Question 121
Let $f: \mathbb{N} \to \mathbb{R}_{\gt 0}$ be a given increasing function that takes positive values, where $\mathbb{N}$ is the set of all positive integers and $\mathbb{R}_{\gt 0}$ the set of all positive real numbers.
For any pair $(m, n)$ of positive integers, we call it *disobedient* if $f(mn) \neq f(m)f(n)$. For any positive integer $m$, we call it *ultra-disobedient* if for any nonnegative integer $N$, there are always infinitely many positive integers $n$ satisfying that $(m, n), (m, n + 1), \dots, (m, n + N)$ are all *disobedient* pairs.
Show that if there exists some *disobedient* pair, then there exists some *ultra-disobedient* positive integer.
For any pair $(m, n)$ of positive integers, we call it *disobedient* if $f(mn) \neq f(m)f(n)$. For any positive integer $m$, we call it *ultra-disobedient* if for any nonnegative integer $N$, there are always infinitely many positive integers $n$ satisfying that $(m, n), (m, n + 1), \dots, (m, n + N)$ are all *disobedient* pairs.
Show that if there exists some *disobedient* pair, then there exists some *ultra-disobedient* positive integer.