Taiwan IMO Selection Camp

For such an $f$ as described in the problem, if the constant term of $f$ is 0, then we may consider $g(x) = f(x)/x$, which also satisfies the conditions of the problem; therefore we may assume that the constant term of $f$ is nonzero, and it suffices to prove that in this case $f$ is a constant polynomial.

Let $P = \{p$ is a prime $|$ there exists a positive integer $n$ such that $f(n)$ is a multiple of $p$, but $n$ is not a multiple of $p\}$. We claim that $P$ contains at most 2011 primes.

If not, then by contradiction, let $p_1, \cdots, p_{2012}$ be distinct primes in $P$, and let positive integers $n_1, \cdots, n_{2012}$ respectively satisfy $p_i|f(n_i)$ and $p_i \nmid n_i$.

By the Chinese Remainder Theorem, there exists a positive integer $n$ such that $n \equiv n_i \pmod{p_i}$, $i = 1, \cdots, 2012$,

since $n - n_i | f(n) - f(n_i)$. We know that $p_1p_2\cdots p_{2012} | f(n)$. At the same time, by definition $n$ is coprime to $p_1p_2\cdots p_{2012}$, a contradiction. Therefore $P$ contains at most 2011 primes; denote $P = \{p_1, \cdots, p_k\}$.

Let $f(x) = a + xg(x)$, that is, $a$ is the nonzero constant term of $f$, and $g$ is some polynomial with integer coefficients. Suppose $f$ is not a constant polynomial, then $g$ is nonzero. Now let $m = 3a^2p_1p_2 \cdots p_k$. Then $f(m) = a(1 + 3ap_1p_2 \cdots p_k g(m))$, and it is clear that $1 + 3ap_1p_2 \cdots p_k g(m)$ is coprime to $m$, and
$$|1 + 3ap_1p_2 \cdots p_k g(m)| \ge |3ap_1p_2 \cdots p_k g(m)| - 1 \ge 3 - 1 > 1.$$
Therefore $1 + 3ap_1p_2 \cdots p_k g(m)$ has at least one prime factor that $m$ does not have, contradicting the definition of $P$; hence $f$ must be a constant polynomial. This completes the proof.