该递推方程对应的特征方程为
\[
x^2-6x+1=0
\]
解得
\[
x_{1,2}=3\pm 2\sqrt{2}
\]
所以
\[
a_n=A_1 x_1^n+A_2 x_2^n=A_1\cdot(3+2\sqrt{2})^n+A_2\cdot(3-2\sqrt{2})^n
\]
这里 $A_1,A_2$ 为待定常数.
由所给条件 $a_0=2$ 和 $a_1=10$, 得
\[
\begin{cases}
2&=A_1+A_2,\\
10&=A_1\cdot(3+2\sqrt{2})+A_2\cdot(3-2\sqrt{2}).
\end{cases}
\]
由第一式, $A_2=2-A_1$, 代入第二式, 得
\[
\begin{split}
&10=A_1\cdot(3+2\sqrt{2})+(2-A_1)\cdot(3-2\sqrt{2})\\
\Rightarrow\ &10=(3+2\sqrt{2})A_1-(3-2\sqrt{2})A_1+2(3-2\sqrt{2})\\
\Rightarrow\ &10=4\sqrt{2}A_1+6-4\sqrt{2}\\
\Rightarrow\ &4=4\sqrt{2}(A_1-1)\\
\Rightarrow\ &A_1=1+\frac{1}{\sqrt{2}}
\end{split}
\]
从而 $A_2=2-A_1=1-\frac{1}{\sqrt{2}}$. 于是我们得到 $a_n$ 的表达式,
\[
a_n=(1+\frac{1}{\sqrt{2}})(3+2\sqrt{2})^n+(1-\frac{1}{\sqrt{2}})(3-2\sqrt{2})^n
\]
注意到
\[
\begin{aligned}
3+2\sqrt{2}&=(\sqrt{2}+1)^2\\
3-2\sqrt{2}&=(\sqrt{2}-1)^2
\end{aligned}
\]
故
\[
\begin{split}
a_n&=(1+\frac{\sqrt{2}}{2})(3+2\sqrt{2})^n+(1-\frac{\sqrt{2}}{2})(3-2\sqrt{2})^n\\
&=\frac{2+\sqrt{2}}{2}\cdot(\sqrt{2}+1)^{2n}+\frac{2-\sqrt{2}}{2}\cdot(\sqrt{2}-1)^{2n}\\
&=\frac{3+2\sqrt{2}+1}{4}\cdot(\sqrt{2}+1)^{2n}+\frac{3-2\sqrt{2}+1}{4}\cdot(\sqrt{2}-1)^{2n}\\
&=\frac{1}{4}\cdot\biggl[\Bigl((\sqrt{2}+1)^2+1\Bigr)\cdot(\sqrt{2}+1)^{2n}+\Bigl((\sqrt{2}-1)^2+1\Bigr)\cdot(\sqrt{2}-1)^{2n}\biggr]\\
&=\frac{1}{4}\cdot\biggl[(\sqrt{2}+1)^{2n+2}+(\sqrt{2}+1)^{2n}+(\sqrt{2}-1)^{2n+2}+(\sqrt{2}-1)^{2n}\biggr]\\
&=\frac{1}{4}\cdot\biggl[\Bigl((\sqrt{2}+1)^{2n+2}+2(\sqrt{2}+1)^{n+1}\cdot(\sqrt{2}-1)^{n+1}+(\sqrt{2}-1)^{2n+2}\Bigr)+\Bigl((\sqrt{2}+1)^{2n}-2(\sqrt{2}+1)^{n+1}\cdot(\sqrt{2}-1)^{n+1}+(\sqrt{2}-1)^{2n}\Bigr)\biggr]\\
&=\frac{1}{4}\cdot\biggl[\Bigl((\sqrt{2}+1)^{n+1}+(\sqrt{2}-1)^{n+1}\Bigr)^2+\Bigl((\sqrt{2}+1)^{n}-(\sqrt{2}-1)^{n}\Bigr)^2\biggr]\\
&=\biggl(\frac{(\sqrt{2}+1)^{n+1}+(\sqrt{2}-1)^{n+1}}{2}\biggr)^2+\biggl(\frac{(\sqrt{2}+1)^{n}-(\sqrt{2}-1)^{n}}{2}\biggr)^2\\
&=A_n^2+B_n^2
\end{split}
\]
由上面的计算过程, $a_n$ 也可以写为
\[
a_n=\biggl(\frac{(\sqrt{2}+1)^{n+1}-(\sqrt{2}-1)^{n+1}}{2}\biggr)^2+\biggl(\frac{(\sqrt{2}+1)^{n}+(\sqrt{2}-1)^{n}}{2}\biggr)^2
\]
特别的, 当 $n$ 为偶数时,
令 $A_n=\frac{(\sqrt{2}+1)^{n+1}-(\sqrt{2}-1)^{n+1}}{2}$, $B_n=\frac{(\sqrt{2}+1)^{n}+(\sqrt{2}-1)^{n}}{2}$. 此时易见 $A_n, B_n$ 为正整数.
当 $n$ 为奇数时,
令 $A_n=\frac{(\sqrt{2}+1)^{n+1}+(\sqrt{2}-1)^{n+1}}{2}$, $B_n=\frac{(\sqrt{2}+1)^{n}-(\sqrt{2}-1)^{n}}{2}$. 此时亦可证 $A_n, B_n$ 为正整数.
总之, 证明了 $a_n$ 可表示为两个自然数的平方和.