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 $${\lambda }_1=\frac{7+\sqrt{45}}{2}={\left(\frac{3+\sqrt{5}}{2}\right)}^2,{\lambda }_2=\frac{7-\sqrt{45}}{2}={\left(\frac{3-\sqrt{5}}{2}\right)}^2,$$ ${\lambda }_1$ 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, $$\begin{array}{rl}{a}_n+a_{n+1}+2& ={\lambda }_1^{n-1}{C}_1({\lambda }_1+{\lambda }_1^2)+{\lambda }_2^{n-1}{C}_2({\lambda }_2+{\lambda }_2^2)+2\\ & ={\lambda }_1^{n-1}+{\lambda }_2^{n-1}+2\\ & ={\left({\left(\frac{3+\sqrt{5}}{2}\right)}^{n-1}\right)}^2+{\left({\left(\frac{3-\sqrt{5}}{2}\right)}^{n-1}\right)}^2+2\\ & ={\left[{\left(\frac{3+\sqrt{5}}{2}\right)}^{n-1}+{\left(\frac{3-\sqrt{5}}{2}\right)}^{n-1}\right]}^2=x_n^2,\end{array}$$ 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 $x_1=2,x_2=3,x_n=3x_{n-1}-x_{n-2}$, $n\ge 3$. $\square$