 3a^2}\right), x_2 = \frac{1}{2}\left(-a + \sqrt{4 - 3a^2}\right)\), or vice versa.