Exercise 4.3.22* ($\sum\limits_{ m \in \mathbb{N}^*:\, \phi(m) = n} \mu(m) = 0$)

Proof. By Problem 4.3.39, we know that $ℱ(n)$ is a finite set, so the sum $S_n=\underset{m\in ℱ(n)}{\sum <!--\; FUNCTION\; APPLICATION\; -->}\mu (m)$ is finite.

$ℱ(1)=\{1,2\}$, so $S_1=\mu (1)+\mu (2)=0$. We suppose now that $n>1$. We note that

where

(For instance, $𝒢(6)=\{7,14\}$.)

If $m\in 𝒢(n)$, then $m$ is square-free, and $m>1$ thus

where $p_i$ is a prime number for every $i$, so $p_1=2$ if and only if $m$ is even.

Moreover, if $m\in 𝒢(n)$ is odd, then $2m=2p_1{p}_2\cdots p_k$ is square-free, and

so $2m\in 𝒢(n)$.

Conversely, if $m\in 𝒢(n)$ is even, then $m=2m^{\prime }$, where $m^{\prime }$ is odd and square-free, and $\varphi (m)=\varphi (m^{\prime })$, so $m^{\prime }\in 𝒢(n)$.

Therefore we can group the terms of $𝒢(n)$ by pairs $\{m,2m\}$, where $m$ is odd, and

When $m$ is odd,

Thus $S=0$.

For every positive integer $n$,

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