Clearly for all $ m,n \in \mathbb{Z}$, $ 41|f(a+41m)-f(a)$ and $ 49|f(b+49n)-f(b)$ but since $ 41|f(a)$ and $ 49|f(b)$, then $ 41|f(a+41m)$ and $ 49|f(b+49n)$ for all integers $ m$ and $ n$. Since $ (41,49)=1$, then there exists $ m,n$ such that $ a+41m=b+49n=c$, and thus $ 2009|f(c)$ in this case.