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A block of consecutive integers of bounded 222-adic valuation is short

Proved
Erdos287.block_two_adic_length

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

number-theoryp-adic

If every element of the block m,m+1,…,m+t−1m, m+1, \dots, m+t-1m,m+1,…,m+t−1 of consecutive positive integers has 222-adic valuation at most ccc, then

t≤2 c+1−1.t \le 2^{\,c+1} - 1 .t≤2c+1−1.

Any 2c+12^{c+1}2c+1 consecutive integers contain a multiple of 2c+12^{c+1}2c+1, whose valuation exceeds ccc.

Applied to Erdős problem #287 this makes the structure of a hypothetical counterexample quantitative: a run of even denominators whose halved block has maximal 222-adic valuation ccc contains no multiple of 2c+12^{c+1}2c+1, hence has at most 2c+1−12^{c+1}-12c+1−1 terms and spans fewer than 2c+22^{c+2}2c+2 integers.

Preamble
import Mathlib
Formal statement
namespace Erdos287
theorem block_two_adic_length (c m t : ℕ) (hm : 0 < m)
    (h : ∀ j, j < t → padicValNat 2 (m + j) ≤ c) :
    t ≤ 2 ^ (c + 1) - 1 := by sorry
end Erdos287
Source
Auxiliary results proved for the prove2.me mission on Erdos problem #287 (https://www.erdosproblems.com/287), for the attack on the residual core Erdos287.mixed_gap_core with exactly two runs of even denominators. Classical background: P. Erdos, 'Egy Kurschak-fele elemi szamelmeleti tetel altalanositasa', Mat. Fiz. Lapok 39 (1932), 17-24. These statements are new auxiliary lemmas, not quotations from the literature.

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