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.
 A: @Ian, the solutions to the quadratic are:
$-cn-n^2+3n$ and $cn+n^2-c-3n+4$ which are both integers. The second solution gives, for $n=1,2,\ldots$, the assignment $k(n)= 2, c+2, 2c+4$, which are all positive indices.
A: 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 $a_1=1$.
A: For $n=1$, $a_1a_2=c=a_2$.
In the following, $n\ge 2$.
You already have $a_n=(n-1)c+(n-2)^2$.
Then,
$$\begin{align}&a_na_{n+1}=a_k\\\\&\iff ((n-1)c+(n-2)^2)(nc+(n-1)^2)=(k-1)c+(k-2)^2\\\\&\iff  k^2+(c-4)k-(c^2n^2-c^2n+2cn^3-7cn^2+7cn+n^4-6n^3+13n^2-12n)=0\\\\&\iff k=\frac{-c+4\pm\sqrt{\Delta}}{2}\end{align}$$
where
$$\begin{align}\Delta&=(c-4)^2+4(c^2n^2-c^2n+2cn^3-7cn^2+7cn+n^4-6n^3+13n^2-12n)\\\\&=4n^4+(8c-24)n^3+(4c^2-28c+52)n^2+(-4c^2+28c-48)n+(c-4)^2\\\\&=(2n^2+(2c-6)n-c+4)^2\end{align}$$
It follows from this that
$$a_na_{n+1}=a_k\iff k=n(n-3)+c(n-1)+4,-n(n+c-3)$$
Since the latter is non-positive for $n\ge 2$, 
$$a_na_{n+1}=a_k\iff k=n(n-3)+c(n-1)+4$$
where $n(n-3)+c(n-1)+4$ is a positive integer.
