MathDB
$n^2/2^n \neq m^2/2^m$ for Every $m$

Source: Problem 2, China National High School Mathematics League 2023

September 10, 2023
number theorychang 09

Problem Statement

For some positive integer nn, nn is considered a <spanclass=′latex−bold′>unique</span><span class='latex-bold'>unique</span> number if for any other positive integer m≠nm\neq n, {2nn2}≠{2mm2}\{\dfrac{2^n}{n^2}\}\neq\{\dfrac{2^m}{m^2}\} holds. Prove that there is an infinite list consisting of composite unique numbers whose elements are pairwise coprime.
Back to ProblemsView on AoPS