On the powerful Möbius function

We define the Möbius function as follows :
$$\mu : \mathbb{N}^* \longrightarrow \{-1,0,1\},\qquad \mu(n)= \begin{cases} 1 & \text{si } n=1,\\ 0 & \text{si } \exists p\in\mathbb{P},\ p^2\mid n,\\ (-1)^k & \text{si } n=p_1p_2\cdots p_k,\quad p_1,\ldots,p_k\in\mathbb{P}\text{ distinct}. \end{cases}$$Show that $\forall n \in \mathbb{N}, \: \exists k \in \mathbb{N}$ such that
$$\displaystyle\sum_{\substack {d|n \\ d \geq 1} } \mu(n)^{2} =2^{k(n)}$$and explain what is k in terms of n.