Exercise 4.3.19 ($1/\phi(n) = \frac{1}{n} \sum_{d\mid n} \mu(d)^2/\phi(d)$)

Question

Show that $1/\phi(n) = \frac{1}{n} \sum_{d\mid n} \mu(d)^2/\phi(d)$ for all positive integers $n$.

Richard Ganaye · Accepted Answer

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
Consider the functions $F$ and $G$ defined on $\N^*$ by
$$F(n) = \frac{n}{\phi(n)},\qquad G(n) = \sum_{d \mid n} \frac{\mu^2(d)}{\phi(d)}.$$
We know that $\phi(n) 
e 0$ for all positive integers $n$. Since 
$$\frac{1}{F(n)} = \frac{\phi(n)}{n} = \prod_{p \mid n} \left( 1 - \frac{1}{p} ight),$$
$1/F = \mathrm{id}/\phi$ is a multiplicative function, thus $F$ is also multiplicative.

Next, $\mu$ and $\phi$ are multiplicative, thus $\mu^2/\phi$ is multiplicative. By Theorem 4.8, $G$ is multiplicative.

Moreover $F(1) = G(1) = 1$, and if $p$ is prime, and $\alpha \geq 1$,
\begin{align*}
F(p^\alpha) &= \frac{p^\alpha}{p^\alpha - p^{\alpha-1}} \
&= \frac{p}{p-1},
\end{align*}
and
\begin{align*}
G(p^\alpha) &= \sum_{i=0}^\alpha \frac{\mu(p^i)^2}{\phi(p^i)}\
&= 1 + \frac{\mu(p)^2}{\phi(p)}\
&= 1 + \frac{1}{p-1}\
&=\frac{p}{p-1},
\end{align*}
thus $F(p^\alpha) = G(p^\alpha)$ for all $\alpha \geq 0$. Since $F$ and $G$ are multiplicative, $F = G$, so for all positive integers $n$,
$$\frac{1}{\phi(n)} = \frac{1}{n}  \sum_{d \mid n} \frac{\mu^2(d)}{\phi(d)}.$$
\end{proof}

Exercise 4.3.19 ($1/\phi(n) = \frac{1}{n} \sum_{d\mid n} \mu(d)^2/\phi(d)$)

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Exercise 4.3.19 ($1/\phi(n) = \frac{1}{n} \sum_{d\mid n} \mu(d)^2/\phi(d)$)

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