What fundamental identity of the Möbius function makes Möbius inversion work — that is, allows f(n) to be recovered from g(n) = Σ_{d|n} f(d)?
Aμ is completely multiplicative: μ(mn) = μ(m)μ(n) for all m, n
BΣ_{d|n} μ(d) = 1 if n = 1 and 0 otherwise, collapsing all cross-terms in the inversion
Cμ(n) only takes the values −1, 0, and 1, keeping the inversion numerically bounded
Dμ is the unique arithmetic function satisfying μ(p) = −1 for every prime p
The inversion formula works because when you substitute g and rearrange, you get sums of the form Σ_{d|n} μ(d), which equal 1 only when n = 1 and 0 otherwise. This collapses every cross-term and isolates f(n). Knowing that μ only takes values in {−1, 0, 1} (option C) or that it is multiplicative (option A) does not by itself give the cancellation needed for inversion.
Question 2 Multiple Choice
A student computes μ(12) by reasoning: '12 = 2 × 2 × 3 has three prime factors (counting multiplicity), so μ(12) = (−1)³ = −1.' What is the correct value of μ(12), and what flaw is in the student's reasoning?
Aμ(12) = −1; the student counted prime factors correctly and applied the formula correctly
Bμ(12) = 0; since 12 = 2² × 3 contains a squared prime factor, 12 is not squarefree and μ(12) = 0 by definition
Cμ(12) = 1; the student should count only distinct primes (2 and 3), giving (−1)² = 1
Dμ(12) = −1; but the student's reasoning is flawed because the formula only uses distinct primes
The Möbius function assigns μ(n) = 0 whenever n is divisible by any perfect square greater than 1. Since 12 = 2² × 3 is divisible by 4 = 2², μ(12) = 0. The formula μ(n) = (−1)^k applies only to squarefree n — numbers with no repeated prime factor. Option C is also wrong: distinct-prime counting would give μ(12) = (−1)² = 1, but that too is incorrect because the squarefree check comes first.
Question 3 True / False
The Möbius function μ satisfies μ(mn) = μ(m)μ(n) whenever gcd(m, n) = 1. This means μ is mostly multiplicative.
TTrue
FFalse
Answer: False
μ is multiplicative (the weaker condition holding only for coprime inputs) but NOT completely multiplicative. Complete multiplicativity would require μ(mn) = μ(m)μ(n) for ALL m, n without any coprimality restriction. This fails: μ(4) = 0 (since 4 = 2² is not squarefree), but μ(2)·μ(2) = (−1)(−1) = 1. Multiplicativity and complete multiplicativity are distinct properties, and confusing them leads to errors when computing μ on prime powers.
Question 4 True / False
For every prime p, μ(p) = −1.
TTrue
FFalse
Answer: True
Every prime p is squarefree (it has no squared prime factor) and is the product of exactly k = 1 distinct prime. Therefore μ(p) = (−1)¹ = −1. This is one of the simplest and most consistent facts about μ — it always assigns −1 to primes.
Question 5 Short Answer
Explain why Möbius inversion is described as the 'deconvolution' operation for Dirichlet convolution, and give one example of a function relationship it can 'undo.'
Think about your answer, then reveal below.
Model answer: Dirichlet convolution of two arithmetic functions f and g is (f * g)(n) = Σ_{d|n} f(d)g(n/d). The divisor-sum transform g(n) = Σ_{d|n} f(d) is exactly the Dirichlet convolution of f with the constant function 1: g = f * 1. Möbius inversion recovers f by convolving g with μ, since μ is the Dirichlet inverse of 1 (they satisfy 1 * μ = ε, where ε(n) = [n=1]). A canonical example: since Σ_{d|n} φ(d) = n, Möbius inversion immediately gives φ(n) = Σ_{d|n} μ(n/d)·d.
The key insight is that the divisor-sum g = f * 1 is an operation that loses information about f by averaging over divisors. Möbius inversion reverses this by convolving with μ, the multiplicative inverse of 1 in the Dirichlet ring. Without knowing this algebraic structure, deriving formulas like φ(n) = Σ μ(n/d)·d requires an ad hoc calculation; with it, the formula is immediate.