MathDB
gcd(m,n)=d and gcd(m,4n+1)=1

Source: Mediterranean Mathematical Olympiad 2020 P1 MMC

September 23, 2020
number theorygreatest common divisor

Problem Statement

Determine all integers m≥2m\ge2 for which there exists an integer n≥1n\ge1 with gcd⁡(m,n)=d\gcd(m,n)=d and gcd⁡(m,4n+1)=1\gcd(m,4n+1)=1.
Proposed by Gerhard Woeginger, Austria
Back to ProblemsView on AoPS