设数列 $\{a_n\}$ 满足：$a_{1}=a_{2}=1, a_{n}=7 a_{n-1}-a_{n-2}, n \geqslant 3$，证明：对于每个 $n \in \mathbf{N}^{*}, a_{n}+a_{n+1}+2$ 皆为完全平方数．
【难度】
【出处】
2011中国东南数学奥林匹克试题
【标注】
【答案】
略
【解析】
易求得数列起初的一些项为：$1,1,6,41,281,1926,\cdots$
注意到，$a_1+a_2+2=2^2,a_2+a_3+2=3^2,a_3+a_4+2=7^2,a_4+a_5+2=18^2,\cdots,$，构作数列 $\{x_n\}:x_1=2,x_2=3,x_n=3x_{n-1}-x_{n-2},n\geqslant 3$，则对每个 $n\in N^{\ast},x_n$ 为正整数．
我们来证明：对于每个 $n\in N^{\ast}$，皆有：$a_n+a_{n+1}+2=x_n^2$．
引理：数列 $\{x_n\}$ 满足：对于每个 $k\in N^{\ast},x_kx_{k+2}-x_{k+1}^2=5$．
引理证明：令 $f(k)=x_kx_{k+2}-x_{k+1}^2$，则
$\begin{aligned}
&f(k)-f(k-1)\\
&=(x_kx_{k+2}-x_{k+1}^2)-(x_{k-1}x_{k+1}-x_k^2)\\
&=(x_kx_{k+2}+x_k^2)-(x_{k+1}^2+x_{k-1}x_{k+1})\\
&=x_k(x_{k+2}+x_k)-x_{k+1}(x_{k+1}+x_{k-1})\\
&=3x_kx_{k+1}-3x_{k+1}x_k=0
\end{aligned}$
所以 $f(k)=f(k-1)$，于是 $f(k)=f(k-1)=f(k-2)=\cdots=f(1)=x_1x_3-x_2^2-5$．
回到本题，对 $n$ 归纳，据数列 $\{a_n\}$ 的定义，
$\begin{aligned}
a_1+a_2+2=4=x_1^2\\
a_2+a_3+2=9=x_2^2\\
\end{aligned}$
若结论直至 $n(n\geqslant 2)$ 皆已成立，则对于 $n+1$，有
$\begin{aligned}
&a_{n+1}+a_{n+2}+2\\
&=(7a_n-a_{n-1})+(7a_{a+1}-a_n)+2\\
&=7(a_n+a_{n+1}+2)-(a_{n-1}+a_n+2)-10\\
&=7x_n^2-x_{n-1}^2-10\\
&=(3x_n)^2-x_{n-1}^2-2x_n^2-10\\
&=(3x_n-x_{n-1})(3x_n+x_{n-1})-2x_n^2-10\\
&=x_{n+1}(x_{n+1}+2x_{n-1})-2x_n^2-10\\
&=x_{n+1}^2+2(x_{n-1}x_{n+1}-x_n^2-5)=x_{n+1}^2\\
\end{aligned}$
即在 $n+1$ 时结论也成立．故本题得证．
注意到，$a_1+a_2+2=2^2,a_2+a_3+2=3^2,a_3+a_4+2=7^2,a_4+a_5+2=18^2,\cdots,$，构作数列 $\{x_n\}:x_1=2,x_2=3,x_n=3x_{n-1}-x_{n-2},n\geqslant 3$，则对每个 $n\in N^{\ast},x_n$ 为正整数．
我们来证明：对于每个 $n\in N^{\ast}$，皆有：$a_n+a_{n+1}+2=x_n^2$．
引理：数列 $\{x_n\}$ 满足：对于每个 $k\in N^{\ast},x_kx_{k+2}-x_{k+1}^2=5$．
引理证明：令 $f(k)=x_kx_{k+2}-x_{k+1}^2$，则
$\begin{aligned}
&f(k)-f(k-1)\\
&=(x_kx_{k+2}-x_{k+1}^2)-(x_{k-1}x_{k+1}-x_k^2)\\
&=(x_kx_{k+2}+x_k^2)-(x_{k+1}^2+x_{k-1}x_{k+1})\\
&=x_k(x_{k+2}+x_k)-x_{k+1}(x_{k+1}+x_{k-1})\\
&=3x_kx_{k+1}-3x_{k+1}x_k=0
\end{aligned}$
所以 $f(k)=f(k-1)$，于是 $f(k)=f(k-1)=f(k-2)=\cdots=f(1)=x_1x_3-x_2^2-5$．
回到本题，对 $n$ 归纳，据数列 $\{a_n\}$ 的定义，
$\begin{aligned}
a_1+a_2+2=4=x_1^2\\
a_2+a_3+2=9=x_2^2\\
\end{aligned}$
若结论直至 $n(n\geqslant 2)$ 皆已成立，则对于 $n+1$，有
$\begin{aligned}
&a_{n+1}+a_{n+2}+2\\
&=(7a_n-a_{n-1})+(7a_{a+1}-a_n)+2\\
&=7(a_n+a_{n+1}+2)-(a_{n-1}+a_n+2)-10\\
&=7x_n^2-x_{n-1}^2-10\\
&=(3x_n)^2-x_{n-1}^2-2x_n^2-10\\
&=(3x_n-x_{n-1})(3x_n+x_{n-1})-2x_n^2-10\\
&=x_{n+1}(x_{n+1}+2x_{n-1})-2x_n^2-10\\
&=x_{n+1}^2+2(x_{n-1}x_{n+1}-x_n^2-5)=x_{n+1}^2\\
\end{aligned}$
即在 $n+1$ 时结论也成立．故本题得证．
答案
解析
备注
