, 2008

$f(x)$ can be any polynomial of the form
$$p a_p \binom{x}{p} + a_{p-1} \binom{x}{p-1} + a_{p-2} \binom{x}{p-2} + \dots + a_1 \binom{x}{1} + a_0,$$
where $a_0, a_1, \dots, a_p$ are arbitrary integers with $a_p \neq 0$.

Note that we can always write
$$f(x) = c_p \binom{x}{p} + c_{p-1} \binom{x}{p-1} + \cdots + c_1 \binom{x}{1} + c_0$$
for some rational coefficients $c_0, c_1, \dots, c_p$, since $\binom{x}{p}, \binom{x}{p-1}, \dots, \binom{x}{0}$ have different degrees. By putting $x = 0$, we get $c_0 = f(0) \in \mathbb{Z}$. By putting $x = 1$, we get $c_1 + c_0 = f(1) \in \mathbb{Z}$, and hence $c_1 \in \mathbb{Z}$. Similarly, by putting $x = 2, 3, \dots, p$, we find that all $c_j$'s are integers. Conversely, if all $c_j$'s are integers, it is obvious that $f(m) \in \mathbb{Z}$ for any $m \in \mathbb{Z}$.
Now, putting $x = p + 1$ and using $\binom{p+1}{k} = \binom{p}{k} + \binom{p}{k-1}$, we obtain
$$f(p+1) - f(1) = \sum_{k=0}^{p} c_k \binom{p+1}{k} - (c_1 + c_0) = \sum_{k=1}^{p} c_k \left[ \binom{p}{k} + \binom{p}{k-1} \right] - c_1.$$
Since $p \mid \binom{p}{r}$ whenever $1 \le r \le p-1$, we have
$$f(p+1) - f(1) \equiv c_p \pmod{p}.$$
Therefore, $p \mid f(p+1) - f(1)$ if and only if $p \mid c_p$. This shows $f(x)$ can be any polynomial of the form mentioned at the beginning.