sum squared möbius function 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?

Big hint: The left side counts something, and the right side counts the same thing via an inclusion-exclusion argument.
