3.3.25 Given the sequence $\left\{c_{n}\right\}$ satisfies: $c_{0}=1, c_{1}=0, c_{2}=2005$, and
$$
c_{n+2}=-3 c_{n}-4 c_{n-1}+2008, n=1,2,3, \cdots \text {. }
$$

Let $a_{n}=5\left(c_{n+2}-c_{n}\right)\left(502-c_{n-1}-c_{n-2}\right)+4^{n} \times 2004 \times 501(n=2,3, \cdots)$. Is $a_{n}$ a perfect square for $n>2$? Explain your reasoning.
From $c_{n+2}-c_{n}=-4\left(c_{n}+c_{n-1}\right)+2008$ we know
$$
a_{n}=-20\left(c_{n}+c_{n-1}-502\right)\left(502-c_{n-1}-c_{n-2}\right)+4^{n+1} \times 501^{2} \text {. }
$$

Let $d_{n}=c_{n}-251$, then $d_{0}=-250, d_{1}=-251, d_{2}=-1754, d_{n+2}=-3 d_{n}-4 d_{n-1}$,
$$
a_{n}=20\left(d_{n}+d_{n-1}\right)\left(d_{n-1}+d_{n-2}\right)+4^{n+1} \times 501^{2} .
$$

Let $w_{n}=d_{n}+d_{n-1}$, then $w_{1}=-501, w_{2}=1503, w_{n+2}=w_{n+1}-4 w_{n}$,
$$
a_{n}=20 w_{n} w_{n-1}+4^{n+1} \times 501^{2} .
$$

Let $w_{n}=501 T_{n}$, then $T_{1}=-1, T_{2}=3, T_{n+2}=T_{n+1}-4 T_{n}$, define $T_{0}=1$,
$$
a_{n}=501^{2} \times 2^{2}\left(5 T_{n} T_{n-1}+4^{n}\right) .
$$

Let $\alpha, \beta$ be the two roots of $x^{2}-x+4=0$, then $\alpha+\beta=1, \alpha \beta=4$,
$$
\begin{aligned}
T_{n}-\alpha T_{n-1} & =\beta\left(T_{n-1}-\alpha T_{n-2}\right)=\cdots=\beta^{n-1}\left(T_{1}-\alpha T_{0}\right)=\beta^{n-1}(\alpha-1), \\
T_{n}-\beta T_{n-1} & =\alpha^{n-1}(\beta-1),\left(T_{n}-\alpha T_{n-1}\right)\left(T_{n}-\beta T_{n-1}\right) \\
& -(\alpha \beta)^{n-1}(\alpha-1)(\beta-1)=4^{n},
\end{aligned}
$$

i.e., $T_{n}^{2}-T_{n} T_{n-1}+4 T_{n-1}^{2}=4^{n}$. Therefore,
$$
a_{n}=501^{2} \times 2^{2}\left(5 T_{n} T_{n-1}+T_{n}^{2}-T_{n} T_{n-1}+4 T_{n-1}^{2}\right)=1002^{2}\left(T_{n}+2 T_{n-1}\right)^{2} .
$$

Comment: In this problem, the substitution method is used three times to simplify the expression of $a_{n}$. By using the characteristic equation of $\{T_{n}\}$, we obtain the relation $T_{n}^{2}-T_{n} T_{n-1}+4 T_{n-1}^{2}=4^{n}$, which is the key to completing the square of the expression for $a_{n}$.