📂Number Theory

The Moebius Function in Analytic Number Theory

Definition ¹

For a prime number $p_{1} , \cdots , p_{k}$, let’s express a natural number $n$ as follows. The arithmetic function $\mu$ defined as such is called the Möbius function. $$\mu (n) := \begin{cases} 1 &, n=1 \\ (-1)^{k} &, a_{1} = \cdots = a_{k} = 1 \\ 0 & , \text{otherwise} \end{cases}$$

Basic Properties

• [1] Möbius series: It’s an identity $I$. In other words, $$\sum_{d \mid n } \mu (d) = I(n)$$
• [2] Multiplicativity: For all $m, n \in \mathbb{N}$ that satisfy $\gcd (m,n) = 1$, $\mu (mn) = \mu (m) \mu (n)$

Description

$$\begin{matrix} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \mu (n) & 1 & -1 & -1 & 0 & -1 & 1 & -1 & 0 & 0 & 1 \\ \sum_{d \mid n} \mu (d) & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}$$ The Möbius function, simply put, is a function that only cares about numbers that are not multiplied by the same prime number $2$ or more times. If a prime number is not multiplied more than $2$ times, its sign only changes depending on whether the prime numbers are used an even or odd number of times. However, this is just an explanation of the definition. The Möbius function itself may not have an intuitive meaning, but it is a major function that appears endlessly throughout number theory, especially in analytic number theory.

Proof

[1]

Let’s say $n = p_{1}^{a_{1}} \cdots p_{k}^{a_{k}} > 1$, then according to the binomial theorem, $$\begin{align*} & \sum_{d \mid n } \mu (d) \\ =& \mu (1) \\ & + \mu (p_{1}) + \cdots + \mu (p_{k}) \\ & + \mu (p_{1}p_{2}) + \cdots + \mu (p_{k-1}p_{k}) \\ & \vdots \\ & + \mu (p_{1}p_{2} \cdots p_{k-1}p_{k}) \\ =& 1 \\ & + \underbrace{(-1) + \cdots + (-1)}_{k} \\ & + \underbrace{1 + \cdots + 1}_{k(k-1)/2} \\ & \vdots \\ & + (-1)^{k} \\ =& 1 + \binom{k}{1} (-1) + \binom{k}{2}(-1)^{2} + \cdots + \binom{k}{k} (-1)^{k} \\ =& [1 + (-1)]^{k} \end{align*}$$

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[2]

If either $m$ or $n$ has a prime square as a divisor, then from the definition of $\mu$, $\mu (mn) = 0$ and therefore $\mu (m) \mu (n) = 0$, then $$\mu (mn) = \mu (m) \mu (n)$$ Hence, let’s assume $m$ and $n$ are in the form where each prime is multiplied only once. $$m = p_{1} \cdots p_{s} \\ n = q_{1} \cdots q_{t}$$ Then, $$\mu (mn) = (-1)^{s + t} = (-1)^{s} (-1)^{t} = \mu (m) \mu (n)$$

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