## Asymptotic formula for square free integers \begin{aligned} \displaystyle\sum_{n \leq x} \displaystyle\sum_{d^2 | n} \mu(d) & = \displaystyle\sum_{d \leq \sqrt{x}} \mu(d)\left\lfloor \frac{x}{d^2} \right\rfloor \\ & = x\displaystyle\sum_{d \leq \sqrt{x}} \frac{\mu(d)}{d^2} + O(\sqrt{x}) \\ & = x \frac{6}{\pi^2} + O(x\displaystyle\sum_{d > \sqrt{x}} \frac{1}{d^2} + \sqrt{x}) \\ & = x \frac{6}{\pi^2} + O((1 + \sqrt{x}) + \sqrt{x}) \\ & = x \frac{6}{\pi^2} + O(\sqrt{x}). \end{aligned}

## A derivation of a Riemann zeta function identity

Yesterday, I saw the following Riemann zeta function identity: $\displaystyle\sum_{n=1}^{\infty} \frac{\sigma_a(n)\sigma_b(n)}{n^s} = \frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}$.

I took some time to try to derive it myself and to my great pleasure, I succeeded.

Eventually, I realized that it suffices to show that $\{(dd_a, dd_b, d^2 n) : d_a | n, d_b | n : d, d_a, d_b, n \in \mathbb{Z}\}$

and $\{(dd_a, dd_b, n) : dd_a d_b | n : d, d_a, d_b, n \in \mathbb{Z}\}$

are equal as multisets. As sets, they are both representations of the set of $3$-tuples of positive integers such that the third is a multiple of the least common multiple of the first two. In the latter one, the frequency of $(a,b,c)$ is the number of $d$ that divides both $a$ and $b$ such that $ab | cd$. In the other one, if we write $(a,b,c)$ as $(d_1 d_2 a', d_1 d_2 b', c)$ where $\mathrm{gcd}(a', b') = 1$, the $ab | cd$ condition equates to $d_1^2 d_2 a'b' | c$, which corresponds to the number of $d_1$ dividing $a$ and $b$ and such that $d_1^2 | c$ and with that, $d_2a', d_2b'$ both dividing $d_1^2 / c$, which is the frequency of $(a,b,c)$ via the former representation.

The coefficients $\{a_n\}$ of the Dirichlet series of the LHS of that identity can be decomposed as follows: $a_n = \displaystyle\sum_{d^2 | n, d_a | \frac{n}{d^2}, d_b | \frac{n}{d^2}} (dd_a)^a (dd_b)^b$.

The coefficients $\{b_n\}$ of the Dirichlet series of the RHS of that identity are $b_n = \displaystyle\sum_{dd_a d_b | n} (dd_a)^a (dd_b)^b$.

Observe how both are equivalent in that via the multiset equivalence proved above, $n$ determines the same multiset of $(dd_a, dd_b)$ for both and across that, the values of the same function $(dd_a)^a (dd_b)^b$ are summed. Hence the two series are equal.