Liouville Function, Summatory of

We introduced the Liouville function here.  The summatory function of ${\displaystyle f\left(n\right)}$ is defined as

In this post, we examine the summatory function of the Liouville function.


Question
Prove that

Solution:-
   Note that for ${\displaystyle n=p^a}$ where p is a prime number,
   ${\displaystyle \sum _{d|n}\lambda \left(d\right)}$  =   ${\displaystyle \sum _{j=0}^a\lambda \left(p^j\right)}$  =   ${\displaystyle \sum _{j=0}^a{(-1)}^j}$
=   ${\displaystyle \frac{1\cdot (1-{(-1)}^{a+1})}{1-(-1)}}$  =   ${\displaystyle \frac{1-{(-1)}^{a+1}}{2}}$
=   ${\displaystyle 1_{\text{even}}\left(a\right)}$  =   ${\displaystyle \{\begin{array}{ll}1& \text{if}a\text{is even}\\ 0& \text{if}a\text{is odd}\end{array}}$       ("Even indicator function")

Hence for ${\displaystyle n=p_1^{a_1}{p}_2^{a_2}...p_k^{a_k}}$
   ${\displaystyle \sum _{d|n}\lambda \left(d\right)}$  =   ${\displaystyle \sum _{i_1,i_2,...,i_k}\lambda (p_1^{i_1}{p}_2^{i_2}...p_k^{i_k})}$ =   ${\displaystyle \sum _{i_1,i_2,...,i_k}\lambda \left(p_1^{i_1}\right)\lambda \left(p_2^{i_2}\right)...\lambda \left(p_k^{i_k}\right)}$
=   ${\displaystyle \sum _{i_1=0}^{a_1}\lambda \left(p_1^{i_1}\right)\sum _{i_2=0}^{a_2}\lambda \left(p_2^{i_2}\right)...\sum _{i_1=0}^{a_k}\lambda \left(p_k^{i_k}\right)}$
=   ${\displaystyle 1_{\text{even}}\left(a_1\right)1_{\text{even}}\left(a_2\right)...1_{\text{even}}\left(a_k\right)}$ 
=   ${\displaystyle \{\begin{array}{ll}1& \text{if}\forall i{a}_i\in 2\mathbb{Z}\\ 0& \text{otherwise}\end{array}}$      =   ${\displaystyle \{\begin{array}{ll}1& \text{if}n\text{is a perfect square}\\ 0& \text{otherwise}\end{array}}$

Challenge: Find ${\displaystyle \sum _{D^2|n}\mu \left(\frac{n}{D^2}\right)}$, where ${\displaystyle \mu \left(\right)}$ is the Möbius function.  Solution here.