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Claim. P. Erdős and M. Rosenfeld, The factor-difference set of integers, Acta Arith. 79 (1997), no. 4, 353--359 (card), Proposition 4.1 and the remark after it. Order the factorizations $n=a_ib_i$ with $a_i\ge b_i$ by $d_i=a_i-b_i$, starting at $i=0$. The sums $a_i+b_i$ are distinct integers of size at least $2\sqrt n$, so $a_i+b_i\ge2\sqrt n+i$ and $d_i\ge2n^{1/4}\sqrt i$. Hence $a_i>\sqrt n+n^{1/4}\sqrt i$ for $i\ge1$, and every $n$ has at most $1+c^2$ divisors in $[\sqrt n,\sqrt n+cn^{1/4}]$ for each $c>0$. For $\epsilon\ge1/4$ the interval $(n^{1/2},n^{1/2}+n^{1/2-\epsilon})$ lies in $[\sqrt n,\sqrt n+n^{1/4}]$, so every $n$ has at most two divisors there, and the answer to Problem 886 is yes for these $\epsilon$. The journal record gives only the year, so this page carries the first of January.

Covers. Every $\epsilon\ge1/4$, the endpoint included; nothing for $0<\epsilon<1/4$.

Depends on. No page of this wiki.

Acceptance. Refereed: Acta Arithmetica (the paper thanks its referee). The site labels the problem OPEN, so its commentary is not acceptance.