# 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?

$$\frac{\zeta(s)}{\zeta(2s) } = \sum_{n=1}^{\infty}\frac{ \mu^2(n)}{n^{s}}$$

Perhaps this can help?

• 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... Commented May 4, 2012 at 17:56

• If I let the sets $A_j=\{n \leq x \: : p_j^2 |n \}$ where $j$ spans all primes until $p_j \leq \sqrt{x}$ then the cardinality of non square numbers smaller then $x$ is : $$|\cup A_j|=\sum_j \left \lfloor \frac{n}{p_j^2}\right \rfloor -\sum_{i<j} \left \lfloor \frac{n}{p_i^2 p_j^2}\right \rfloor+…$$ $$=\sum_{d\leq \sqrt{x}} \left \lfloor \frac{n}{d^2}\right \rfloor (-\mu(d))$$ Thus the number of square free numbers should be: $x+\sum_{d\leq \sqrt{x}} \left \lfloor \frac{n}{d^2}\right \rfloor (\mu(d))$. Why do I have this extra $x$ term? Commented May 26 at 6:56
• Actually ignore me the sum should take place over $d>1$. That is why there is $x$. It may be incorporated to $d=1$. :) Commented May 26 at 8:20
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.