Exercise 4.2.20 (Liouville's lambda function)

Proof.

(a)

Let $n,m$ be positive integers. We write $$n=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_l^{{\alpha }_l},m=p_1^{{\beta }_1}{p}_2^{{\beta }_2}\cdots p_l^{{\beta }_l},\mathit{nm}=p_1^{{\gamma }_1}{p}_2^{{\gamma }_2}\cdots p_l^{{\gamma }_l}$$

the prime decompositions of $n,m,\mathit{nm}$, where ${\alpha }_i\ge 0,{\beta }_i\ge 0,{\gamma }_i\ge 0$, and ${\gamma }_i={\alpha }_i+{\beta }_i$, $i=1,2,\dots ,l$.

Then

$$\mathrm{\Omega }(n)=\underset{i=1}{\overset{l}{\sum }}{\alpha }_i,\mathrm{\Omega }(m)=\underset{i=1}{\overset{l}{\sum }}{\beta }_i,$$

and

$$\begin{array}{llll}\hfill \mathrm{\Omega }(\mathit{nm})& =\underset{i=1}{\overset{l}{\sum }}{\gamma }_i& \hfill & \\ \hfill & =\underset{i=1}{\overset{l}{\sum }}{\alpha }_i+\underset{i=1}{\overset{l}{\sum }}{\beta }_i& \hfill & \\ \hfill & =\mathrm{\Omega }(n)+\mathrm{\Omega }(m).& \hfill & \end{array}$$

Therefore

$$\begin{array}{llll}\hfill \lambda (\mathit{nm})& ={(-1)}^{\mathrm{\Omega }(\mathit{nm})}& \hfill & \\ \hfill & ={(-1)}^{\mathrm{\Omega }(n)+\mathrm{\Omega }(m)}& \hfill & \\ \hfill & ={(-1)}^{\mathrm{\Omega }(n)}{(-1)}^{\mathrm{\Omega }(m)}& \hfill & \\ \hfill & =\lambda (n)\lambda (m).& \hfill & \end{array}$$

This shows that $\lambda$ is totally multiplicative.

(b)

We define $F(n)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{d\mid n}\lambda (n)$. Since $\lambda$ is multiplicative, by Theorem 4.4, $F$ is also multiplicative.

Using $\lambda (p^a)={(-1)}^a$ for $p$ prime and $a\in \mathbb{N}$, we obtain for every $\alpha \in \mathbb{N}$,

$$\begin{array}{llll}\hfill F(p^{\alpha })& =\underset{d\mid p^{\alpha }}{\sum }\lambda (d)& \hfill & \\ \hfill & =\underset{a=0}{\overset{\alpha }{\sum }}\lambda (p^a)& \hfill & \\ \hfill & =\underset{a=0}{\overset{\alpha }{\sum }}{(-1)}^a& \hfill & \\ \hfill & & \hfill \end{array}$$

Therefore

$$F(p^{\alpha })=\frac{1-{(-1)}^{\alpha }}{2}=\{\begin{array}{cc}1\hfill & \text{if }\alpha \text{ is even},\hfill \\ 0\hfill & \text{otherwise.}\hfill \end{array}$$

If $n=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_l^{{\alpha }_l}$ is any positive integer, since $F$ is multiplicative,

$$\begin{array}{llll}\hfill F(n)& =\frac{1-{(-1)}^{{\alpha }_1}}{2}\frac{1-{(-1)}^{{\alpha }_2}}{2}\cdots \frac{1-{(-1)}^{{\alpha }_l}}{2}& \hfill & \\ \hfill & =\{\begin{array}{cc}1\hfill & \text{if }{\alpha }_1,\dots ,{\alpha }_n\text{ are even},\hfill \\ 0\hfill & \text{otherwise.}\hfill \end{array}& \hfill & \end{array}$$

Since $n$ is a perfect square if and only if all exponents ${\alpha }_i$ are even, we obtain

$$\underset{d\mid n}{\sum }\lambda (d)=\{\begin{array}{cc}1\hfill & \text{if }n\text{ is a perfect square}\hfill \\ 0\hfill & \text{otherwise.}\hfill \end{array}$$