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

Question

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?

Answer 1

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.

Answer 2

Some thoughts, too long to fit a comment.

Let $c(n)$ denote the number of divisors $d|n$ such that gcd$(d,\frac{n}{d})=1$.

It is easy to show that $f(x)= \pm n$ has exactly $2c(n)$ solutions.

Then, after rearranging (we can since the series is positive), the sum becomes

$$ \sum_n \frac{2c(n)}{n^2} \,.$$

Now I claim that $c(n)=2^{f(n)}$ where $f(n)$ is the number of prime divisors of $n$. This can be proven easily by either observing that if $p|n$ then $p$ either divides $d$ OR $\frac{n}{d}$ but not both, or by observing that $c(n)$ is a multiplicative function.

Thus the problem asks to calculate

$$2\sum_n \frac{2^{f(n)}}{n^2} \,.$$

Here is where I have no idea on how to proceed further.

One idea would be to transform the series into an Euler product over the primes. Another would be to try to use the multiplicative of $c$, since this looks like the ideas behind Dirichclet L-series might help (I have absolutely no idea about that though).