Prove that the solution set of $a_n \not= n$, $n \in \mathbb{N}$ is finite. 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

Let $n > 20080$ (large enough).   
Let $ b_n = a_n - n$.
We want to show that for some $N$, for any $ n > N$, $ b_n = 0 $.   
Claim: $b_{n+2} \leq \frac{ b_{n+1} + \frac{2b_{n+1} + b_n + 2010 } { 20080 }} {2}    $

 Proof: $ (a_{n+2})^2 \leq n + b_n + (n+2)(n+ 1+b_{n+1}) + 2008 $
$= n^2 + n (4 + b_{n+1} ) + 2b_{n+1} + b_n + 2010 $
$ < n^2 + n (4 + b_{n+1} + \frac{ 2b_{n+1} + b_n + 2010 } { 20080 }) $
$< ( n + 2+\frac{b_{n+1} + \frac{ 2b_{n+1} + b_n + 2010 } { 20080 }} {2} )^2 $

Corollary: There is an $N_1$ such that for any $n \geq N_1$, $b_n \leq 0$.   
Given that you've shown $ b_n \geq 0$, we conclude that for $ n > N$, $b_n = 0$.
