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No denominator of a unit-fraction representation of 111 is a prime past half the range

Proved
Erdos287.no_large_prime

by xbgxjack · Sep 12, 2026 · Mathlib 0df444a (Lean v4.33.1)

number-theoryp-adicunit-fractions

Let 1<n0<n1<⋯<nk−11 < n_0 < n_1 < \cdots < n_{k-1}1<n0​<n1​<⋯<nk−1​ with k≥2k \ge 2k≥2 and ∑i1/ni=1\sum_{i} 1/n_i = 1∑i​1/ni​=1. If ppp is a prime with nk−1<2pn_{k-1} < 2pnk−1​<2p — that is, ppp exceeds half of the largest denominator — then ppp is not one of the denominators.

The reason is that ppp would then be the only multiple of ppp among the denominators (the next multiple 2p2p2p is already larger than nk−1n_{k-1}nk−1​), so the maximal ppp-adic valuation would be attained exactly once, which is impossible for a sum equal to 111.

Combined with Bertrand's postulate this says that every representation of 111 by distinct unit fractions must skip a prime in the top half of its range: there is always a prime p∈(nk−1/2, nk−1]p \in (n_{k-1}/2,\, n_{k-1}]p∈(nk−1​/2,nk−1​] and no such prime can occur as a denominator.

Preamble
import Mathlib
Formal statement
namespace Erdos287
theorem no_large_prime (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)
    (p : ℕ) (hp : Nat.Prime p) (hlt : f (k - 1) < 2 * p) :
    ∀ i, i < k → f i ≠ p := by sorry
end Erdos287
Source
Auxiliary results proved for the prove2.me mission on Erdős problem #287 (https://www.erdosproblems.com/287). Classical background: P. Erdős, "Egy Kürschák-féle elemi számelméleti tétel általánosítása", Mat. Fiz. Lapok 39 (1932), 17–24; J. Kürschák, Mat. és Fiz. Lapok 27 (1918), 299–300. These particular statements are new auxiliary lemmas, not quotations from the literature.

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