As usual, let $\mathbb{Z}[x]$ denote the set of single-variable polynomials in $x$ with integer coefficients. Find all functions $\theta: \mathbb{Z}[x] \rightarrow \mathbb{Z}$ such that for any polynomials $p, q \in$ $\mathbb{Z}[x]$,  
- $\theta(p+1)=\theta(p)+1$, and  
- if $\theta(p) \neq 0$ then $\theta(p)$ divides $\theta(p \cdot q)$.
The answer is $\theta: p \mapsto p(c)$, for each choice of $c \in \mathbb{Z}$. Obviously these work, so we prove these are the only ones. In what follows, $x \in \mathbb{Z}[x]$ is the identity polynomial, and $c=\theta(x)$. 【 First solution (Merlijn Staps). Consider an integer $n \neq c$. Because $x-n \mid p(x)-p(n)$, we have 
$$ \theta(x-n) \mid \theta(p(x)-p(n)) \Longrightarrow c-n \mid \theta(p(x))-p(n) . $$
On the other hand, $c-n \mid p(c)-p(n)$. Combining the previous two gives $c-n \mid \theta(p(x))-p(c)$, and by letting $n$ be large we conclude $\theta(p(x))-p(c)=0$, so $\theta(p(x))=p(c)$.