Möbius Inversion with Liouville Function

This article concerns Möbius Inversion with the Liouville Function.  For ${\displaystyle n=p_1^{a_1}{p}_2^{a_2}...p_k^{a_k}}$, the number of distinct prime factors of n is

Question
(i)  Prove that

(ii) Hence show that

Solution:-
(i) Note that ${\displaystyle \mu \left(d\right)=0}$ for divisors d containing any prime power of index 2 or higher.  We only need to consider divisors of the form ${\displaystyle d=p_{i_1}{p}_{i_2}...p_{i_h}}$, where ${\displaystyle h=0}$ is interpreted as ${\displaystyle d=1}$.
    ${\displaystyle \sum _{d|n}\mu \left(d\right)\lambda \left(d\right)}$
= ${\displaystyle \sum _{h=0}^k\sum _{p_{i_1},p_{i_2},...,p_{i_h}}\mu (p_{i_1}{p}_{i_2}...p_{i_h})\lambda (p_{i_1}{p}_{i_2}...p_{i_h})}$
= ${\displaystyle \sum _{h=0}^k\sum _{p_{i_1},p_{i_2},...,p_{i_h}}{(-1)}^h{(-1)}^{1+1+...+1}}$   (where 1+1+...+1 has h copies of 1)
= ${\displaystyle \sum _{h=0}^k\sum _{p_{i_1},p_{i_2},...,p_{i_h}}{(-1)}^{2h}}$ = ${\displaystyle \sum _{h=0}^k\sum _{p_{i_1},p_{i_2},...,p_{i_h}}1}$ = ${\displaystyle \sum _{h=0}^k\left(\begin{array}{c}k\\ h\end{array}\right)}$
= ${\displaystyle 2^k}$  = ${\displaystyle 2^{\omega \left(n\right)}}$
(ii)  Recall that ${\displaystyle \lambda \left(\frac{n}{d}\right)=\frac{\lambda \left(n\right)}{\lambda \left(d\right)}=\lambda \left(n\right)\lambda \left(d\right)}$
     From the result in part (i) we multiply both sides by ${\displaystyle \lambda \left(n\right)}$ to get
                  ${\displaystyle \sum _{d|n}\mu \left(d\right)\lambda \left(n\right)\lambda \left(d\right)=\lambda \left(n\right)2^{\omega \left(n\right)}}$
                  ${\displaystyle \sum _{d|n}\mu \left(d\right)\lambda \left(\frac{n}{d}\right)=\lambda \left(n\right)2^{\omega \left(n\right)}}$
    Treating this as the "Möbius Inversed" formula, the "original" formula is
                  ${\displaystyle \sum _{d|n}\lambda \left(d\right)2^{\omega \left(d\right)}=\lambda \left(n\right)}$
Hence
     ${\displaystyle \sum _{d|n}\lambda \left(\frac{n}{d}\right)2^{\omega \left(d\right)}}$ =   ${\displaystyle \sum _{d|n}\lambda \left(n\right)\lambda \left(d\right)2^{\omega \left(d\right)}}$ =   ${\displaystyle \lambda \left(n\right)\cdot \sum _{d|n}\lambda \left(d\right)2^{\omega \left(d\right)}}$
=   ${\displaystyle \lambda \left(n\right)\cdot \lambda \left(n\right)}$ =   ${\displaystyle \lambda {\left(n\right)}^2}$ =   ${\displaystyle 1}$

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