Problem Statement
Let $h(x)$ count the number of integers $1\leq a<b<x$ such that $(a,b)=1$ and $\sigma(a)=\sigma(b)$, where $\sigma$ is the sum of divisors function.
Is it true that $h(x)>x^{2-o(1)}$?
Is it true that $h(x)>x^{2-o(1)}$?
Categories:
Number Theory
Progress
Erdős [Er74b] proved that $\limsup h(x)/x= \infty$, and claimed a similar proof for this problem. A complete proof that $h(x)/x\to \infty$ was provided by Pollack and Pomerance [PoPo16].A similar question can be asked if we replace the condition $(a,b)=1$ with the condition that $a$ and $b$ are squarefree. Weisenberg suggests another variant, with the condition that there are no proper factors $u\mid a$ and $v\mid b$ such that $\sigma(u)=\sigma(v)$ and $(u,a/u)=(v,b/v)=1$, which is the weakest restriction one can impose that is still strong enough to eliminate trivial duplicates.
Source: erdosproblems.com/824 | Last verified: January 16, 2026