As a proof of the second part of part(b) of this question :
Liouville function and perfect square
I have the solution given below:
But I can not see how this solution explains the case when $n =5 \times 3^2 $
Could anyone explain this for me please? and if the solution does not deal with this case can anyone give me hints towards the correct solution?
First, let's recall the definition of the liouville function,$\lambda(n)=(-1)^{\Omega(n)}$ where $\Omega(n)$ is the number of prime factors counted with multiplicity. Note that $\lambda$ is completely multiplicative.
We want to know what is $\lambda\ast 1(n)=\sum_{d|n}\lambda(d)$ (Here, $\ast$ denotes the convolution of arithmetic functions)
Since $\lambda$ and $1$ are multiplicative, then $\lambda\ast 1$ is also multiplicative, so it's enough to know the values of $\lambda\ast 1$ at the prime powers.
Note $$\lambda\ast 1(p^{k})=\sum_{d|p^{k}}\lambda(d)=\sum_{j=0}^k\lambda(p^j)=\sum_{j=0}^k(-1)^j$$ From here, convince yourself that $\lambda\ast 1(p^{k})=1$ if $k$ is even and $\lambda\ast 1(p^{k})=0$ if $k$ is odd.
Conclude that $\lambda\ast 1(n)=1$ if $n$ is square and $0$ otherwise.
Edit: If you don't know the fact that convolution of two multiplicative functions is multiplicative, alternatively we can prove directly that $n\mapsto \sum_{d|n}\lambda_f(d)$ is a multiplicative function.
Indeed, let $n=p_1^{k_1}\dots p_r^{k_r}$. Since $\lambda$ is (completely) multiplicative you have \begin{align*}\sum_{d|n}\lambda(d)&=\sum_{(j_1,\dots,j_r)\leq (k_1,\dots,k_r)}\lambda(p_1^{j_1}\dots p_r^{j_r})\\ &=\sum_{(j_1,\dots,j_r)\leq (k_1,\dots,k_r)}\lambda(p_1^{j_1})\dots \lambda(p_r^{j_r})\\ &=\left(\sum_{j_1\leq k_1}\lambda(p_1^{j_1})\right)\dots \left(\sum_{j_r\leq k_r}\lambda(p_r^{j_r})\right) \\ &=\left(\sum_{d|p_1^{k_1}}\lambda(d)\right)\dots \left(\sum_{d|p_r^{k_r}}\lambda(d)\right) \end{align*}