10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=a_{2}=1, \\
a_{n+1}=\frac{n^{2} a_{n}^{2}+5}{\left(n^{2}-1\right) a_{n-1}}(n>1) .
\end{array}
$$

Question: Does there exist real numbers $x, y (x \neq 0)$, such that $\left\{\frac{b_{n+2}+b_{n}}{b_{n+1}}\right\}$ (where, $b_{n}=(n x+y) a_{n}, n \in \mathbf{Z}_{+}$) is a constant sequence?
10. From the given, we have
$$
\begin{array}{l}
\left(n^{2}-1\right) a_{n-1} a_{n+1}=n^{2} a_{n}^{2}+5 . \\
\text { Also, }\left((n-1)^{2}-1\right) a_{n-2} a_{n}=(n-1)^{2} a_{n-1}^{2}+5,
\end{array}
$$

Subtracting the two equations, we get
$$
\begin{array}{l}
\left(n^{2}-1\right) a_{n-1} a_{n+1}-\left((n-1)^{2}-1\right) a_{n-2} a_{n} \\
=n^{2} a_{n}^{2}-(n-1)^{2} a_{n-1}^{2} \\
\Rightarrow(n-1) a_{n-1}\left((n+1) a_{n+1}+(n-1) a_{n-1}\right) \\
\quad=n a_{n}\left(n a_{n}+(n-2) a_{n-2}\right)(n>2) .
\end{array}
$$

The sequence $\left\{\frac{b_{n+2}+b_{n}}{b_{n+1}}\right\}$ being a constant sequence is equivalent to
$$
\frac{b_{n+1}+b_{n-1}}{b_{n}}=\frac{b_{n}+b_{n-2}}{b_{n-1}}(n>2),
$$

which means
$$
\begin{array}{l}
b_{n-1}\left(b_{n+1}+b_{n-1}\right)=b_{n}\left(b_{n}+b_{n-2}\right) . \\
\text { Substituting } b_{n}=(x n+y) a_{n} \text {, and comparing with } \\
(n-1) a_{n-1}\left((n+1) a_{n+1}+(n-1) a_{n-1}\right) \\
=n a_{n}\left(n a_{n}+(n-2) a_{n-2}\right)
\end{array}
$$

By comparing coefficients, we find that $x=1, y=0$ satisfies the given conditions.