Sum over $\mathbb Q^+$

Relevant wiki: Principle of Inclusion and Exclusion Problem Solving

If we allow all positive integers $a,b$ , we find $\sum_{a,b}\frac{1}{a^2b^2}=\zeta(2)^2$ . If we exclude all $a,b$ with a fixed common prime divisor $p$ , we find $\sum_{a,b}\frac{1}{a^2b^2}=(1-\frac{1}{p^4})\zeta(2)^2$ . If we exclude all $a,b$ with $\gcd(a,b)\neq 1$ , we find $\sum_{\gcd(a,b) = 1}\frac{1}{a^2b^2}=\prod_{p}(1-\frac{1}{p^4})\zeta(2)^2=\frac{\zeta(2)^2}{\zeta(4)}=\boxed{2.5}$ , by the inclusion-exclusion principle.

We first note that $F(s)=\sum_{x\in\mathbb Q^+}\frac 1{f(x)^s}=\sum_{\stackrel{a,b=1}{\text{gcd}(a,b)=1}}^\infty\frac 1{(a\cdot b)^s}.$ Moreover for $s>1$ we have that $\begin{aligned} \zeta(s)^2&=\left(\sum_{a=1}^\infty\frac 1{a^s}\right)^2=\sum_{a,b=1}^\infty\frac 1{(a\cdot b)^s}=\sum_{d=1}^\infty\sum_{\stackrel{a,b=1}{\text{gcd}(a,b)=d}}^\infty\frac 1{(a\cdot b)^s}\\ &=\sum_{d=1}^\infty\frac 1{d^{2s}}\cdot\sum_{\stackrel{a,b=1}{\text{gcd}(a,b)=1}}^\infty\frac 1{(a\cdot b)^s}=\zeta(2s)\cdot F(s) \end{aligned}$ where $\zeta(s)$ is Riemann zeta function . Therefore if $s$ is a positive even number $2n$ then the sum converges to $\begin{aligned} F(2n)&=\frac {\zeta(2n)^2}{\zeta(4n)}=\frac {\left((-1)^{n-1}\dfrac {2^{2n-1}B_{2n}\pi^{2n}}{(2n)!}\right)^2}{(-1)^{2n-1}\dfrac {2^{4n-1}B_{4n}\pi^{4n}}{(4n)!}} \\&=\binom{4n}{2n}\frac {B_{2n}^2}{2|B_{4n}|}\in\mathbb Q \end{aligned}$ where $B_k$ is the $k$ -th Bernoulli number (it is rational!). Here there are some values of the function $F(2)=\boxed{\dfrac 52},\, F(4)=\frac 76,\, F(6)=\frac {715}{691},\, F(8)=\frac {7293}{7234}.$

We are looking for $x = \sum_{(\alpha,\beta) = 1} \frac 1{\alpha^2\beta^2}.$

Define $c = \text{gcd}(a,b),\ a = c\alpha,\ b = c\beta$ . Then

$\left(\sum_a \frac 1{a^2}\right)\left(\sum_b \frac 1{b^2}\right) = \sum_{a,b} \frac 1{a^2b^2} = \sum_c \sum_{(\alpha,\beta) = 1} \frac 1{(c\alpha)^2(c\beta)^2} = \left(\sum_c \frac 1{c^4}\right)\left(\sum_{(\alpha,\beta) = 1} \frac 1{\alpha^2\beta^2}\right).$ Thus $\zeta(2)^2 = \zeta(4)x,$ and $x = \frac{\zeta(2)^2}{\zeta(4)} = \frac{(\pi^2/6)^2}{\pi^4/90} = \frac{90}{36} = \boxed{2\tfrac12}.$

I'm not sure if we can consider that as a solution

First , I've observed that when $ab$ become higher, $\frac{1}{f(x)^2}$ become smaller.

So with a simple Python script, I calculated the 50th first terms for a and b

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arr, s = [0.0], 0
for a in range(1,50):
    for b in [d for d in range(1,50) if (a/float(d)) not in arr]:
        arr.append(a/float(b)) # useful to don't compute several times the same number (ex: 1/1, 2/2, ...)
        s+=1/float(((a*b)**2))
print(s)

output: 2.44505876736

So we can conclude that the result is close to $\boxed{2.5}$

Right, I'm a little late to the party

The number of $x$ that satisfies $f(x) = q$ is the number of ways to partition the distinct prime factors of $q$ into 2 ordered groups: The number of such $x$ equals $2^{\omega(q)}$ , where $\omega(n)$ is the number of distinct prime factors of $n$ . Hence, using the substitution $f(x) = n$ , the sum equates: $\large \sum_{n \ge 0}\frac{2^{\omega(n)}}{n^s} = \sum_{n \ge 0}\frac{|\mu|*1(n)}{n^s} = \zeta(s)\frac{\zeta(s)}{\zeta(2s)} = \frac{\zeta(s)^2}{\zeta(2s)}$

Find ratio of

[double sum {(1/n^2)(1/m^2)} from n & m =1 to Inf ] = (π^4/36)

with

[sum {1/n^4} from n=1 to Inf ] = (π^4/90) ,

using WolframAlpha.

Answer=2.5