TITLE: sum squared möbius function
QUESTION [6 upvotes]: I would like to prove the following equality
$$\sum_{n=1}^N \mu^2(n) = \sum_{k=1}^{\sqrt{N}} \mu(k) \cdot \lfloor N / k^2 \rfloor$$
with N a square number.
Can anyone give me a hint?
p.s. I know already that
$$\frac{\zeta(s)}{\zeta(2s) } = \sum_{n=1}^{\infty}\frac{ \mu^2(n)}{n^{s}}$$
Perhaps this can help?

REPLY [3 votes]: Hint:  Since $$\sum_{d|n} \mu(d)=\left\{ \begin{array}{cc}
1 & \text{if }n=1\\
0 & \text{otherwise}
\end{array}\right\} $$ we have that
$$\sum_{d\leq N}\mu(d)^{2}=\sum_{d\leq N}\sum_{k^{2}|d}\mu(k).$$  Now, try switching the order of summation.