Baltic Way 2019

In what follows $d(n)$ denotes the number of positive divisors of a positive integer $n$. We show that the number of pairs is $d(N{)}^2$. Let $m$ be a fixed positive divisor of $N$. We claim that the number of pairs, $(a,b)$, such that $\frac{ab}{a+b}=m$ is $d(m{)}^2$. This becomes clear if we rewrite the equation as $$(a-m)(b-m)=m^2.$$ Since both $a-m$ and $b-m$ are greater than $-m$, it follows that $a-m$ and $b-m$ are both positive divisors of $m^2$, so the number of pairs $(a,b)$ for a fixed $m\mid N$ is $d(m^2)$, since we have $d(m^2)$ choices of $a$, and the choice of $a$ determines $b$ uniquely. The number of pairs is therefore $$\sum _{m|N}d(m^2)$$ Assume that $N$ has prime factorisation $p_1^{s_1}\cdots p_t^{s_t}$. Since $d(kl)=d(k)d(l)$ when $\text{gcd}(k,l)=1$, it follows that $$\begin{array}{rl}\sum _{m|N}d(m^2)& =\prod _{i=1}^t(d(1)+d(p_i^2)+\cdots +d(p_i^{2s_i}))\\ & =\prod _{i=1}^t(1+3+\cdots +(2s_i+1))\\ & =\prod _{i=1}^t(s_i+1{)}^2\\ & =d(N{)}^2\end{array}$$