Consider the Jordan function $J\_2(n)$ defined by
$$
J\_2(n) = \#\{x \in (\mathbb{Z}/n)^2 \mid ord(x) = n\}
$$
(this is OEIS [A007434](http://oeis.org/search?q=1%2C3%2C8%2C12%2C24%2C24%2C48%2C48&language=english&go=Search)). One can prove the following identity pretty easily:
$$
\sum\_{d \mid n} J\_2(d) \sigma\_1(n/d) = n\sigma\_1(n)
$$
by noting that both sides are multiplicative functions, and so it suffices to check that this holds for $n = p^k$.


One could alternatively look at the Dirichlet generating functions; specifically, let 
$$
F(s) = \sum\_{n=1}^\infty \frac{\sigma\_1(n)}{n^s}
$$
and
$$
G(s) = \sum\_{n=1}^\infty \frac{J\_2(n)}{n^s}
$$
in which case our identity is equivalent to the identity 
$$
F(s)G(s) = F(s-1).
$$
Is there a nice generating function argument for this? This seems like a much nicer statement than hands-on computation, but I don't really know how to prove it.


I should remark that $G(s) = \frac{\zeta(s-2)}{\zeta(s)}$, so this is equivalent to the identity
$$
F(s)\zeta(s-2) = F(s-1)\zeta(s).
$$



Is there a nice generating function proof of the following identity?
The identity $G(s)=\frac{\zeta(s-2)}{\zeta(s)}$ is a corollary of the formula $J\_2(p^e) = p^{2e} - p^{2e-2}$. The last identity from the question is simple because $F(s)=\zeta(s)\zeta(s−1).$


