A sequence defined by iterating over polynomials $f(x)$ is a polynomial with integer coefficients.$a_{1}=f(0)$ ,$a_{2}=f(a_{1})$ and $a_k=f(a_{k-1}) \forall k \geq 3$, I have to show that if $a_{k}=0$ for some $k\geq 3$ then either $a_{1}=0$ or $a_{2}=0$.
My attempt:
Say for example we have $f(f(f(0)))=0$ that means $f(0)$ is a root of $f(f(x))=0$  hence $f(0)$ divides the constant term of $f(f(x))$ which is $f(f(0))$,so we have $f(0)|f(f(0))$ and similarly since,$f(f(0))$ is a root of $f(x)=0$ we also have $f(f(0))$ divides $f(0)$. Then I can say both $f(0)$ and $f(f(0))$ have same absolute value.Now if $f(0)=f(f(0))$ then we have a constant sequence and we are done.But why they can't be unequal?
 A: Let $k\geq3$ be such that $a_k=0$. 
For $a$, $b$ integers, we must have $$a-b\mid f(a)-f(b).$$ This is easy to prove, just by noticing $a-b\mid a^i-b^i$ for all natural $i$. In particular, we have the following chain of divisibilities: $$f(0)-0\mid f^2(0)-f(0)\mid\cdots\mid f^{k+1}(0)-f^k(0),$$ where $f^k$ denotes the $k$-fold composition of $f$. Since $f^k(0)=0$ by hypothesis, $$f^{k+1}(0)-f^k(0)=f(0)-0.$$ Since furthermore, $x\mid y$ implies $|x|\leq|y|$, the absolute values of all of the numbers in our divisibility chain must be equal.
If $f(0)=0$, then $a_1=0$, and we're already done. If not, $f(0)-0$ must have some sign, either positive or negative. In either case, we can take the first $n$ such that $f^n(0)-f^{n-1}(0)$ has a sign opposite to it.
If $n=2$, this implies $f^2(0)=0$, so that $a_2=0$. Otherwise, we would have to have $$f^n(0)=f^{n-2}(0).$$ But this would mean that all values of $f^i(0)$ for $i\geq1$ would be contained in the set $$\{f(0),f^2(0),\ldots,f^{n-1}(i)\},$$ whose values are in a strictly monotone sequence starting from $0$ by assumption. In particular, $a_k\neq0$, a contradiction.
This proves what we wished to prove. $\blacksquare$
