Apostol - Analytic Number Theory, Chapter 3 problem 4a The problem comes from "Introduction to Analytic Number Theory" by Tom M. Apostol, Chapter 3, Problem 4a:
Question: Prove $\sum_{n \le x} \mu(n)[ \frac xn]^2 = \frac{x^2}{\zeta(2)} + O(x \log(x))$
where $\mu(n)$ is the Mobius function, the $\zeta(n)$ is the Riemann zeta function, [x] is the greatest integer less than or equal to real number x, and n is an integer. 
Work done so far:
The author states that $\sum_{n=1}^{\infty}\frac{\mu(n)}{n^2} = \frac{1}{\zeta(2)} = \frac{6}{\pi^2}$ stating the relation will be proved in a later chapter. Using this:
$$\sum_{n=1}^{\infty}\frac{\mu(n)x^2}{n^2} = \frac{x^2}{\zeta(2)} $$
$$\Rightarrow \sum_{n \le x}\mu(n)\frac{x^2}{n^2} + \sum_{x \lt n}\mu(n)\frac{x^2}{n^2} = \frac{x^2}{\zeta(2)}$$ 
$$\Rightarrow \sum_{n \le x}\mu(n)\frac{x^2}{n^2} = \frac{x^2}{\zeta(2)} - \sum_{x \lt n}\mu(n)\frac{x^2}{n^2} \le \frac{x^2}{\zeta(2)} + x^2\sum_{x \lt n}\frac{1}{n^2}$$
$$= \frac{x^2}{\zeta(2)} + x^2 \cdot O(\frac{1}{x}) = \frac{x^2}{\zeta(2)} + O(x)$$
Since for any real number $x = [x] + O(1)$, substitute $\frac xn$ with $([\frac xn] + O(1))$
$$\sum_{n \le x}\mu(n)\frac{x^2}{n^2} = \sum_{n \le x}\mu(n)([\frac{x}{n}]+O(1))^2 = \sum_{n \le x}\mu(n)([\frac{x}{n}]^2+O(x)) $$
$$= \sum_{n \le x}\mu(n)[\frac{x}{n}]^2 + \sum_{n \le x}\mu(n)O(x)$$ 
$$\Rightarrow \sum_{n \le x}\mu(n)[\frac{x}{n}]^2 = \frac{x^2}{\zeta(2)} + O(x) - \sum_{n \le x}\mu(n)O(x) = \frac{x^2}{\zeta(2)} + O(x)(1 + \sum_{n \le x}\mu(n))$$
At this point, it's very tempting to just claim that $(1+\sum_{n \le x}\mu(n)) = O(\log(x))$ but I have a feeling it's not that simple. In general, I sense this proof is fragile given that I'm using a relation that author simply says "we'll prove it later" (he uses this relation in order to prove a theorem however). 
The actual question is, am I going on the right path? Or is there a more elegant and simple proof than this, especially one that doesn't rely on the unproven relation. 
 A: You definitely seem to be on the right path! As it happens, $\sum_{n\le x} \mu(n)$ gets as large as $\sqrt x$ in size infinitely often, so your proposed claim isn't valid.
I suspect the place you'll find extra leverage is by writing $([\frac xn]+O(1))^2$ as $[\frac xn]^2 + O(\frac xn)$ rather than as $[\frac xn]^2+O(x)$.
A: Your claim that $\sum_{n \leqslant x} \mu (n) = O(\log{x})$ might be false or very hard to prove (I don't know whether this claim is falsified). Actually $\sum_{n \leqslant x} \mu (n) = O(x^{1/2+\epsilon})$ is equivalent to Riemann hypothesis.
You'd better use $\left( \left[ \frac{x}{n} \right] + O(1) \right)^2$ as $\left( \frac{x}{n} \right)^{2} + O \left( \frac{x}{n} \right)$ instead of $\left( \frac{x}{n} \right)^{2} + O \left( x \right)$. 
Then we have
$$\begin{align}
\sum_{n \leqslant x} \mu (n) \left[ \frac{x}{n} \right]^{2} &= \sum_{n \leqslant x} \mu (n) \left( \left( \frac{x}{n} \right)^{2} + O \left( \frac{x}{n} \right) \right) \\
& = x^{2} \sum_{n \leqslant x} \frac{\mu (n)}{n^{2}} + O \left( x \sum_{n \leqslant x} \frac{\mu (n)}{n} \right) \\
& = x^{2} \left( \frac{1}{\zeta (2)} + O \left( \frac{1}{x} \right) \right) + O \left( x \sum_{n \leqslant x} \frac{\mu (n)}{n} \right) \\
& = \frac{x^{2}}{\zeta (2)} + O (x) + O(x \log{x}) \\
& = \frac{x^{2}}{\zeta (2)} + O(x \log{x}).
\end{align}$$
And this proves the proposition.
