Questions: Arithmetic Functions and Multiplicativity

Question 1

A multiplicative function f satisfies f(4) = 3 and f(9) = 5. What is f(36)?

• A. 15, because gcd(4, 9) = 1 and f(36) = f(4) · f(9)
• B. 8, because f(36) = f(4) + f(9)
• C. Cannot be determined without knowing f on all prime powers
• D. 15, but only if f is also completely multiplicative

Answer: (A)

Since 36 = 4 · 9 and gcd(4, 9) = 1 (they share no prime factors: 4 = 2² and 9 = 3²), multiplicativity directly gives f(36) = f(4) · f(9) = 3 · 5 = 15. The coprimality condition is satisfied, so the standard definition applies. No additional condition is needed — multiplicativity suffices here because the inputs are coprime.

Question 2

A function f satisfies f(p²) = f(p)² for every prime p. Does this follow from multiplicativity alone?

• A. Yes — multiplicativity gives f(p · p) = f(p) · f(p) for all primes p
• B. No — multiplicativity only applies when gcd(m, n) = 1, but gcd(p, p) = p ≠ 1
• C. Yes — prime powers are a special case where the coprimality condition is waived
• D. No — this property requires knowing f on all integers, not just primes

Answer: (B)

Multiplicativity requires gcd(m, n) = 1. For p · p, gcd(p, p) = p ≠ 1, so the rule does not apply. Therefore f(p²) ≠ f(p)² in general for a merely multiplicative function. Euler's totient illustrates this: φ(p²) = p² − p, but φ(p)² = (p−1)², and these are different. For p = 5: φ(25) = 20 but φ(5)² = 16. Complete multiplicativity would give f(p²) = f(p)², but multiplicativity alone does not.

Question 3

A multiplicative function is entirely determined by its values on prime powers p^k.

• T. True
• F. False

Answer: True

By the Fundamental Theorem of Arithmetic, every positive integer n factors uniquely as n = p₁^{a₁} · p₂^{a₂} · ... · pₖ^{aₖ} where the prime power factors are pairwise coprime. Multiplicativity then gives f(n) = f(p₁^{a₁}) · f(p₂^{a₂}) · ... · f(pₖ^{aₖ}). Knowing f(p^k) for every prime p and every exponent k therefore determines f everywhere. This is what makes multiplicative functions computationally tractable.

Question 4

A substantially multiplicative function satisfies f(mn) = f(m)f(n) primarily when gcd(m, n) = 1.

• T. True
• F. False

Answer: False

This describes merely multiplicative, not completely multiplicative. A completely multiplicative function satisfies f(mn) = f(m)f(n) for ALL positive integers m and n, with no coprimality restriction. Complete multiplicativity is strictly stronger: every completely multiplicative function is multiplicative, but not vice versa. The identity function f(n) = n is completely multiplicative; Euler's totient φ is only multiplicative.

Question 5

Explain why the coprimality condition in the definition of multiplicativity is mathematically essential. What goes wrong if you naively apply f(p²) = f(p)·f(p) for a merely multiplicative function?

Model answer: The coprimality condition ensures that when we write n = m · k with gcd(m, k) = 1, the prime factorizations of m and k are completely disjoint — no prime is shared. Multiplicativity factors f over this disjoint union. If m and k share a prime factor, their product 'double-counts' that prime, and the multiplicativity rule is invalid. For f(p²) = f(p · p), gcd(p, p) = p ≠ 1, so the rule cannot be applied. Concretely, for Euler's totient: φ(p²) = p² − p (direct computation), but φ(p) · φ(p) = (p−1)². These are equal only when p−1 = p, which never happens.

This is why the values f(p^k) on prime powers must be established separately (via their own formulas or definitions), and only then can multiplicativity be used to compute f at composite numbers from coprime prime-power building blocks. The function factors over coprime pairs, not arbitrary pairs.