2. Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=1, b_{0}=0$, and $\left\{\begin{array}{l}a_{n+1}=7 a_{n}+6 b_{n}-3, \\ b_{n+1}=8 a_{n}+7 b_{n}-4,\end{array}\right.$
$n=0,1,2, \cdots$. Prove that $a_{n}(n=0,1,2, \cdots)$ is a perfect square.
Answer
$a_{n+1}=7 a_{n}+6 b_{n}-3 \Rightarrow b_{n}=\frac{a_{n+1}-7 a_{n}+3}{6}$, substitute into $b_{n+1}=8 a_{n}+7 b_{n}-4$
get $\frac{a_{n+2}-7 a_{n+1}+3}{6}=8 a_{n}+7 \cdot \frac{a_{n+1}-7 a_{n}+3}{6}-4$
$\Rightarrow a_{n+2}=14 a_{n+1}-a_{n}-6$, set $a_{n+2}+k=14\left(a_{n+1}+k\right)-\left(a_{n}+k\right) \Rightarrow 12 k=-6$
$\Rightarrow k=-\frac{1}{2}$, then let $c_{n}=a_{n}-\frac{1}{2} \Rightarrow c_{n+2}=14 c_{n+1}-c_{n}$.
The characteristic equation of the sequence $\left\{c_{n}\right\}$ is $x^{2}-14 x+1=0 \Rightarrow x_{1}=7-4 \sqrt{3}, x_{2}=7+4 \sqrt{3}$,
thus $c_{n}=\alpha \cdot(7-4 \sqrt{3})^{n}+\beta \cdot(7+4 \sqrt{3})^{n}(n \in \mathbf{N})$.
By $c_{0}=\frac{1}{2}, c_{1}=\frac{7}{2} \Rightarrow\left\{\begin{array}{l}\frac{1}{2}=\alpha+\beta, \\ \frac{7}{2}=\alpha \cdot(7-4 \sqrt{3})+\beta \cdot(7+4 \sqrt{3})\end{array} \Rightarrow \alpha=\beta=\frac{1}{4}\right.$,
thus $c_{n}=\frac{1}{4}(7-4 \sqrt{3})^{n}+\frac{1}{4}(7+4 \sqrt{3})^{n}=\frac{(2-\sqrt{3})^{2 n}}{4}+\frac{(2+\sqrt{3})^{2 n}}{4}$,
therefore $a_{n}=\frac{(2-\sqrt{3})^{2 n}}{4}+\frac{(2+\sqrt{3})^{2 n}}{4}+\frac{1}{2}=\left[\frac{(2-\sqrt{3})^{n}}{2}+\frac{(2+\sqrt{3})^{n}}{2}\right]^{2}$.
Let $x_{n}=\frac{(2-\sqrt{3})^{n}}{2}+\frac{(2+\sqrt{3})^{n}}{2}$, then $x_{0}=1, x_{1}=2, x_{n+2}=4 x_{n+1}-x_{n}$,
thus $x_{n}$ is a positive integer. Therefore, $a_{n}=x_{n}^{2}$ is a perfect square.