Dirichlet Convolution involving mobius function I want to show the sum
$$\sum_{d/n}|\mu(d)|\mu(\frac{n}{d})$$
only has a result not $0$ when $n$ is a perfect square (and then it's $\mu(\sqrt n)$. I think it might have something to do with perfect squares having an odd number of divisors because $d=n/d$, but I don't know how to use that to prove it?

It's a fairly easy observation with standard Dirichlet series and Euler products:
$$\sum_{n=1}^\infty\frac{|\mu(n)|}{n^s}=\prod_p\left(1+\frac{1}{p^s}\right), \qquad \sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\prod_p\left(1-\frac{1}{p^s}\right) $$
$$\implies \sum_{n=1}^\infty\frac{1}{n^s}\sum_{d\mid n}|\mu(d)|\mu\left(\frac{n}{d}\right)=\left[\sum_{n=1}^\infty\frac{|\mu(n)|}{n^s}\right]\left[\sum_{n=1}^\infty\frac{\mu(n)}{n^s}\right]$$
$$=\prod_p\left(1+\frac{1}{p^s}\right)\prod_p\left(1-\frac{1}{p^s}\right)=\prod_p\left(1-\frac{1}{p^{2s}}\right)=\sum_{n=1}^\infty\frac{\mu(n)}{n^{2s}}.$$
