Problem:For fixed positive integers $s, t$, define $a\_n$ as the following.
$a\_1 = s, a\_2 = t$, and $\forall n \ge 1$, $a\_{n+2} = \lfloor \sqrt{a\_n+(n+2)a\_{n+1}+2008} \rfloor$.
Prove that the solution set of $a\_n \not= n$, $n \in \mathbb{N}$ is finite.

I have try prove $a\_{n}\ge n$,We already know $a\_3 \ge 3$ and $a\_4 \ge 4$.
We prove by induction that $a\_n \ge n$ for all $n\ge 3$.
Assume that $a\_n \ge n$ and $a\_{n+1} \ge n+1$. 
Then 
$$(a\_{n+2}+1)^2 > a\_n + (n+2)a\_{n+1} +2008 \ge n+(n+2)(n+1) + 2008 > (n+2)^2$$, so $a\_{n+2} \ge n+2$.
Therefore, we have $a\_n \ge n$ for $n \ge 3$.then I can't Thanks