Given sequence $ \{ c_n \}$ satisfying the conditions that $ c_0\equal{}1$, $ c_1\equal{}0$, $ c_2\equal{}2005$, and  $ c_{n\plus{}2}\equal{}\minus{}3c_n \minus{} 4c_{n\minus{}1} \plus{}2008$, ($ n\equal{}1,2,3, \cdots$). Let $ \{ a_n \}$ be another sequence such that $ a_n\equal{}5(c_{n\plus{}1} \minus{} c_n) \cdot (502 \minus{} c_{n\minus{}1} \minus{} c_{n\minus{}2}) \plus{} 4^n \times 2004 \times 501$, ($ n\equal{}2,3, \cdots$).

Is $ a_n$ a perfect square for every $ n > 2$?
1. Given the sequence \( \{ c_n \} \) with initial conditions \( c_0 = 1 \), \( c_1 = 0 \), \( c_2 = 2005 \), and the recurrence relation \( c_{n+2} = -3c_n - 4c_{n-1} + 2008 \) for \( n = 1, 2, 3, \ldots \).

2. Define another sequence \( \{ a_n \} \) such that 
   \[
   a_n = 5(c_{n+1} - c_n) \cdot (502 - c_{n-1} - c_{n-2}) + 4^n \times 2004 \times 501 \quad \text{for} \quad n = 2, 3, \ldots
   \]

3. To simplify the recurrence relation, let \( d_n = c_n - 251 \). Then the recurrence relation for \( d_n \) becomes:
   \[
   d_{n+2} = -3d_n - 4d_{n-1}
   \]

4. We can express \( c_{n+2} - c_n \) as:
   \[
   c_{n+2} - c_n = -4(c_n + c_{n-1}) + 2008
   \]

5. Substituting \( c_{n+2} - c_n \) into the expression for \( a_n \), we get:
   \[
   a_n = -20(c_n + c_{n-1} - 502)(502 - c_{n-1} - c_{n-2}) + 4^{n+1} \cdot 501^2
   \]

6. Using \( d_n = c_n - 251 \), we have:
   \[
   a_n = 20(d_n + d_{n-1})(d_{n-1} + d_{n-2}) + 4^{n+1} \cdot 501^2
   \]

7. Let \( w_n = d_n + d_{n-1} \). Then the recurrence relation for \( w_n \) is:
   \[
   w_{n+2} = w_{n+1} - 4w_n
   \]

8. Therefore, we can rewrite \( a_n \) as:
   \[
   a_n = 20w_n w_{n-1} + 4^{n+1} \cdot 501^2
   \]

9. Assume \( w_n = 501T_n \). Then we have:
   \[
   T_{n+2} = T_{n+1} - 4T_n
   \]

10. Substituting \( w_n = 501T_n \) into the expression for \( a_n \), we get:
    \[
    a_n = 501^2 \cdot 2^2 \cdot (5T_n T_{n-1} + 4^n)
    \]

11. The characteristic equation for \( T_n \) is:
    \[
    x^2 - x + 4 = 0
    \]
    with roots \( \alpha \) and \( \beta \).

12. Using the properties of the roots, we have:
    \[
    \alpha + \beta = 1 \quad \text{and} \quad \alpha \beta = 4
    \]

13. From the recurrence relation, we get:
    \[
    (T_n + 2T_{n-1})^2 = 5T_n T_{n-1} + 4^n
    \]

14. Substituting this into the expression for \( a_n \), we obtain:
    \[
    a_n = (1002(T_n + 2T_{n-1}))^2
    \]

15. Therefore, \( a_n \) is always a perfect square for any \( n > 2 \).

The final answer is \( \boxed{ a_n } \) is always a perfect square for any \( n > 2 \).