Olympiad Function Question

Question

Does there exist a function $f: \mathbb{Z} \to \mathbb{Z}$ such that for each $k= 0$, $1$, $\ldots$, $2016$ and for each $m\in \mathbb{Z}$ the equation $f(x) + kx = m$ has at least one solution $x\in \mathbb{Z}$?

Answer 1

Yes let $x_{k,m} \in \mathbb Z$ be different for each pair $(k,m)$ and define $f(x_{k,m}) = m - k x_{k,m}$. This would even work for all $k,m \in \mathbb Z$ since there exist one-to-one mappings from $\mathbb Z \times \mathbb Z$ to $\mathbb Z$.