Monochromatic sums

DefinitionErdős Problem 484: linearly many monochromatic sums#✓fully proved

This is the statement that there is a real constant \(c > 0\) such that for every integer \(k \ge 1\) there is an \(N_0 \in \mathbb{N}\) with the following property: for every \(N \ge N_0\) and every colouring \(f : \mathbb{N} \to \{0, \dots, k-1\}\), at least \(cN\) of the integers \(n \in \{1, \dots, N\}\) can be written as \(n = a + b\) with \(a, b \in \{1, \dots, N\}\), \(a \ne b\) and \(f(a) = f(b)\).

Only the colours of \(1, \dots, N\) matter, so this is a statement about \(k\)-colourings of \(\{1, \dots, N\}\); \(c\) does not depend on \(k\), while \(N_0\) may.

def monochromaticSumsLinearCount : Prop :=
  ∃ c : ℝ, 0 < c ∧ ∀ k : ℕ, 0 < k → ∃ N₀ : ℕ, ∀ N : ℕ, N₀ ≤ N → ∀ f : ℕ → Fin k,
    c * N ≤ ({n : ℕ | n ∈ Finset.Icc 1 N ∧
      ∃ a ∈ Finset.Icc 1 N, ∃ b ∈ Finset.Icc 1 N, a ≠ b ∧ f a = f b ∧ a + b = n}.ncard : ℝ)
Used by 1

TheoremErdős–Sárközy–Sós theorem on monochromatic sums (Erdős Problem 484)#●ready to prove

There is a real constant \(c > 0\) such that for every integer \(k \ge 1\) there is an \(N_0 \in \mathbb{N}\) such that for every \(N \ge N_0\) and every colouring \(f : \mathbb{N} \to \{0, \dots, k-1\}\), at least \(cN\) of the integers \(n \in \{1, \dots, N\}\) can be written as \(n = a + b\) with \(a, b \in \{1, \dots, N\}\), \(a \ne b\) and \(f(a) = f(b)\).

Erdős, Sárközy and Sós proved that at least \(N/2 - O(N^{1 - 1/2^{k+1}})\) even numbers in \(\{1, \dots, N\}\) are such monochromatic sums, so any \(c < 1/2\) works.

public theorem_wanted monochromaticSumsLinearCount_holds : monochromaticSumsLinearCount
Sources
  • T. F. Bloom, Erdős Problems, Problem #484, https://www.erdosproblems.com/484 (status PROVED (LEAN)); resolution: P. Erdős, A. Sárközy and V. T. Sós, On a conjecture of Roth and some related problems. I (1989), 47-59, erdosproblems.com/484: at least N/2 - O(N^{1 - 1/2^{k+1}}) even numbers are monochromatic sums, https://www.erdosproblems.com/484.
Uses 1
1 fully proved1 ready to prove

The dependency explorer needs JavaScript and its graph data.

Download the complete graph data.