MathDB
Possible values of gcd

Source: Baltic Way 2014, Problem 19

November 11, 2014
number theorygreatest common divisornumber theory proposed

Problem Statement

Let mm and nn be relatively prime positive integers. Determine all possible values of gcd⁡(2m−2n,2m2+mn+n2−1).\gcd(2^m - 2^n, 2^{m^2+mn+n^2}- 1).
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