, 2011

Obviously, all functions in the answer satisfy the condition of the problem. We will show that there are no other functions satisfying that condition.
Let $f$ be a function satisfying the given condition. For each integer $n$, the function $g$ defined by $g(x)=f(x)+n$ also satisfies the same condition. Therefore, by subtracting $f(0)$ from $f(x)$ we may assume that $f(0)=0$.
For any prime $p$, the condition on $f$ with $(x, y)=(p, 0)$ states that $f(p)$ divides $p^{n}$. Since the set of primes is infinite, there exist integers $d$ and $\varepsilon$ with $0 \leq d \leq n$ and $\varepsilon \in\{1,-1\}$ such that for infinitely many primes $p$ we have $f(p)=\varepsilon p^{d}$. Denote the set of these primes by $P$. Since a function $g$ satisfies the given condition if and only if $-g$ satisfies the same condition, we may suppose $\varepsilon=1$.
The case $d=0$ is easily ruled out, because 0 does not divide any nonzero integer. Suppose $d \geq 1$ and write $n$ as $m d+r$, where $m$ and $r$ are integers such that $m \geq 1$ and $0 \leq r \leq d-1$. Let $x$ be an arbitrary integer. For each prime $p$ in $P$, the difference $f(p)-f(x)$ divides $p^{n}-x^{n}$. Using the equality $f(p)=p^{d}$, we get
$$p^{n}-x^{n}=p^{r}\left(p^{d}\right)^{m}-x^{n} \equiv p^{r} f(x)^{m}-x^{n} \equiv 0 \quad\left(\bmod p^{d}-f(x)\right)$$
Since we have $r<d$, for large enough primes $p \in P$ we obtain
$$\left|p^{r} f(x)^{m}-x^{n}\right|<p^{d}-f(x) .$$
Hence $p^{r} f(x)^{m}-x^{n}$ has to be zero. This implies $r=0$ and $x^{n}=\left(x^{d}\right)^{m}=f(x)^{m}$. Since $m$ is odd, we obtain $f(x)=x^{d}$.