MathDB

4

Part of 2021 Francophone Mathematical Olympiad

Problems(2)

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

Source: 2021 Francophone MO Juniors p4

4/3/2021
Let N⩾1\mathbb{N}_{\geqslant 1} be the set of positive integers. Find all functions f ⁣:N⩾1→N⩾1f \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.
number theorygreatest common divisorleast common multiplefunctional equationfunctionalFrancophone
n = (f(2n)-f(n) )(2 f(n) - f(2n) )

Source: 2021 Francophone MO Seniors p4

4/3/2021
Let N≥1\mathbb{N}_{\ge 1} be the set of positive integers. Find all functions f ⁣:N≥1→N≥1f \colon \mathbb{N}_{\ge 1} \to \mathbb{N}_{\ge 1} such that, for all positive integers mm and nn:
(a) n=(f(2n)−f(n))(2f(n)−f(2n))n = \left(f(2n)-f(n)\right)\left(2 f(n) - f(2n)\right), (b)f(m)f(n)−f(mn)=(f(2m)−f(m))(2f(n)−f(2n))+(f(2n)−f(n))(2f(m)−f(2m))f(m)f(n) - f(mn) = \left(f(2m)-f(m)\right)\left(2 f(n) - f(2n)\right) + \left(f(2n)-f(n)\right)\left(2 f(m) - f(2m)\right), (c) m−nm-n divides f(2m)−f(2n)f(2m)-f(2n) if mm and nn are distinct odd prime numbers.
number theoryfunctional equationfunctionalalgebraFrancophone