hello, i think no, i have calculated $ a_3=5(c_4-c_3)(502-c_3-c_2)+4^3\cdot 2004\cdot501$ where $ c_3=2004$ $ c_4=-4007$ and so is $ a_3=109428921=3^3\cdot7\cdot167\cdot3467$ and this is not a perfect square. Sonnhard.

Shobber has made a mistake. It should be :${a_{n}=5(c_{n+2}-c_{n})\cdot (502-c_{n-1}-c_{n-2})+4^{n}\times 2004\times 501}$ (*) Easy to get that ${c_{n+2}-251=-3(c_{n}-251)-4(c_{n-1}-251)}$ Denote ${d_{n}=c_n-251}$ So ${d_{n+2}=-3d_n-4d_{n-1}}$ We also can get $c_{n+2}-c_n=-4(c_n+c_{n-1})+2008$ put it into (*) we get $a_n=-20(c_n+c_{n-1}-502)(502-c_{n-1}-c_{n-2})+4^{n+1}\cdot501^2=20(d_n+d_{n-1})(d_{n-1}+d_{n-2})+4^{n+1}\cdot501^2$ Denote ${w_n=d_n+d_{n-1}}$ So ${w_{n+2}=w_{n+1}-4w_{n}}$ So ${a_n=20w_nw_{n-1}+4^{n+1}\cdot501^2}$ Then Asume ${w_n=501T_n}$ We also have ${T_{n+2}=T_{n+1}-4T_{n}}$ So ${a_n=501^2\cdot2^2\cdot(5T_nT_{n-1}+4^n)}$- (#) Since ${T_{n+2}=T_{n+1}-4T_{n}}$ let ${\alpha \beta}$ be two roots of the equation ${x^2-x+4=0}$ We know: 1. ${T_n-\alpha T_{n-1}=\beta(T_{n-1}-\alpha T_{n-2})=......=-{\beta}^{n-1}(\alpha-1)}$ 2. ${T_n-\beta T_{n-1}=\alpha(T_{n-1}-\beta T_{n-2})=......=-{\alpha}^{n-1}(\beta-1)}$ (1)*(2): we get : ${T_n-(\alpha+\beta)T_{n-1}T_{n}+\alpha\beta{T_{n-1}}^2=(\alpha\beta)^{n-1}(\alpha\beta-(\alpha+\beta)+1)}$ Easy to get :${\alpha+\beta=1}$ , ${\alpha\beta=4}$ So: ${(T_n+2T_{n-1})^2=5T_nT_{n-1}+4^n}$ Put it into (#) We finally get :${a_n=(1002(T_n+2T_{n-1}))^2}$ So ${a_n}$ is always a perfect square forany $n>2$.Done!