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gcd(n, (n-m)/gcd(n,m)) = 1 for all m < n

Source: 2012 Indonesia Round 2.5 TST 1 Problem 4

May 10, 2012
number theorygreatest common divisormodular arithmeticnumber theory proposed

Problem Statement

Determine all integer n>1n > 1 such that gcd⁡(n,n−mgcd⁡(n,m))=1\gcd \left( n, \dfrac{n-m}{\gcd(n,m)} \right) = 1 for all integer 1≤m<n1 \le m < n.
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