Solve the following system of equations, in which $a$ is a given number satisfying $|a| > 1$:

$\begin{matrix} x_{1}^2 = ax_2 + 1 \\ x_{2}^2 = ax_3 + 1 \\ \ldots \\ x_{999}^2 = ax_{1000} + 1 \\ x_{1000}^2 = ax_1 + 1 \\ \end{matrix}$
To solve the given system of equations, we start by analyzing the equations:

\[
\begin{cases}
x_1^2 = ax_2 + 1 \\
x_2^2 = ax_3 + 1 \\
\vdots \\
x_{999}^2 = ax_{1000} + 1 \\
x_{1000}^2 = ax_1 + 1
\end{cases}
\]

Given that \( |a| > 1 \), we can assume without loss of generality that \( a > 1 \) by transforming \( a \rightarrow -a \) and \( x_i \rightarrow -x_i \).

### Step 1: Analyzing the Range of \( x_i \)
We need to determine the possible values of \( x_i \). Let's consider the equation \( x_i^2 = ax_{i+1} + 1 \). Since \( a > 1 \), we can infer that:

\[
x_i^2 - 1 = ax_{i+1}
\]

This implies that \( x_i^2 - 1 \) must be positive if \( x_{i+1} \) is positive, and negative if \( x_{i+1} \) is negative. Therefore, we have two cases to consider:

1. \( x_i \geq 1 \) for all \( i \)
2. \( 0 > x_i > -1 \) for all \( i \)

### Step 2: Case 1 - \( x_i \geq 1 \)
If \( x_i \geq 1 \) for all \( i \), then we can assume that all \( x_i \) are equal. Let \( x_i = x \) for all \( i \). Then the system reduces to:

\[
x^2 = ax + 1
\]

Solving this quadratic equation:

\[
x^2 - ax - 1 = 0
\]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[
x = \frac{a \pm \sqrt{a^2 + 4}}{2}
\]

Since \( x \geq 1 \), we take the positive root:

\[
x = \frac{a + \sqrt{a^2 + 4}}{2}
\]

### Step 3: Case 2 - \( 0 > x_i > -1 \)
If \( 0 > x_i > -1 \) for all \( i \), we can similarly assume that all \( x_i \) are equal. Let \( x_i = x \) for all \( i \). Then the system reduces to:

\[
x^2 = ax + 1
\]

Solving this quadratic equation:

\[
x^2 - ax - 1 = 0
\]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[
x = \frac{a \pm \sqrt{a^2 + 4}}{2}
\]

Since \( 0 > x > -1 \), we take the negative root:

\[
x = \frac{a - \sqrt{a^2 + 4}}{2}
\]

### Step 4: Verifying the Solutions
We need to verify that these solutions satisfy the original system. For \( x = \frac{a + \sqrt{a^2 + 4}}{2} \), we have:

\[
x^2 = \left( \frac{a + \sqrt{a^2 + 4}}{2} \right)^2 = \frac{a^2 + 2a\sqrt{a^2 + 4} + (a^2 + 4)}{4} = \frac{2a^2 + 4 + 2a\sqrt{a^2 + 4}}{4} = \frac{a^2 + 4}{2} + \frac{a\sqrt{a^2 + 4}}{2}
\]

\[
ax + 1 = a \left( \frac{a + \sqrt{a^2 + 4}}{2} \right) + 1 = \frac{a^2 + a\sqrt{a^2 + 4}}{2} + 1
\]

Both expressions are equal, confirming the solution.

For \( x = \frac{a - \sqrt{a^2 + 4}}{2} \), we have:

\[
x^2 = \left( \frac{a - \sqrt{a^2 + 4}}{2} \right)^2 = \frac{a^2 - 2a\sqrt{a^2 + 4} + (a^2 + 4)}{4} = \frac{2a^2 + 4 - 2a\sqrt{a^2 + 4}}{4} = \frac{a^2 + 4}{2} - \frac{a\sqrt{a^2 + 4}}{2}
\]

\[
ax + 1 = a \left( \frac{a - \sqrt{a^2 + 4}}{2} \right) + 1 = \frac{a^2 - a\sqrt{a^2 + 4}}{2} + 1
\]

Both expressions are equal, confirming the solution.

The final answer is \( \boxed{ x_i = \frac{a + \sqrt{a^2 + 4}}{2} } \) or \( x_i = \frac{a - \sqrt{a^2 + 4}}{2} \).