Number of ordered pair $(a,b)$ of natural numbers such that ${ab\over a+b}=n$ (n $\in \mathbb{N}$) is $a_n$ then find , $\lim_{n \to \infty} {a_n\over n^2}$.
MY attempt : Suppose the case where $n=2^k$ for some $k \in \mathbb{N}$ . Then , ${ab\over a+b}=n$ is equivalent to $(a-2^k)(b-2^k)=2^{2k}$ . In this case $a_n=4k$ and thus $\lim_{n \to \infty} {a_n\over n^2}=0$
Now I have an intuition (from observations) that ${a_n\over n^2}$ would be maximum when $a_n$ is a power of 2 , and if bigger factors (like $3,5,etc$) are included then the ratio ${a_n\over n^2}$ would decrease. And thus $\lim_{n \to \infty} {a_n\over n^2}=0$ for any $n$
But I am not able to think of a solid mathematical proof for this .
Could someone please help me solve this ?
Thanks!
As already remarked in the comments, the integer solutions of $\frac{ab}{a+b}=n$ are the integer solutions of $(a-n)(b-n)=n^2$, so they are given by $(a,b)=\left(d+n,\frac{n^2}{d}+n\right)$ for any integer divisor $d$ of $n^2$.
The wanted limit is zero since the following Dirichlet series is clearly convergent: $$ \sum_{n\geq 1}\frac{\tau(n^2)}{n^2}=\prod_p\left(1+\frac{\tau(p^2)}{p^2}+\frac{\tau(p^4)}{p^4}+\ldots\right)=\prod_{p}\left(1+\frac{3}{p^2}+\frac{5}{p^4}+\ldots\right) $$ $$ \sum_{n\geq 1}\frac{\tau(n^2)}{n^2}=\prod_{p}\frac{p^2(p^4-1)}{(p^2-1)^3} =\frac{\zeta(2)^3}{\zeta(4)}=\frac{5\pi^2}{12}. $$