Exercise 4.3.7 ($\sum_{\delta \mid n} \mu(\delta) d(\delta) = (-1)^{\omega(n)}$)

Proof.

(a)

We define for every positive integer $n$, $$F(n)=\underset{\delta \mid n}{\sum }\mu (\delta )d(\delta ),G(n)={(-1)}^{\omega (n)}.$$

We show first that $F$ and $G$ are multiplicative functions.

• We suppose that $n\wedge m=1$. Since $\omega (\mathit{nm})=\omega (n)+\omega (m)$,

  $$G(\mathit{nm})={(-1)}^{\omega (\mathit{nm})}={(-1)}^{\omega (n)+\omega (m)}={(-1)}^{\omega (n)}{(-1)}^{\omega (m)}=G(n)G(m).$$

• We know that $\mu$ and $d$ are multiplicative functions, thus their product $\mathit{\mu d}$ is also multiplicative. By Theorem 4.8, $F$ is multiplicative.

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

$$\begin{array}{llll}\hfill F(p^{\alpha })& =\underset{\delta \mid p^{\alpha }}{\sum }\mu (\delta )d(\delta )& \hfill & \\ \hfill & =\underset{i=0}{\overset{\alpha }{\sum }}\mu (p^i)d(p^i)& \hfill & \\ \hfill & =\mu (1)d(1)+\mu (p)d(p)& \hfill & \\ \hfill & =1-2& \hfill & \\ \hfill & =-1& \hfill & \\ \hfill & =G(p^{\alpha }).& \hfill & \end{array}$$

Therefore $F=G$. So for all positive integer $n$,

$$\underset{\delta \mid n}{\sum }\mu (\delta )d(\delta )={(-1)}^{\omega (n)}.$$

(b)

Here we define $$H(n)=\underset{d\mid n}{\sum }\mu (d)\sigma (d).$$

The same argument given in part (a) shows that $H$ is a multiplicative function.

Moreover $H(1)=1$, and if $\alpha >1$,

$$\begin{array}{llll}\hfill H(p^{\alpha })& =\underset{i=0}{\overset{\alpha }{\sum }}\mu (p^i)\sigma (p^i)& \hfill & \\ \hfill & =\sigma (1)-\sigma (p)& \hfill & \\ \hfill & =-p.& \hfill & \end{array}$$

If $n=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_l^{{\alpha }_l}$, then

$$\begin{array}{llll}\hfill H(n)& =H(p_1^{{\alpha }_1})H(p_2^{{\alpha }_2})\cdots H(p_l^{{\alpha }_l})& \hfill & \\ \hfill & ={(-1)}^l{p}_1{p}_2\cdots p_l& \hfill & \\ \hfill & ={(-1)}^{\omega (n)}\underset{p\mid n}{\prod }p.& \hfill & \end{array}$$

Therefore, for all positive integers $n$,

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