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]