Exercise 4.3.13 ($\sum_{d \mid n} d \mu(d) = (-1)^{\omega(n)} \phi(n) \frac{s(n)}{n}$ where $s(n) = \sum_{p \mid n} p$)

Proof. If $n=p_1^{a_1}{p}_2^{a_2}\cdots p_l^{a_l}$ and $m=q_1^{b_1}{q}_2^{b_2}\cdots q_r^{b_r}$ are relatively prime, where $p_1,\dots ,p_l,q_1,\dots q_r$ are distinct prime, then $\mathit{nm}=p_1^{a_1}{p}_2^{a_2}\cdots p_l^{a_l}{q}_1^{b_1}{q}_2^{b_2}\cdots q_r^{b_r}$ is the decomposition of $\mathit{nm}$ in primes, so

so $s$ is a multiplicative function.

We define for all positive integers $n$,

Since $\mu ,\varphi ,s$ and $n\mapsto {(-1)}^{\omega }(n)$, $n\mapsto n$ are multiplicative functions, so are $F$ and $G$.

Moreover $F(1)=G(1)=1$, and if $\alpha \ge 1$,

and

Since $F(p^{\alpha })=G(p^{\alpha })$ for all $\alpha \ge 0$, where $F$ and $G$ are multiplicative, we obtain $F=G$.

For all positive integers $n$,

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