Let $f(\frac ab)=ab$, where $\frac ab$ is irreducible, in $\mathbb Q^+$. What is $\sum_{x\in\mathbb Q^+}\frac 1{f(x)^2}$? 
Let $f(\frac ab)=ab$, where $\frac ab$ is irreducible, in $\mathbb Q^+$. What is $\sum_{x\in\mathbb Q^+}\frac 1{f(x)^2}$?

Club challenge problem. I don't think it's possible to do with only high school calculus. Help, please?

Note first that $f$ is multiplicative over the rationals in some sense.  More precisely, if $\frac{a}{b}$ and $\frac{c}{d}$ are in reduced form and $(a,d) = (b,c) = 1$, we have that $f(\frac{ac}{bd}) = f(\frac{a}{b}) \cdot f(\frac{c}{d})$.
You have to justify this next step, but (if the sum behaves nicely) we can thus split it up into an infinite product using unique factorization of the rationals:
$$
\begin{align*}
\sum_{x \in \mathbb{Q}^+} \frac{1}{f(x)^2} &=
\prod_{p \text{ prime}} \left(\cdots + \frac{1}{f(p^{-2})^2} + \frac{1}{f(p^{-1})^2} + \frac{1}{f(1)^2} + \frac{1}{f(p^1)^2} + \frac{1}{f(p^2)^2} + \cdots\right)
\\
&= \prod_{p \text{ prime}} \left( 1 + \frac{2}{p^2} + \frac{2}{p^4} + \frac{2}{p^6} + \cdots \right)
\\
&= \prod_{p \text{ prime}} \left( \frac{2}{1 - \frac{1}{p^2}} - 1 \right)
= \prod_{p \text{ prime}} \left( \frac{p^2 + 1}{p^2 - 1} \right)
\end{align*}
$$
I admit I am unsure of how to evaluate this product, but this seems helpful.
Update: This product evaluates to $\frac52$!  See the comments for an explanation.
