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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?

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Notice that $\mu^2(n)$ is always $0$ or $1$ depending on wether $n$ is squarefree or not. So you can easily interpret the left-hand side... – Joel Cohen May 4 '12 at 17:56
up vote 4 down vote accepted

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

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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.

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