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f(n+p) != f(n) for any n and prime p

Source: Balkan MO 2010, Problem 4

May 4, 2010
functionnumber theorygreatest common divisorrelatively primetotient functionnumber theory proposed

Problem Statement

For each integer nn (n≥2n \ge 2), let f(n)f(n) denote the sum of all positive integers that are at most nn and not relatively prime to nn. Prove that f(n+p)≠f(n)f(n+p) \neq f(n) for each such nn and every prime pp.
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