liouville lambda function $\lambda(n):=(-1)^{\Omega(n)} , \Omega(n):= \sum_{p \in \mathbb P} v_p(n) , n \in \mathbb N$
I need to proof that
$\lambda$ is completely multiplicative and furthermore  $\sum_{d|n} \lambda(d)$= is $1$ , if $n$ is a square number and $0$ otherwise.
secondly i need to show the identity of $\lambda(n)=\sum_{d \in \mathbb N ,d^2|n} \mu(\frac{n}{d^2})$
proof:
$\lambda(nm):=(-1)^{\Omega(nm)}=(-1)^{\sum_{p \in \mathbb P} v_p(nm)}=(-1)^{\sum_{p \in \mathbb P} v_p(n)+\sum_{p \in \mathbb P} v_p(m)}=(-1)^{\Omega(n)+\Omega(m)}=(-1)^{\Omega(n)}*(-1)^{\Omega(m)}=\lambda(n)*\lambda(m)$ .Therefore $\lambda$ is completely multiplicative  ,since gcd(m,n)=d.
now I proof : $\sum_{d|n} \lambda(d)$= is $1$ , if $n$ is a square number and $0$ otherwise.
proof:
since $d \in \mathbb N$ you may define $d:=p_1^{v_1}*...*p_n^{v_n}$
$\sum_{d|n} \lambda(d):=\sum_{d|n} (-1)^{\Omega(d)}=\sum_{d|n} (-1)^{\Omega(d)}=\sum_{d|n} (-1)^{\Omega(v_1+..+v_n)}  $
now I am stuck.. someone may help me?
greetings Neil
 A: $\Omega(nm) =\Omega(n)+\Omega(m)$ so that $\lambda(n) = (-1)^{\Omega(n)}$ is (completely) multiplicative. 
Thus $f(n) = \sum_{d | n} \lambda(d)$ is multiplicative too, and it suffices to look at $n= p^k$ ie. $$f(p^k) = \sum_{d | p^k} \lambda(d)=\sum_{m=0}^k \lambda(p^m)=\sum_{m=0}^k (-1)^m = 1_{k \text{ is even}}$$ $$\implies f(n) = \prod_{p^k \|n} f(p^k) = \prod_{p^k \|n} 1_{k \text{ is even}}= 1_{n \text{ is square}}$$

Equivalently when you have a multiplicative function,  look at its Dirichlet series $$\sum_{n=1}^\infty \lambda(n)n^{-s}= \prod_p (1+\sum_{k = 1}^\infty \lambda(p^k)p^{-sk}) = \prod_p (1+\sum_{k = 1}^\infty (-1)^k p^{-sk}) = \prod_p \frac{1}{1+p^{-s}}$$
$$\sum_{n=1}^\infty f(n) n^{-s} = (\sum_{n=1}^\infty \lambda(n)n^{-s})(\sum_{m=1}^\infty m^{-s}) = \zeta(s)\prod_p \frac{1}{1+p^{-s}} = \prod_p \frac{1}{1-p^{-2s}} = \sum_{n=1}^\infty n^{-2s}$$
A: If $\lambda(n)=\sum_{d \in \mathbb N ,d^2|n} \mu(\frac{n}{d^2})$ is where you are stuck, then the proof goes as follows-
We know
$$f=\lambda\ast\mu$$
where $f$ is the square indicator function.
So,
$$\lambda=f\ast\mu=\sum_{d\vert n}f(d)\mu\left(\frac nd\right)=\sum_{d^2\vert n}\mu\left(\frac n{d^2}\right)$$
hence completing the proof.
