Austria 2010

Assume that $m$ divides $f(n)$ for some integer $n$ and some $2\le m\le 2010$. As $f(1)=2011$ and $2011$ is a prime number, $m$ cannot be a divisor of $f(1)$, so we may restrict ourselves to the case $n\ne 1$. In this case, we can write $f(n)$ as $$f(n)=\frac{n^{2011}-1}{n-1}.$$ Let $p$ be a prime divisor of $m$. Then we have $p\mid f(n)\mid n^{2011}-1$, which results in $$n^{2011}\equiv 1(\mathrm{mod}p).(1)$$ This immediately implies that $n$ and $p$ are coprime. By (1), the order ${\text{ord}}_p(n)$ of $n$ modulo $p$, i.e., the smallest positive exponent $k$ such that $n^k\equiv 1(\mathrm{mod}p)$, is a divisor of $2011$. As $2011$ is prime, this results in ${\text{ord}}_p(n)\in \{1,2011\}$. If ${\text{ord}}_p(n)=1$, we have $n\equiv 1(\mathrm{mod}p)$, which results in $0\equiv f(n)\equiv 2011(\mathrm{mod}p)$ (by the original definition of $f(n)$). This implies $p\mid 2011$ and therefore $p=2011$, contradiction. We conclude that ${\text{ord}}_p(n)=2011$. As ${\text{ord}}_p(n)$ always divides $\phi (p)=p-1$ by Fermat's theorem, we conclude that $p-1$ is a multiple of $2011$. This is a contradiction to $1<p<2011$. qed