📂Number Theory

Arithmetic Functions' Multiplicative Properties

Definition ¹

1. For $\forall n \in \mathbb{N}$, if an arithmetic function $f$ that is not $f(n) = 0$ satisfies the following, it is called a multiplicative function. $$f(mn) = f(m) f(n) \qquad,\gcd(m,n)=1$$
2. If a multiplicative function satisfies the following condition, it is called a completely multiplicative function. $$f(mn) = f(m) f(n) \qquad,m,n \in \mathbb{N}$$

Basic Properties

• [1]: If $f$ is multiplicative, then $f(1) = 1$.
• [2]: That $f$ is a multiplicative function is equivalent to for all primes $p_{1} , \cdots , p_{r}$ and all $a_{1} , \cdots, a_{r} \in \mathbb{N}$, $f \left( p_{1}^{a_{1}} \cdots p_{r}^{a_{r}} \right) = f \left( p_{1}^{a_{1}} \right) \cdots f \left( p_{r}^{a_{r}} \right)$.
• [3]: If $f$ is multiplicative, for primes $p$ that satisfy $p \mid n$, $$\sum_{d \mid n} \mu (d) f(d) = \prod_{p \mid n} \left( 1 - f(p) \right)$$
• [4]: If $f$ is multiplicative, then $f$ being a completely multiplicative function is equivalent to for all primes $p$ and all $a \in \mathbb{N}$, $f \left( p^{a} \right) = \left[ f(p) \right]^{a}$.
• [5]: If $f$ is multiplicative, then $f$ being a completely multiplicative function is equivalent to the inverse of $f$ with respect to Dirichlet convolution $f^{-1}$ being represented as follows. $$f^{-1} (n) = \mu (n) f (n)$$
• [6]: If $f$ is completely multiplicative, then $F(n) := \sum_{d \mid n} f(d)$ is multiplicative.

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• $\mu$ is the Möbius function.

Description

Naturally, being completely multiplicative implies being multiplicative.

An arithmetic function being multiplicative is akin to the ‘independent’ condition spoken of across mathematics. Since function values can be thought of in smaller fragments, it naturally harbors rich mathematical properties.

Theorem [1] is derived very simply but implies $f(1) \ne 0$, which becomes a critical condition for the set of multiplicative functions to form a group.

Proof

[1]

Since $f(1) = f( 1 \cdot 1) = f(1) f(1)$, $f(1) = 1$ must hold.

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[2]

$(\Rightarrow)$

Since $f$ is multiplicative and always $\gcd \left( p^{a} , q^{b} \right) = 1$ for different primes $p \ne q$, $$\begin{align*} f \left( p_{1}^{a_{1}} \cdots p_{r}^{a_{r}} \right) =& f \left( p_{1}^{a_{1}} \right) f \left( p_{2}^{a_{2}} \cdots p_{r}^{a_{r}} \right) \\ =& f \left( p_{1}^{a_{1}} \right) f \left( p_{2}^{a_{2}} \right) f \left( p_{3}^{a_{3}} \cdots p_{r}^{a_{r}} \right) \\ =& f \left( p_{1}^{a_{1}} \right) \cdots f \left( p_{r}^{a_{r}} \right) \end{align*}$$

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$(\Leftarrow)$

If we set $m = p_{1}^{a_{1}} \cdots p_{r}^{a_{r}}$, $n = q_{1}^{b_{1}} \cdots q_{s}^{b_{s}}$, and $\gcd(m,n) =1$, then $$\begin{align*} f(mn) =& f \left( p_{1}^{a_{1}} \cdots p_{r}^{a_{r}} q_{1}^{b_{1}} \cdots q_{s}^{b_{s}} \right) \\ =& f \left( p_{1}^{a_{1}} \right) \cdots f \left( p_{r}^{a_{r}} \right) f \left( q_{1}^{b_{1}} \right) \cdots f \left( q_{s}^{b_{s}} \right) \\ =& f \left( p_{1}^{a_{1}} \cdots p_{r}^{a_{r}} \right) f \left( q_{1}^{b_{1}} \cdots q_{s}^{b_{s}} \right) \\ =& f(m)f(n) \end{align*}$$

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[3]

Strategy: The properties of the Möbius function must be used.

Definition of the Möbius function: Given a prime $p_{1} , \cdots , p_{k}$, let a natural number $n$ be represented as $n = p_{1}^{a_{1}} \cdots p_{k}^{a_{k}}$. The arithmetic function $\mu$ defined as follows is called the Möbius function. $$\mu (n) := \begin{cases} 1 &, n=1 \\ (-1)^{k} &, a_{1} = \cdots = a_{k} = 1 \\ 0 & , \text{otherwise} \end{cases}$$ Multiplicativity: For all $m, n \in \mathbb{N}$ that satisfy $\gcd (m,n) = 1$, $\mu (mn) = \mu (m) \mu (n)$

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$$g(n) := \sum_{d \mid n} \mu (d) f(d)$$ And let’s say $m$ satisfies $g(m,n) = 1$. Then, for $d = ab$ that satisfies $a \mid m$ and $b \mid n$, $\gcd(a,b) = 1$ also holds, so we can consider sigma separately. Also, since $\mu$ is multiplicative and $f$ is multiplicative by the premise, $$\begin{align*} g(mn) =& \sum_{ab \mid mn} \mu (ab) f(ab) \\ =& \sum_{a \mid m \\ b \mid n} \mu (ab) f(ab) \\ =& \sum_{a \mid m \\ b \mid n} \mu (a) \mu (b) f(a) f(b) \\ =& \sum_{a \mid m} \mu (a) f(a) \sum_{b \mid n} \mu (b) f(b) \\ =& g(m)g(n) \end{align*}$$

Therefore, $g$ is also multiplicative. Then, according to Theorem [2], $g \left( p_{1}^{a_{1}} \cdots p_{r}^{a_{r}} \right) = g \left( p_{1}^{a_{1}} \right) \cdots g \left( p_{r}^{a_{r}} \right)$, so we only need to calculate $g \left( p^{a} \right)$ for a fixed prime $p$. According to Theorem [1], $f(1)=1$ and $\mu (1) = 1$, and for higher orders $k \ge 2$, all $\mu (p^{k}) = 0$, $$\begin{align*} g \left( p^{a} \right) =& \sum_{d \mid p^{a}} \mu (d) f(d) \\ =& \mu (1) f(1) + \mu (p) f(p) \\ =& 1 \cdot 1 - 1 \cdot f(p) \\ =& 1 - f(p) \end{align*}$$ Summarizing for natural numbers $n = p_{1}^{a_{1}} \cdots p_{r}^{a_{r}}$ results in $$g(n) = \prod_{p \mid n} g \left( p^{a} \right) = \prod_{p \mid n} \left( 1 - f(p) \right)$$

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[4]

$(\Rightarrow)$

Since $f$ is completely multiplicative, $$f \left( p^{a} \right) = f(p) \left( p^{a-1} \right) = \left[ f(p) \right]^{2} \left( p^{a-2} \right) = \cdots = \left[ f \left( p \right) \right]^{a}$$

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$(\Leftarrow)$

Let’s consider that for a prime $p_{1} , \cdots , p_{s}$, an integer $a_{1} , \cdots, a_{s} , b_{1} , \cdots , b_{s} \in \mathbb{N}_{0}$, and a natural number $c_{1}, \cdots , c_{t} \in \mathbb{N}$, any $m$ and $n$ are represented as follows. $$m = p_{1}^{a_{1}} \cdots p_{t}^{a_{t}} \\ n = p_{1}^{b_{1}} \cdots p_{t}^{b_{t}} \\ mn = p_{1}^{c_{1}} \cdots p_{t}^{c_{t}}$$ If $m$ and $n$ do not have prime $p_{i}$ as a divisor, then $a_{i} = 0$ and $b_{i} = 0$, respectively. Since $f$ is multiplicative, $$\begin{align*} f(mn) =& f \left( p_{1}^{c_{1}} \cdots p_{t}^{c_{t}} \right) \\ =& f \left( p_{1}^{c_{1}}) \cdots f( p_{t}^{c_{t}} \right) \\ =& \left[ f ( p_{1}) \right]^{c_{1}} \cdots \left[ f ( p_{t}) \right]^{c_{t}} \\ =& \left[ f ( p_{1}) \right]^{a_{1}} \left[ f ( p_{1}) \right]^{b_{1}} \cdots \left[ f ( p_{t}) \right]^{a_{t}} \left[ f ( p_{t}) \right]^{b_{t}} \\ =& \left[ f ( p_{1}) \right]^{a_{1}} \cdots \left[ f ( p_{t}) \right]^{a_{t}} \left[ f ( p_{1}) \right]^{b_{1}} \cdots \left[ f ( p_{t}) \right]^{b_{t}} \\ =& \left[ f ( p_{1}^{a_{1}}) \right] \cdots \left[ f ( p_{t}^{a_{t}}) \right] \left[ f ( p_{1}^{b_{1}}) \right] \cdots \left[ f ( p_{t}^{b_{t}}) \right] \\ =& f \left( p_{1}^{a_{1}} \cdots p_{t}^{a_{t}} \right) f \left( p_{1}^{b_{1}} \cdots p_{t}^{b_{t}} \right) \\ =& f(m)f(n) \end{align*}$$

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[5]

$(\Rightarrow)$

Let’s say $g(n) := \mu (n) f(n)$.

Since $f$ is completely multiplicative and $\displaystyle \sum_{d \mid n } \mu (d) = I(n)$, $$ \begin{align*} (gf)(n) =& \sum_{d \mid n} \mu (d) f (d) f \left( {{ n } \over { d }} \right) \\ =& f(n) \sum_{d \mid n} \mu (d) \\ =& f(n) I(n) \end{align} $$ In Theorem [1], since $f(1) = 1$ and the identity function $I$ is $I(n) = 0$ in $n > 1$, $f(n) I(n) = I(n)$. Summarizing, $(g \ast\ f) (n) = I (n)$, so $g = f^{-1}$.

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$(\Leftarrow)$

For any prime $p$ and natural number $a \in \mathbb{N}$, let’s set $n = p^{a}$.

If $f^{-1} (n) := \mu (n) f (n)$ then $$\begin{align*} \sum_{d \mid n} \mu (d) f (d) f \left( {{ n } \over { d }} \right) =& \sum_{d \mid n} f^{-1} (d) f \left( {{ n } \over { d }} \right) \\ =& (f^{-1} \ast\ f ) (n) \\ =& I(n) \\ =& 0 \end{align*}$$

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

According to the definition of the Möbius function, if $k \ge 2$ then $\mu (p^{k} ) = 0$, $$\mu (1) f(1) f \left( p^{a} \right) + \mu (p) f(p) f \left( p^{a-1} \right) + 0 + \cdots + 0 = 0$$ Since the premise is that $f$ is a multiplicative function, according to Theorem [1], $f(1) = 1$. Meanwhile, the Möbius function is $\mu (1) = 1$ and $\mu (p) = -1$, therefore, $$1 \cdot 1 \cdot f \left( p^{a} \right) + (-1) \cdot f(p) f \left( p^{a-1} \right) = 0$$ Summarizing, $$f \left( p^{a} \right) = f(p) f \left( p^{a-1} \right)$$ When recursively resolved, $f \left( p^{a} \right) = f(p)^{a}$, therefore, by Theorem [4], $f$ is completely multiplicative.

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[6]

If we assume $\gcd (m,n) = 1$, we can distinguish the divisors of $m$ from those of $n$. For convenience, let’s say $a_{1} , \cdots, a_{M}$ are the divisors of $m$, and $b_{1} , \cdots , b_{N}$ are the divisors of $n$. $$\begin{align*} F(mn) =& \sum_{d \mid mn} f(d) \\ =& f(1) + f(a_{1}) + \cdots + f (a_{M}) + f(b_{1}) + \cdots + f(b_{N}) \\ & + f(a_{1}b_{N}) + \cdots + f(a_{M} b_{1}) + \cdots + f(a_{1} a_{2} b_{N}) + \cdots + f(mn) \\ =& \sum_{a \mid m} f(a) \sum_{b \mid n} f(b) \\ =& F(m) F(n) \end{align*}$$

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