\[\frac{f(m) - f(n)}{f(m-n)}=k\]Setting n = 0, \[f(m) | f(0)\]Setting m = 0, \[f(-n)| f(n)\]Setting m = 0, n = -n, \[f(n)| f(-n)\]\[f(n) = f(-n)\]Setting m = 2n, \[f(n) |f(2n)\]\[f(2n) \ge f(n)\]Setting n as -n \[f(m+n)|(f(m) - f(n))\]\[f(m) - f(n) \ge f(m+n)\] Assume \[f(m+n) > f(m)\], then\[f(m+n) > f(m) - f(n)\]However, this contradicts to \[f(m) - f(n) \ge f(m+n)\], which means \[f(m+n) \] is either smaller or equal to f(m).
Assume \[f(m+n) < f(m)\]then setting m = n, \[f(2n) < f(n)\], which contradicts to \[f(2n) \ge f(n)\], Thus \[f(m+n) = f(m)\] Substitute this back to the original equation\[\frac{f(m+n)-f(n)}{f(m)}\]Thus, we proved \[f(m) | f(n)\]