If you mean \[f_1 = 1, f_2 = c, f_{n+1} = 2f_n - f_{n-1} + 2 \quad (n \geq 2).\] then $f_n=n^2+(n-1)(c-4)$ and $f_kf_{k+1}=(k^2+k)^2+(c-4)[2k^3+k^2-k-1+k(k-1)(c-4)]=[k^2+k+(c-4)(k-\frac 12)]^2+(c-4)...=r^2+r(c-4)-c+4$ $D=(c-4)^2+4[[(k^2+(k-\frac 12)(c-3)]^2-4(c-4)(k^2+k)+...$

Thanks, edited.

With simple induction, we show that $f_n=(n-1)c+(n-2)^2$ The base cases of $n=1,2$ are obvious. Assume it is true for $n=k, k-1$. Then, $f_{k+1}=2(k-1)c+2(k-2)^2-(k-2)c-(k-3)^2+2$ $=(2k-2-k+2)c+(2k^2-8k+8-k^2+6k-9+2)=kc+(k-1)^2$, so our induction is complete. Note also that $f_kf_{k+1}=[(k-1)c+(k-2)^2][kc+(k-1)^2]$ This can be rearranged to $[(k-1)c+(k^2-3k+3)]c+[(k-1)c+(k^2-3k+3)-1]^2=f_{(k-1)c+(k^2-3k+3)}$ Thus, for all $c,k$, there exists $r=(k-1)c+(k^2-3k+3) \in \mathbb N$ such that $f_kf_{k+1}=f_r$