Wiki
Wiki

Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

Updated

Problem 887

../

claims/: The 2 claim pages of Problem 887, one per claimant's result; the problem's standing derives from them.


Statement. Is there an absolute constant KK such that, for every C>0C>0, if nn is sufficiently large then nn has at most KK divisors in (n1/2,n1/2+Cn1/4)(n^{1/2},n^{1/2}+C n^{1/4}).

Status. The site labels the problem OPEN. The answer is yes for perfect squares by Chan's bound for squares and for a class of almost squares by Chan's bound for almost squares; the general case is open.

Source. erdosproblems.com/887, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #887, https://www.erdosproblems.com/887.

References.

  • [Ch14] Chan, Tsz Ho, Factors of a perfect square. Acta Arith. (2014), 141-143.
  • [Ch15] Chan, Tsz Ho, Factors of almost squares and lattice points on circles. Int. J. Number Theory (2015), 1701-1708.
  • [ErRo97] Erdős, Paul and Rosenfeld, Moshe, The factor-difference set of integers. Acta Arith. (1997), 353-359.

Formalization. Statement in formal-conjectures.

Current assessment

The site labels the problem OPEN (page last edited 10 April 2026). The answer is yes for perfect squares, with K=5K=5 (Chan 2014), and for the numbers (N−a)(N+b)(N-a)(N+b) with 0≤a≤b≤e(log⁡n)2/70\le a\le b\le e^{(\log n)^{2/7}}, with K=18K=18 (Chan 2015); the general case is open. Erdős and Rosenfeld's bound of 1+C21+C^2 divisors (card) depends on CC, so it gives no absolute KK. Their Proposition 4.2 gives infinitely many nn with four divisors within about 8n1/48n^{1/4} of n\sqrt n. The site's commentary places these four divisors in (n1/2,n1/2+n1/4)(n^{1/2},n^{1/2}+n^{1/4}), but the remark after their Proposition 4.1 allows at most two divisors there; the site's commentary on Problem 886 prints 16n1/416n^{1/4}. A note posted in the site's thread on 6 July 2026 claims infinitely many nn with five divisors in (n,n+31n1/4)(\sqrt n,\sqrt n+31n^{1/4}). That would show any admissible KK is at least 55 and answer the side question whether four is best possible, but it settles no instance of the question, so it has no claim page. Letendre's preprint (card), cited in the thread, bounds the count only in windows of length n1/4−δn^{1/4-\delta}, shorter than this question's, and settles no instance.

Linked library material

These entries are derived from explicit links on library pages. They are navigation only and do not by themselves record mathematical progress.