How to evaluate $\sum_{\gcd (p,q)=1} \frac{1}{p^2q^2}$?

How do find the following sum $$ \sum_{\gcd (p,q)=1} \frac{1}{p^2q^2} $$

Solution 1:

Let's look at the more general case,

$$\sum_{\gcd (p,q) = d} \frac{1}{p^2q^2}.$$

$\gcd (p,q) = d$ means we have $p = d\cdot a$ and $q = d\cdot b$ with $\gcd(a,b) = 1$, so we obtain

$$\sum_{\gcd(p,q) = d} \frac{1}{p^2q^2} = \sum_{\gcd(a,b) = 1} \frac{1}{(da)^2(db)^2} = \frac{1}{d^4}\sum_{\gcd(a,b) = 1} \frac{1}{a^2b^2}.$$

Now, any pair of positive integers occurs in exactly one of the sets $A_d = \{ (k,m) : \gcd(k,m) = d\}$, hence

\begin{align} \left(\sum_{d=1}^\infty \frac{1}{d^4}\right)\left(\sum_{\gcd(p,q) = 1} \frac{1}{p^2q^2}\right) &= \sum_{d=1}^\infty \sum_{\gcd(a,b) = d} \frac{1}{a^2b^2}\\ &= \sum_{a,b \in \mathbb{N}\setminus\{0\}} \frac{1}{a^2b^2}\\ &= \left(\sum_{n=1}^\infty \frac{1}{n^2}\right)^2. \end{align}

Thus we obtain

$$\sum_{\gcd(p,q) = 1}\frac{1}{p^2q^2} = \frac{\left(\sum_{n=1}^\infty \frac{1}{n^2}\right)^2}{\sum_{d=1}^\infty \frac{1}{d^4}} = \frac{\zeta(2)^2}{\zeta(4)} = \frac{\pi^4/36}{\pi^4/90} = \frac{90}{36} = \frac{5}{2}.$$