Question: Prove that $$S(n) = \sum_{d=1}^n \mu(d) \left[ \frac{n}{d^2}\right],$$ where $\displaystyle \left[\frac{n}{d^2}\right]$ denotes the largest integer that does not exceed $\displaystyle \frac{n}{d^2}$. $S(n)$ counts the number of "square-free" integers less than or equal to $n$. That is, it counts the number of integers less than or equal to $n$ that are not divisible by a square integer. $\mu$ is the Mobius function, or $$\mu(n) = \begin{cases}1 & \text{if } n=1,\\ 0 & \text{if } p^2 | n \text{ for some prime $p$},\\ (-1)^r & \text{if } n=p_1p_2\ldots p_r \text{ for distinct primes $p_1,\ldots, p_r$}. \end{cases}$$We are given that $$S(n) = \sum_{j=1}^n \sum_{d^2 | j} \mu(d).$$
Immediately one sees that as soon as $d^2 > n$, $\displaystyle \left[ \frac{n}{d^2}\right] = 0$ and the terms beyond this point vanish. Therefore, assuming such an $f$ exists, if $f^2 = n$, then the sum above becomes $$\sum_{d=1}^n \mu(d) \left[ \frac{n}{d^2}\right] = \sum_{d=1}^f \mu(d) \left[\frac{n}{d^2}\right].$$ I am a little stuck, and am not sure where to go after this point...
Hint: Well, you have that $$S(n)=\sum _{j=1}^n\sum _{d^2|j}\mu (d),$$ which is adding a bunch of $\mu (d).$ In principle, $d\leq n$ so $S(n)=\sum _{d=1}^n\mu (d)|\{m\in [n]:d^2|m\}|,$ where $[n]=\{1,2,\cdots ,n\}$ this is basically counting how many times you add $ \mu (d)$ in the sum. Now, can you cout the number of $m\in [n]$ such that $d^2|m$? It will be like $d^2,d^2*2,d^2*3,\cdots ,d^2*k$ where $k$ is such that $d^2k\leq n$ but $d^2(k+1)>n.$ So what is $k$ in terms of $d$ and $n$?