Problem 1781

Assume that there are only finitely many primes $p_{1}, p_{2}, \ldots, p_{m}$ that divide some function value produced of $f$.
There are infinitely many positive integers $a$ such that $v_{p_{i}}(a)>v_{p_{i}}(f(1))$ for all $i=1,2, \ldots, m$, e.g. $a=\left(p_{1} p_{2} \ldots p_{m}\right)^{\alpha}$ with $\alpha$ sufficiently large. Pick any such $a$. The condition of the problem then yields $a \mid (f(a+1)-f(1))$. Assume $f(a+1) \neq f(1)$. Then we must have $v_{p_{i}}(f(a+1)) \neq v_{p_{i}}(f(1))$ for at least one $i$. This yields $v_{p_{i}}(f(a+1)-f(1))=\min \left\{v_{p_{i}}(f(a+1)), v_{p_{i}}(f(1))\right\} \leq v_{p_{1}}(f(1))<v_{p_{i}}(a)$. But this contradicts the fact that $a \mid (f(a+1)-f(1))$.
Hence we must have $f(a+1)=f(1)$ for all such $a$.
Now, for any positive integer $b$ and all such $a$, we have $(a+1-b) \mid (f(a+1)-f(b))$, i.e., $(a+1-b) \mid (f(1)-f(b))$. Since this is true for infinitely many positive integers $a$ we must have $f(b)=f(1)$. Hence $f$ is a constant function, a contradiction. Therefore, our initial assumption was false and there are indeed infinitely many primes $p$ dividing $f(c)$ for some positive integer $c$.

Assume that there are only finitely many primes $p_{1}, p_{2}, \ldots, p_{m}$ that divide some function value of $f$. Since $f$ is not identically $1$, we must have $m \geq 1$.
Then there exist non-negative integers $\alpha_{1}, \ldots, \alpha_{m}$ such that
$$f(1)=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{m}^{\alpha_{m}}$$
We can pick a positive integer $r$ such that $f(r) \neq f(1)$. Let
$$M=1+p_{1}^{\alpha_{1}+1} p_{2}^{\alpha_{2}+1} \ldots p_{m}^{\alpha_{m}+1} \cdot (f(r)+r)$$
Then for all $i \in \{1, \ldots, m\}$ we have that $p_{i}^{\alpha_{i}+1}$ divides $M-1$ and hence by the condition of the problem also $f(M)-f(1)$. This implies that $f(M)$ is divisible by $p_{i}^{\alpha_{i}}$ but not by $p_{i}^{\alpha_{i}+1}$ for all $i$ and therefore $f(M)=f(1)$.
Hence
$$\begin{aligned} M-r & >p_{1}^{\alpha_{1}+1} p_{2}^{\alpha_{2}+1} \ldots p_{m}^{\alpha_{m}+1} \cdot (f(r)+r)-r \\ & \geq p_{1}^{\alpha_{1}+1} p_{2}^{\alpha_{2}+1} \ldots p_{m}^{\alpha_{m}+1}+(f(r)+r)-r \\ & >p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{m}^{\alpha_{m}}+f(r) \\ & \geq |f(M)-f(r)| \end{aligned}$$
But since $M-r$ divides $f(M)-f(r)$ this can only be true if $f(r)=f(M)=f(1)$, which contradicts the choice of $r$.