Evaluating $\sum_{\gcd\left(m,n\right)=1}\frac{1}{m^2n^2}$ I was wondering how one would evaluate the sum $$\sum_{\gcd\left(m,n\right)=1}\frac{1}{m^2n^2}.$$
The first thought that came to mind to to try something like this: $$\sum_{\gcd\left(m,n\right)=1}\frac{1}{m^2n^2}=2\sum_{\substack{\gcd\left(m,n\right)=1,\\m<n}}\frac{1}{m^2n^2}=2\sum_{k=1}^{\infty}\frac{f\left(k\right)}{k^2},$$
where $f\left(k\right)$ counts the number of ways that $k=mn$ for $m<n$ relatively prime. However, I can't see how I would proceed. Is this even the right way to be thinking about this?
In general, what techniques should I look into when looking at problems like this, and which books should I read to learn about these and similar techniques? Also, if this function $f$ has been studied before (and if it has a more standard notation/name), what should I search for to learn more?
 A: We have $$\sum_{d\ge 1} \sum_{\gcd(m,n)=d} \frac{1}{m^2 n^2}
= \zeta(2)^2$$
because we may classify all pairs $(m,n)$ according to their GCD.
On the other hand, this is also
$$\sum_{d\ge 1} \frac{1}{d^4} \sum_{\gcd(m,n)=1} \frac{1}{m^2 n^2}.$$
because when we divide the pair by their GCD we get a pair with GCD one.
Hence $$\sum_{\gcd(m,n)=1} \frac{1}{m^2 n^2} = \frac{\zeta(2)^2}{\zeta(4)}
= \frac{5}{2}.$$
Addendum. Trying to answer this question quickly yesterday I missed one of its salient features. Suppose $q(n)$ gives the number of ordered factorizations of $n$ into two factors that are co-prime to each other. This means if $p^v$ is a (maximal) prime power in the prime factorization of $n$ it has to go into one of the two factors at the full multiplicity (exponent $v$). Therefore we have
$$q(n) = 2^{\omega(n)}$$
where $\omega(n)$ is the number of distinct primes dividing $n$ (not counted according to multiplicity), because there are two possibilities for $p^v$ (first factor and second factor).
It was shown recently at this MSE link that $$\Lambda(s) = \sum_{n\ge 1} \frac{2^{\omega(n)}}{n^s}
=\frac{\zeta(s)^2}{\zeta(2s)}.$$
The result then follows by setting $s=2.$
A: Using the fact that $$\sum_{d|n}\mu(d)=\begin{cases}
1 & \text{if }n=1\\
0 & \text{otherwise}
\end{cases}$$ where $\mu(d)$ is the mobius function, we can write $$\sum_{\gcd(m,n)=1}\frac{1}{m^{2}n^{2}}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^{2}}\frac{1}{n^{2}}\sum_{d|m,d|n}\mu(d).$$ Switching the order of summation so that we sum over $d$ first, this becomes $$\sum_{d=1}^{\infty}\mu(d) \sum_{m=1}^{\infty}\frac{1}{(dm)^{2}}\sum_{n=1}^{\infty}\frac{1}{(dn)^{2}},$$ and this may be split up to find that your sum equals $$\sum_{d=1}^\infty \frac{\mu(d)}{d^4}\sum_{m=1}^\infty \frac{1}{m^2}\sum_{m=1}^\infty \frac{1}{n^2}=\frac{\zeta(2)^2}{\zeta(4)}=\frac{5}{2}.$$
