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Found problems: 1

2023 Taiwan TST Round 1, A
Tags: Taiwan , cringe
Let $f:\mathbb{N}\to\mathbb{R}_{>0}$ be a given increasing function that takes positive values. For any pair $(m,n)$ of positive integers, we call it [i]disobedient[/i] if $f(mn)\neq f(m)f(n)$. For any positive integer $m$, we call it [i]ultra-disobedient[/i] if for any nonnegative integer $N$, there are always infinitely many positive integers $n$ satisfying that $(m,n), (m,n+1),\ldots,(m,n+N)$ are all disobedient pairs. Show that if there exists some disobedient pair, then there exists some ultra-disobedient positive integer. [i] Proposed by usjl[/i]