Square Number



Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly.
Parse $2 \pm \sqrt{3}=\left(\frac{\sqrt{6} \pm \sqrt{2}}{2}\right)^{2}$, so
$$
\begin{aligned}
& \frac{1}{6}\left((2+\sqrt{3})^{2 n-3}+(2-\sqrt{3})^{2 n-3}+2\right) \\
- & \frac{1}{6}\left(\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)^{2 n-3}+\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)^{2 n-3}\right)^{2}
\end{aligned}
$$

Let $\alpha=\frac{\sqrt{6}+\sqrt{2}}{2}$, then $\alpha^{-1}=\frac{\sqrt{6}-\sqrt{2}}{2}$. Suppose $a_{n}=\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)^{2 n-3}=\frac{c_{n} \sqrt{6}+d_{n} \sqrt{2}}{2}$, where $c_{n}$ and $d_{n}$ are rational numbers, then $a_{1}-\frac{\sqrt{6}-\sqrt{2}}{2}$, i.e., $c_{1}=1, d_{1}=-1$ are integers. Assuming $c_{n}$ and $d_{n}$ are integers, then
$$
\begin{aligned}
a_{n+1} & =a_{n} \cdot\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)^{2}=\frac{c_{n} \sqrt{6}+d_{n} \sqrt{2}}{2}(2+\sqrt{3}) \\
& =\frac{\left(2 c_{n}+d_{n}\right) \sqrt{6}+\left(2 d_{n}+3 c_{n}\right) \sqrt{2}}{2},
\end{aligned}
$$

i.e., $c_{n+1}=2 c_{n}+d_{n}, d_{n+1}=2 d_{n}+3 c_{n}$ are also integers.
Therefore, $\frac{1}{6}\left(a_{n}+a_{n}^{-1}\right)^{2}=\frac{1}{6}\left(c_{n} \sqrt{6}\right)^{2}=c_{n}^{2}$ is a perfect square.
Note: $c_{1}=c_{2}=1, c_{n+1}=4 c_{n}-c_{n-1}$. The sequence in this problem is the $\left\{x_{n}\right\}$ from the previous problem.