Sum of Möbius function over Complementary Divisors of Square-Divisors

While considering the summatory function of the Liouville Function, we found that

which takes the value of 1 when n is a perfect square and 0 otherwise.  The following question was given as a challenge and we present its solution in this article.

Question
Find

Solution:-
From
     ` \sum_{d|n}\lambda(d)=\bb{1}_\text{PerfectSquare}(n) `
using the Möbius Inversion formula, we have
     ` \sum_{d|n}\mu(\frac{n}{d})\bb{1}_\text{PerfectSquare}(d)=\lambda(n) `
The summands will be zero except when ` d=D^2 `.  Hence we get
     ` \sum_{D^2|n}\mu(\frac{n}{D^2}) =\lambda(n) `
[End]

Liouville Function, Summatory of

We introduced the Liouville function here.  The summatory function of ` f(n) ` is defined as

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


Question
Prove that

Solution:-
   Note that for ` n=p^a ` where p is a prime number,
   ` \sum_{d|n}\lambda(d) `  =   ` \sum_{j=0}^a \lambda(p^j) `  =   ` \sum_{j=0}^a (-1)^j `
=   ` \frac{1\cdot(1-(-1)^{a+1})}{1-(-1)}`  =   ` \frac{1-(-1)^{a+1}}{2} `
=   ` \bb{1}_\text{even}(a) `  =   ` {(1,\text{if }a\text{ is even}),(0,\text{if }a\text{ is odd}):} `       ("Even indicator function")

Hence for ` n=p_1^{a_1}p_2^{a_2}...p_k^{a_k} `
   ` \sum_{d|n}\lambda(d) `  =   ` \sum_{i_1,i_2,...,i_k}\lambda(p_1^{i_1}p_2^{i_2}...p_k^{i_k}) ` =   ` \sum_{i_1,i_2,...,i_k}\lambda(p_1^{i_1})\lambda(p_2^{i_2})...\lambda(p_k^{i_k}) `
=   ` \sum_{i_1=0}^{a_1}\lambda(p_1^{i_1})\sum_{i_2=0}^{a_2}\lambda(p_2^{i_2})...\sum_{i_1=0}^{a_k}\lambda(p_k^{i_k}) `
=   ` \bb{1}_\text{even}(a_1)\bb{1}_\text{even}(a_2)...\bb{1}_\text{even}(a_k) ` 
=   ` {(1,\text{if }\forall i\text{ }a_i\in2ZZ),(0,\text{otherwise}):} `      =   ` {(1,\text{if }n\text{ is a perfect square}),(0,\text{otherwise}):} `

Challenge: Find ` \sum_{D^2|n}\mu(\frac{n}{D^2}) `, where ` \mu() ` is the Möbius function.  Solution here.