Liouville Function - Introduction

The Liouville function, after Joseph Liouville, is defined for ${\displaystyle n=p_1^{a_1}{p}_2^{a_2}...p_k^{a_k}}$ as

interpreted so that ${\displaystyle \lambda \left(1\right)=1}$.  It is quite obvious that

This, coupled with the fact that the Liouville function is completely multiplicative (see below), we have

 

Question
Prove that the Liouville function is completely multiplicative i.e.

for all integers m and n (not necessarily coprime).
Solution:-
   Let ${\displaystyle p_1,p_2,...,p_k}$ be the list of primes appearing in m and/or n. Write ${\displaystyle m=p_1^{a_1}{p}_2^{a_2}...p_k^{a_k}}$ and ${\displaystyle n=p_1^{b_1}{p}_2^{b_2}...p_k^{b_k}}$, in which any of the non-negative integers ${\displaystyle a_1,a_2,...,a_k }$ and ${\displaystyle b_1,b_2,...,b_k }$ can be zero.  Then
                     ${\displaystyle mn=p_1^{a_1+b_1}{p}_2^{a_2+b_2}...p_k^{a_k+b_k}}$
Consequently
   ${\displaystyle \lambda \left(mn\right)={(-1)}^{(a_1+b_1)+(a_2+b_2)+...+(a_k+b_k)}}$
            ${\displaystyle ={(-1)}^{(a_1+a_2+...+a_k)+(b_1+b_2+...+b_k)}}$
            ${\displaystyle ={(-1)}^{a_1+a_2+...+a_k}{(-1)}^{b_1+b_2+...+b_k}}$
            ${\displaystyle =\lambda \left(m\right)\lambda \left(n\right)}$
[End]


More about the Liouville function here, and a result involving Möbius Inversion function here.