Write the sum in terms of the Riemann zeta function

Question

I believe it is a question from JHMT. Write the sum $\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{gcd(a,b)}{(a+b)^3}$ in terms of Riemann zeta function. The answer should be $-Z(2)+\frac{Z(2)^2}{Z(3)}$where $Z(s)$ represents the zeta function. I have no idea how to obtain this answer.

Here is some of my attempts.

Let $r=gcd(a,b), a=a_0r, b=b_0r$ and $gcd(a_0,b_0)=1$

Then

$\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{gcd(a,b)}{(a+b)^3}$ = $\sum_{r=1}^\infty \frac{1}{r^2} \sum_{(a_0,b_0)=1}\frac{1}{(a_0+b_0)^3}$

And I don't know how to go further, maybe using the Möbius function?

Answer 1

Notice that

$$ \sum_{d|n}\varphi(d)=n, $$

so we have (all sums are over positive integers)

$$ \begin{aligned} S &=\sum_a\sum_b{\gcd(a,b)\over(a+b)^3}=\sum_d\sum_{a\equiv0\pmod d}\sum_{b\equiv0\pmod d}{\varphi(d)\over(a+b)^3} \\ &=\sum_d{\varphi(d)\over d^3}\sum_\mu\sum_\nu{1\over(\mu+\nu)^3}. \end{aligned} $$

To evaluate the sum over $\mu$ and $\nu$. Notice that the number of positive solutions to $\mu+\nu=k$ is precisely $k-1$, so

$$ \sum_\mu\sum_\nu{1\over(\mu+\nu)^3}=\sum_k{k-1\over k^3}=\zeta(2)-\zeta(3). $$

Finally, using the identity $\sum_n\varphi(n)n^{-s}=\zeta(s-1)/\zeta(s)$, we conclude that

$$ S={\zeta(2)\over\zeta(3)}\cdot[\zeta(2)-\zeta(3)]={\zeta(2)^2\over\zeta(3)}-\zeta(2). $$