for $m=n=1$ we have $f(x)=2x+1$ as an example Assume $mn\neq 1$ Suppose such an $f$ exists. Pick a natural $t$ for which $f(t)\neq 0$ Set $a=u-t-1, b = u^2-u+t$ Where $u$ is an arbitrary integer larger than $t+1$ $mf(a)+nf(b) = mf(u-t-1)+nf(u^2-u+t) \equiv mf(-t-1)+nf(t) (mod u)$ We may assume that $mf(-t-1)+nf(t) = k(say)$ is non-zero since both $mf(-t-1)+nf(t)$ and $nf(-t-1)+mf(t)$ cannot be simultaneously zero Since $a+b+1 = u^2$ and $mf(a)+nf(b) \equiv k (mod u)$ we may choose $u$ s.t. $k$ is coprimme to $u$ So $(m,n)=(1,1)$ is the only solution