Questions: Möbius Function and Möbius Inversion

Question 1

What is μ(12)?

• A. 1, because 12 has two distinct prime factors (2 and 3)
• B. −1, because 12 has three prime factors counting multiplicity (2, 2, and 3)
• C. 0, because 12 = 2² · 3 contains a squared prime factor
• D. −1, because μ(6) = 1 and 12 = 2·6 so signs alternate

Answer: (C)

The Möbius function is 0 whenever n has any prime factor appearing to a power greater than 1. Since 12 = 2² · 3, the prime 2 appears squared, so μ(12) = 0. The tempting distractor (option A) counts distinct prime factors correctly — there are two, 2 and 3 — but ignores the squarefreeness requirement. The key rule: μ(n) can only be ±1 when n is squarefree.

Question 2

The Möbius inversion formula recovers f from g when g(n) = Σ_{d|n} f(d). Which identity is the algebraic engine that makes this inversion work?

• A. μ is a multiplicative arithmetic function
• B. Σ_{d|n} μ(d) = 1 for all n ≥ 1
• C. Σ_{d|n} μ(d) = 1 if n = 1 and 0 otherwise
• D. The Dirichlet series for μ(n) converges for Re(s) > 1

Answer: (C)

The identity Σ_{d|n} μ(d) = [n=1] is the orthogonality condition that makes inversion work. When you substitute g into the inversion formula and expand the double sum, cross terms vanish because Σ_{d|n} μ(d) = 0 for all n > 1, leaving only f(n). In Dirichlet convolution language, μ is the inverse of the constant function 1, so μ * 1 = ε (the multiplicative identity). Option A is true but insufficient: multiplicativity makes μ tractable, but the actual cancellation mechanism is the orthogonality identity.

Question 3

If n is a product of exactly three distinct primes, then μ(n) = −1.

• T. True
• F. False

Answer: True

If n = p·q·r for distinct primes p, q, r, then n is squarefree (no prime appears twice), so μ(n) = (−1)³ = −1. This is correct. The key is that 'product of k distinct primes' already implies squarefree — if the primes are distinct, none can appear twice. Compare with μ(12) = 0: 12 has two distinct prime factors but is not squarefree because 2 appears squared.

Question 4

Möbius inversion only works when f is a multiplicative function — it fails for arbitrary arithmetic functions.

• T. True
• F. False

Answer: False

The Möbius inversion formula f(n) = Σ_{d|n} μ(n/d)g(d) holds for any arithmetic function f — multiplicativity is not required. It is a general inversion result over the divisibility lattice. The formula follows from the single orthogonality identity Σ_{d|n} μ(d) = [n=1], which holds regardless of what f looks like. What multiplicativity helps with is efficient computation and nice Dirichlet series structure, not the inversion identity itself.

Question 5

Explain in your own words why the Möbius function μ can 'undo' a divisor sum. What property of μ makes the inversion formula work?

Model answer: The key property is the orthogonality identity Σ_{d|n} μ(d) = 1 if n = 1 and 0 if n > 1. In Dirichlet convolution language, μ * 1 = ε (where 1 is the constant function and ε is the identity). When g = 1 * f (meaning g(n) = Σ_{d|n} f(d)), convolving both sides with μ gives μ * g = μ * 1 * f = ε * f = f. The cancellation works because all the cross terms in the double sum vanish due to the orthogonality identity, leaving only f(n).

This is the Dirichlet convolution perspective: μ is the multiplicative inverse of the constant function 1 in the ring of arithmetic functions under Dirichlet convolution. The sign-oscillation of μ (values +1, −1, 0) is precisely calibrated to cancel accumulated contributions across all divisors — an 'orthogonality' analogous to Fourier series over the divisibility lattice.