Example 3 Given the sequence $\left\{c_{n}\right\}$ satisfies
$$
\begin{array}{l}
c_{0}=1, c_{1}=0, c_{2}=2005, \\
c_{n+2}=-3 c_{n}-4 c_{n-1}+2008(n=1,2, \cdots) . \\
\text { Let } a_{n}=5\left(c_{n+2}-c_{n}\right)\left(502-c_{n-1}-c_{n-2}\right)+
\end{array}
$$
$4^{n} \times 2004 \times 501(n=1,2, \cdots)$. Is $a_{n}$ a perfect square when $n>2$?
Solve: Convert the linear non-homogeneous recurrence relation
$$
c_{n+2}=-3 c_{n}-4 c_{n-1}+2008
$$

into a linear homogeneous recurrence relation
$$
c_{n+2}-251=-3\left(c_{n}-251\right)-4\left(c_{n-1}-251\right) \text {, }
$$

where 251 is the root of the equation $x=-3 x-4 x+2008$. Let $d_{n}=c_{n}-251$. Then,
$$
\begin{array}{l}
d_{0}=-250, d_{1}=-251, d_{2}=1754, \\
d_{n+2}=-3 d_{n}-4 d_{n-1}(n=1,2, \cdots) .
\end{array}
$$

Thus, $c_{n+2}-c_{n}=d_{n+2}-d_{n}=-4\left(d_{n}+d_{n-1}\right)$,
$$
502-c_{n-1}-c_{n-2}=-\left(d_{n-1}+d_{n-2}\right) \text {. }
$$

Therefore, $a_{n}=20\left(d_{n}+d_{n-1}\right)\left(d_{n-1}+d_{n-2}\right)+$
$$
4^{n+1} \times 501^{2} \text {. }
$$

Let $d_{n}+d_{n-1}=501 t_{n}$, then
$$
a_{n}=501^{2} \times 2^{2}\left(5 t_{n} t_{n-1}+4^{n}\right) \text {. }
$$

From $d_{n}+d_{n-1}=501 t_{n}$ and equation (1), we get
$$
\begin{array}{l}
t_{1}=-1, t_{2}=3, \\
t_{n+2}=t_{n+1}-4 t_{n}(n=1,2, \cdots) .
\end{array}
$$

Thus, by mathematical induction, for all $n \in \mathbf{N}_{+}$, $t_{n}$ is an integer.
Define $t_{0}$ such that $t_{2}=t_{1}-4 t_{0}$, i.e.,
$$
t_{0}=\frac{1}{4}\left(t_{1}-t_{2}\right)=-1 \text {. }
$$

Then, by the proposition $(1) \Leftrightarrow(3)$, we have
$$
t_{n+1}^{2}-t_{n+1} t_{n}+4 t_{n}^{2}=4^{n}\left(t_{1}^{2}-t_{0} t_{2}\right)=4^{n+1} \text {. }
$$

Therefore, $t_{n}^{2}-t_{n} t_{n-1}+4 t_{n-1}^{2}=4^{n}$, and
$$
\begin{array}{l}
a_{n}=501^{2} \times 2^{2}\left(5 t_{n} t_{n-1}+4^{n}\right) \\
=501^{2} \times 2^{2}\left(t_{n}^{2}+4 t_{n} t_{n-1}+4 t_{n-1}^{2}\right) \\
=501^{2} \times 2^{2}\left(t_{n}+2 t_{n-1}\right)^{2}
\end{array}
$$

is a perfect square.