Show that $\displaystyle \sum_{d^2 | n} \mu (d) = |\mu (n)|$. where $\mu$ is the Mobius function.
I am not sure on how to continue this problem. My starting point is to show that if $n = n_1^2 n_2$ then $d^2 | n \Leftrightarrow d| n_1$ ($n _2$ is square-free). If this is correct, how would I conclude?
Your starting point is correct, but you have to show it. You may continue like this: $$\sum_{d^2 | n} \mu (d)=\sum_{d | n_1} \mu (d)= \begin{cases} 1, & \text{if $n_1=1$} \\ 0, & \text{if $n_1>1$} \end{cases} $$ using the most important property of the Möbius function. So the sum is $1$ if $n$ is squarefree, and $0$ otherwise, exactly as $|\mu(n)|$.