MathDB
GCD (f(m),n ) + LCM (m,f(n)) = LCM(f(m),n ) + GCD (m,f(n))

Source: 2021 Francophone MO Juniors p4

April 3, 2021
number theorygreatest common divisorleast common multiplefunctional equationfunctionalFrancophone

Problem Statement

Let N1\mathbb{N}_{\geqslant 1} be the set of positive integers. Find all functions f ⁣:N1N1f \colon \mathbb{N}_{\geqslant 1} \to \mathbb{N}_{\geqslant 1} such that, for all positive integers mm and nn: GCD(f(m),n)+LCM(m,f(n))=GCD(m,f(n))+LCM(f(m),n).\mathrm{GCD}\left(f(m),n\right) + \mathrm{LCM}\left(m,f(n)\right) = \mathrm{GCD}\left(m,f(n)\right) + \mathrm{LCM}\left(f(m),n\right).
Note: if aa and bb are positive integers, GCD(a,b)\mathrm{GCD}(a,b) is the largest positive integer that divides both aa and bb, and LCM(a,b)\mathrm{LCM}(a,b) is the smallest positive integer that is a multiple of both aa and bb.