phi gcd >= gcd phi
Source: Indonesia IMO 2010 TST, Stage 1, Test 4, Problem 4
November 12, 2009
number theorygreatest common divisorinequalitiesnumber theory proposed
Problem Statement
Prove that for all integers and , the inequality
\dfrac{\phi(\gcd(2^m \plus{} 1,2^n \plus{} 1))}{\gcd(\phi(2^m \plus{} 1),\phi(2^n \plus{} 1))} \ge \dfrac{2\gcd(m,n)}{2^{\gcd(m,n)}}
holds.
Nanang Susyanto, Jogjakarta