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
     ${\displaystyle \sum _{d|n}\lambda \left(d\right)=1_{\text{PerfectSquare}}\left(n\right)}$
using the Möbius Inversion formula, we have
     ${\displaystyle \sum _{d|n}\mu \left(\frac{n}{d}\right)1_{\text{PerfectSquare}}\left(d\right)=\lambda \left(n\right)}$
The summands will be zero except when ${\displaystyle d=D^2}$.  Hence we get
     ${\displaystyle \sum _{D^2|n}\mu \left(\frac{n}{D^2}\right)=\lambda \left(n\right)}$
[End]