<s>[INST] Sequence : $a_{n+1}=2a_n-a_{n-1}+2$ Let $c$ be a positive integer. The sequence $a_1, a_2, \ldots$ is defined by $a_1=1, a_2=c$ and  
$a_{n+1}=2a_n-a_{n-1}+2$ for all $n \geq 2$.
Prove that for each $n \in \mathbb{N}$ there exists $k \in \mathbb{N}$ such that $a_na_{n+1} = a_k$.

My attempt :
Trying with small numbers, I see that $a_n=(n-1)c+(n-2)^2$ and will prove by induction.
$a_1 = 1 , a_2 = c+0 = c$ is true.
Suppose that $a_k=(k-1)c+(k-2)^2$ and $a_{k+1}=kc+(k-1)^2$ are true.
$a_{k+2} = 2a_{k+1}+a_k+2 = 2[kc+(k-1)^2]-[(k-1)c+(k-2)^2]+2=(k+1)c+k^2$ 
so $a_n=(n-1)c+(n-2)^2$ is true.
$a_na_{n+1} = [(n-1)c+(n-2)^2][nc+(n-1)^2] $ 
$= n(n-1)c^2 + (n-1)^3c + n(n-2)^2c + (n^2-3n+2)^2 = (k-1)c+(k-2)^2$
Please suggest how to solve this equation.
 [/INST] Since
$$
(a_{n+1}-a_n)-(a_n-a_{n-1})=2\tag1
$$
we have
$$
a_{n+1}-a_n=a_2-a_1+2n-2\tag2
$$
Thus,
$$
\begin{align}
a_{n+1}
&=a_1+(a_2-a_1)n+n^2-n\\
&=a_1+(a_2-a_1-1)n+n^2\tag3
\end{align}
$$
Using $(3)$, we can get by expansion
$$
a_na_{n+1}=a_{n^2+(a_2-a_1-2)n+2a_1-a_2+2}\tag4
$$
That is,
$$
\bbox[5px,border:2px solid #C0A000]{k=n^2+(a_2-a_1-2)n+2a_1-a_2+2}\tag5
$$
This does not require $a_1=1$.
</s>