Number-theoretic functions take integers as input and spit out integers. The big ones — tau (divisor count), sigma (divisor sum), and phi (Euler's totient) — are all multiplicative, meaning you can compute them from prime factorizations by multiplying per-prime values. We unify everything under the multiplicative function framework, then introduce the Möbius function and its surprising role.
A function $f$ is multiplicative if whenever gcd(m, n) = 1:
\[ f(mn) = f(m) \cdot f(n) \]
Completely multiplicative: $f(mn) = f(m)f(n)$ for ALL m, n (no gcd condition).
Why this matters: if you know the prime factorization $n = \prod p_i^{a_i}$, then for any multiplicative function:
\[ f(n) = \prod f(p_i^{a_i}) \]
Just compute f on each prime power and multiply! This is why prime factorization is the universal key.
Also written d(n) or τ(n). Counts positive divisors of n.
\[ \tau(n) = \prod_{i=1}^{k} (a_i + 1) \quad \text{where } n = \prod p_i^{a_i} \]
On prime powers: $ \tau(p^a) = a + 1 $
Multiplicative: Yes, τ is multiplicative (but not completely).
Example: τ(12) = τ\(((2^{2} \\times 3^{1})\) = (2+1)(1+1) = 6. Divisors: {1,2,3,4,6,12}\) ✓
Sum of all positive divisors of n.
\[ \sigma(n) = \prod_{i=1}^{k} \frac{p_i^{a_i + 1} - 1}{p_i - 1} \]
On prime powers: $ \sigma(p^a) = 1 + p + p^2 + ... + p^a = \frac{p^{a+1}-1}{p-1} $ (geometric series)
Multiplicative: Yes, σ is multiplicative.
Example: σ(12) = σ\(((2^{2}) \\times\)\) σ\(((3^{1}) = (1+2+4)(1+3) = 7 \\times\) 4 = 28\) ✓
Counts integers from 1 to n that are coprime to n.
\[ \varphi(n) = n \prod_{p|n} \left(1 - \frac{1}{p}\right) \]
On prime powers: $ \varphi(p^a) = p^a - p^{a-1} = p^{a-1}(p-1) = p^a(1 - 1/p) $
Multiplicative: Yes, φ is multiplicative.
Example: φ(12) = φ\(((2^{2}) \\times\)\) φ\(((3^{1}) = 2 \\times\) 2 = 4. Coprime to 12: {1, 5, 7, 11}\) ✓
Classify n based on σ(n) vs 2n (or equivalently, sum of proper divisors vs n):
Euclid-Euler Theorem: Every even perfect number has the form $ 2^{p-1}(2^p - 1) $ where $2^p - 1$ is a Mersenne prime. No odd perfect numbers are known.
Defined as:
Examples:
Multiplicative: Yes, μ is multiplicative.
If $ g(n) = \sum_{d|n} f(d) $ for all n, then:
\[ f(n) = \sum_{d|n} \mu(d) \cdot g\left(\frac{n}{d}\right) \]
This is the "inclusion-exclusion of number theory." It lets you invert sum-over-divisors relationships.
Example: since $ n = \sum_{d|n} \varphi(d) $, Möbius inversion gives $ \varphi(n) = \sum_{d|n} \mu(d) \cdot \frac{n}{d} $.
Step 1: Prime factorization.
\(60 = 2^{2} \\times 3^{1} \\times 5^{1}\)
Step 2: σ is multiplicative. Compute σ on each prime power.
σ\((2^{2})\) = 1 + 2 + 4 = 7
σ\((3^{1})\) = 1 + 3 = 4
σ\((5^{1})\) = 1 + 5 = 6
Step 3: Multiply them together.
\[ \sigma(60) = \sigma(2^2) \times \sigma(3) \times \sigma(5) = 7 \times 4 \times 6 = 168 \]
Verify: Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Sum = 1+2+3+4+5+6+10+12+15+20+30+60 = 168 ✓
Answer: 168
Step 1: Prime factorization of 100.
\(100 = 2^{2} \\times 5^{2}\)
Step 2: φ is multiplicative. Compute on each prime power.
φ\((2^{2}) = 2^{2} - 2^{1}\) = 4 - 2 = 2
φ\((5^{2}) = 5^{2} - 5^{1}\) = 25 - 5 = 20
Step 3: Multiply.
\[ \varphi(100) = \varphi(2^2) \times \varphi(5^2) = 2 \times 20 = 40 \]
Alternative formula check:
φ\((100) = 100 \\times (1 - \\frac{1}{2}) \\times (1 - \\frac{1}{5}) = 100 \\times \\frac{1}{2} \\times\) ⅘ \(= 100 \\times\) ⅖ = 40 ✓
Intuition: Half the numbers are odd (not divisible by 2), and 4/5 aren't divisible by 5. \(\\frac{1}{2} \\times\) ⅘ = ⅖ of 100 = 40 numbers.
Answer: 40
Step 1: Prime factorization of 30.
\(30 = 2 \\times 3 \\times 5\)
Step 2: Check if square-free.
All exponents are 1. No squared primes. Square-free: yes.
Step 3: Count prime factors.
Three distinct prime factors (2, 3, 5).
Step 4: Apply Möbius definition.
\(Square-free + odd number of primes \\to\) μ(n) = -1
\[ \mu(30) = -1 \]
More practice:
Answer: -1