China Southeastern Mathematical Olympiad

It is well known that the solution of the sequence can be obtained by solving two geometric sequences. We get $a_n = C_1\lambda_1^n + C_2\lambda_2^n$, where
$$\lambda_1 = \frac{7 + \sqrt{45}}{2} = \left( \frac{3 + \sqrt{5}}{2} \right)^2, \quad \lambda_2 = \frac{7 - \sqrt{45}}{2} = \left( \frac{3 - \sqrt{5}}{2} \right)^2,$$
$\lambda_1$ and $\lambda_2$ are solutions of the equation $\lambda^2 - 7\lambda + 1 = 0$, $a_1 = 1 = C_1\lambda_1 + C_2\lambda_2$, $a_2 = 1 = C_1\lambda_1^2 + C_2\lambda_2^2$.
Therefore,
$$\begin{aligned} a_n + a_{n+1} + 2 &= \lambda_1^{n-1}C_1(\lambda_1 + \lambda_1^2) + \lambda_2^{n-1}C_2(\lambda_2 + \lambda_2^2) + 2 \\ &= \lambda_1^{n-1} + \lambda_2^{n-1} + 2 \\ &= \left( \left( \frac{3+\sqrt{5}}{2} \right)^{n-1} \right)^2 + \left( \left( \frac{3-\sqrt{5}}{2} \right)^{n-1} \right)^2 + 2 \\ &= \left[ \left( \frac{3+\sqrt{5}}{2} \right)^{n-1} + \left( \frac{3-\sqrt{5}}{2} \right)^{n-1} \right]^2 = x_n^2, \end{aligned}$$
where $x_n = \left(\frac{3+\sqrt{5}}{2}\right)^{n-1} + \left(\frac{3-\sqrt{5}}{2}\right)^{n-1}$ is the solution of the sequence $\{x_n\}$ of positive integers $x_1 = 2, x_2 = 3, x_n = 3x_{n-1} - x_{n-2}$, $n \ge 3$. $\square$