Proving a formula for the number of square-free integers

Question

$S(n)$ denotes the number of square-free integers not exceeding $n$. I am interested in proving $$S(n)=\sum\limits_{j=1}\limits^{n}\sum\limits_{d^2|j}\mu(d)$$ I have so far been able to prove $$S(n)=\sum\limits_{j=1}\limits^{n}|\mu(j)|$$ and it is known that $\sum\limits_{d|n}\mu(d)=1$ if $n=1$ and $\sum\limits_{d|n}\mu(d)=0$ if $n>1$ however, I am having difficulty moving forward specifically because of the double summation.

Answer 1

You just need to prove that $|\mu(j)|=\sum_{d^2\mid j}\mu(d)$. If $j$ is a squarefree number, both the LHS and the RHS equal one, obviously. Let us assume $j=MN$ with $\gcd(M,N)=1$, $M>1$ and $N$ being squarefree. The RHS just depends on $\sum_{\substack{d\mid M\\ d\text{ squarefree}}}\mu(d) $, which equals zero. And so does the LHS.