Multiplicative Functions

Definition 3.1   A function $f:\mathbb{N}\rightarrow \mathbb{Z}$ is multiplicative if, whenever $m, n\in\mathbb{N}$ and $\gcd(m,n)=1$, we have

$$f(mn) = f(m)\cdot f(n).$$

Recall that the Euler $\varphi$-function is

$$\varphi (n) = \char93 \{a : 1\leq a \leq n$$ and $$\gcd(a,n)=1\}.$$

Proposition 3.2   $\varphi$ is a multiplicative function.

Proof. Suppose that $m, n\in\mathbb{N}$ and $\gcd(m,n)=1$. Consider the map

$$\{c : \,1 \leq c \leq mn \gcd(c,mn)=1\} \xrightarrow{\quad f\quad }$$

$$\{a : 1 \leq a \leq m \gcd(a,m)=1\}\times \{b : 1 \leq b \leq n \gcd(b,n)=1\}\}$$

defined by

$$f(c) = (c$$ mod $$m, \quad{}c$$    mod $$n).$$

The map $f$ is injective: If $f(c)=f(c')$, then $m\mid c-c'$ and $n\mid c-c'$, so, since $\gcd(n,m)=1$, $nm\mid c-c'$, so $c=c'$.

The map $f$ is surjective: Given $a, b$ with $\gcd(a,m)=1$, $\gcd(b,n)=1$, the Chinese Remainder Theorem implies that there exists $c$ with $c\equiv a\pmod{m}$ and $c\equiv b\pmod{n}$. We may assume that $1\leq c\leq nm$, ans since $\gcd(a,m)=1$ and $\gcd(b,n)=1$, we must have $\gcd(c,nm)=1$. Thus $f(c)=(a,b)$.

Because $f$ is a bijection, the set on the left has the same size as the product set on the right. Thus

$$\varphi (mn) = \varphi (m)\cdot \varphi (n).$$

$\qedsymbol$

Example 3.3   The proposition makes it easier to compute $\varphi (n)$. For example,

$$\varphi (12) = \varphi (2^2)\cdot \varphi (3) = 2\cdot 2 = 4.$$

Also, for $n\geq 1$, we have

$$\varphi (p^n) = p^n - \frac{p^n}{p},$$

since $\varphi (p^n)$ is the number of numbers less than $p^n$ minus the number of those that are divisible by $p$. Thus, e.g.,

$$\varphi (389\cdot 11^2) = 388 \cdot (11^2 - 11) = 388\cdot 110 = 42680.$$

The $\varphi$ function is also available in PARI:

  ? eulerphi(389*11^2)
  %15 = 42680

Question 3.4   Is computing $\varphi ($1000 digit number$)$ really easy or really hard?