China Southeastern Mathematical Olympiad

By $a_{n+1}=\frac{a_n^2+(-1{)}^n}{a_{n-1}}$, we have $a_{n+1}{a}_{n-1}=a_n^2+(-1{)}^n$ $(n=2,3,\dots )$, so $$\begin{array}{rl}\frac{a_n-a_{n-2}}{a_{n-1}}& =\frac{a_n{a}_{n-2}-a_{n-2}^2}{a_{n-1}{a}_{n-2}}=\frac{a_{n-1}^2+(-1{)}^{n-1}-a_{n-2}^2}{a_{n-1}{a}_{n-2}}\\ & =\frac{a_{n-1}^2-a_{n-1}{a}_{n-3}}{a_{n-1}{a}_{n-2}}=\frac{a_{n-1}-a_{n-3}}{a_{n-2}}\\ & =\cdots =\frac{a_3-a_1}{a_2}=2,\end{array}$$ that is, $a_n=2a_{n-1}+a_{n-2}$ $(n\ge 3)$, $a_1=1$, $a_2=2$.

Therefore, $a_n=C_1{\lambda }_1^n+C_2{\lambda }_2^n$, ${\lambda }_1+{\lambda }_2=2$, ${\lambda }_1{\lambda }_2=-1$, $a_1=1$, $a_2=2$, $n\in {\mathbb{N}}^{+}$. Then, since ${\lambda }_1{\lambda }_2=-1$ and ${\lambda }_2=2-{\lambda }_1$, we have $$\{\begin{array}{l}1=C_1{\lambda }_1+C_2{\lambda }_2\\ 2=C_1{\lambda }_1^2+C_2{\lambda }_2^2\end{array}\Rightarrow \{\begin{array}{l}{\lambda }_2=C_1{\lambda }_1{\lambda }_2+C_2{\lambda }_2^2\\ 2=C_1{\lambda }_1^2+C_2{\lambda }_2^2\end{array}\Rightarrow \{\begin{array}{l}2-{\lambda }_1=-C_1+C_2{\lambda }_2^2\\ 2=C_1{\lambda }_1^2+C_2{\lambda }_2^2\end{array}\Rightarrow C_1(1+{\lambda }_1^2)={\lambda }_1.$$ Thus, by symmetric condition, we have $C_2(1+{\lambda }_2^2)={\lambda }_2$. So, since $1+{\lambda }_1{\lambda }_2=0$, we have $$\begin{array}{rl}{a}_n^2+a_{n+1}^2& =C_1^2(1+{\lambda }_1^2){\lambda }_1^{2n}+C_2^2(1+{\lambda }_2^2){\lambda }_2^{2n}+\\ & 2C_1{C}_2({\lambda }_1{\lambda }_2{)}^n(1+{\lambda }_1{\lambda }_2)\\ & C_1{\lambda }_1^{2n+1}+C_2{\lambda }_2^{2n+1}=a_{2n+1}.\end{array}$$