How to prove that $b_{2008}\neq 0$ Let the polynomial $f$ be defined as
$$f(x)=a_{m}x^m+a_{m-1}x^{m-1}+\cdots+a_{1}x+a_{0}, \qquad a_{i}\in \Bbb Z \ (i=0,1,2,\cdots,m), \ a_{i}\neq 0.$$
Define the sequence $\{b_{n}\}$ as 
$$b_{1}=0,b_{n+1}=f(b_{n}).$$
Show that
$$b_{2008}\neq 0.$$
My try: let $x_{i},i=1,2,\dots,m$ be the complex roots of the polynomial $f(x)$, then
$$f(x)=a_{m}(x-x_{1})(x-x_{2})\cdots(x-x_{m})$$
Maybe this is Olympic math exam question, this is a problem my frend asked me.
 A: We know that for integers $a$ and $b$ and a polynomial $P$ that $a-b|P(a)-P(b)$. Say that 
$$b_i=P^i(0)=\underbrace{P(P(\dots P(0)\dots ))}_{P \text{ applied $i$ times}}$$
and $b_1=0$. Now, we know that
$$
b_1-b_0|P(b_1)-P(b_0)=b_2-b_1\\
b_{n+1}-b_{n}|P(b_{n+1})-P(b_{n})=b_{n+2}-b_{n+1}
$$
Thus, we see that
$$
b_1-b_0|b_2-b_1|b_3-b_2|\dots |b_{2008}-b_{2007}
$$
Because $a_0\neq 0$, we know that $P(0)\neq 0$, thus $P(0)-0=b_1-b_0\neq 0$. We know now that all differences are $0$ or a multiple of $P(0)=a_0$. Now, suppose $b_{2007}=0$. (I accidently assumed $b_0=0$, so we have to take $2007$ instead, which is odd (and it turns out that is a sufficient condition).)  Then, we get $b_{2008}=P(b_{2007})=P(0)=a_0$. Thus:
$$
a_0=P(0)-0=b_1-b_0|b_2-b_2|\dots|b_{2008}-b_{2007}=P(0)-0=a_0
$$
We now conclude that the absolute values of all differences $|b_{n+1}-b_n|$ must be equal to $P(0)=a_0$. We now get $b_{n+1}=b_n+\pm a_0$. From this, it follows that $2a_0|b_{2n}$ and $2a_0|b_{2n+1}+a_0$. We know that $2007$ is odd, so $2a_0|b_{2007}+a_0$. Therefore, $b_{2007}$ can't be $0$.
This is indeed a typical Math Olympiad question, as I have solve some very similar exercises during my training for the IMO.
A: Observe that $f(p a_0) = a_0 ( p(\text{integer}) + 1)$ and in particular $b_j = p_j a_0$ for some integers $p_j$.
Now $p_{j+1} = p_j(a_m p_j^{m-1} a_0 ^{m-1} + \cdots + a_1) + 1$.
Work modulo 2. Then this maps $0 \mapsto 1$. Also, either $1 \mapsto 0 \mapsto 1$ or $1 \mapsto 1 \mapsto 1$ under two applications. As such, $p_1 \equiv 0 \implies p_2 \equiv 1 \implies p_{2k} \equiv 1$ for all $k$. In particular, $b_{2k} \neq 0$.
