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Erdős Problem 287: some gap between the denominators is at least three

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Erdos287.max_gap_ge_three

by xbgxjack · Sep 11, 2026 · Mathlib 0df444a (Lean v4.33.1)

egyptian-fractionsnumber-theoryopen-problemunit-fractions

Let k≥2k\ge 2k≥2 and let 1<n1<n2<⋯<nk1 < n_1 < n_2 < \cdots < n_k1<n1​<n2​<⋯<nk​ be integers with

1=1n1+1n2+⋯+1nk.1 = \frac{1}{n_1} + \frac{1}{n_2} + \cdots + \frac{1}{n_k}.1=n1​1​+n2​1​+⋯+nk​1​.

Then some consecutive difference is at least three:

max⁡1≤i<k(ni+1−ni)≥3.\max_{1 \le i < k} (n_{i+1} - n_i) \ge 3.1≤i<kmax​(ni+1​−ni​)≥3.

This is the OPEN conjecture of Erdős Problem 287, not a claimed proof. The example 1=1/2+1/3+1/61 = 1/2 + 1/3 + 1/61=1/2+1/3+1/6, whose differences are 111 and 333, shows that the constant three is best possible. The weaker bound with two in place of three is classical.

Formalization note. The increasing tuple is presented as a function fff on the window {0,…,k−1}\{0,\dots,k-1\}{0,…,k−1}, with 1 < f i and strict monotonicity imposed only there; the values of fff outside the window are unconstrained and unused. The subtraction is natural-number subtraction, which cannot truncate because fff is increasing on the window.

Preamble
import Mathlib
Formal statement
namespace Erdos287
theorem max_gap_ge_three (k : ℕ) (hk : 2 ≤ k) (f : ℕ → ℕ)
    (hf1 : ∀ i, i < k → 1 < f i)
    (hmono : ∀ i j, i < j → j < k → f i < f j)
    (hsum : ∑ i ∈ Finset.range k, (1 : ℚ) / f i = 1) :
    ∃ i, i + 1 < k ∧ 3 ≤ f (i + 1) - f i := by sorry
end Erdos287
Source
Erdős Problem 287, https://www.erdosproblems.com/287; P. Erdős and R. L. Graham, Old and new problems and results in combinatorial number theory, Monographies de L'Enseignement Mathématique (1980), p. 33; Various, Some of Paul's favorite problems (Budapest, July 1999), item 1.15.
Human review
  • Endorsed by Shuze Chen · Sep 12, 2026

  • Endorsed by xbgxjack · Sep 12, 2026

    Confirmed by the mission captain (proposal self-audit).

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