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Arithmetic of the rank-rich exclusion: ∣Q∣4≤2132n3|Q|^4\le2^{132}n^3∣Q∣4≤2132n3, 64∣Q∣≥n⇒n≤215664|Q|\ge n\Rightarrow n\le2^{156}64∣Q∣≥n⇒n≤2156 (Erdős 624)

Proved
Erdos624.rank_rich_exclusion_arith

by Ethan Cheung · Oct 24, 2026 · Mathlib 0df444a (Lean v4.33.1)

combinatoricserdos-problems

Pure arithmetic. Let n,sn,sn,s be natural numbers with n>0n>0n>0. If n≤64sn\le 64sn≤64s and s4≤2132n3s^4\le 2^{132}n^3s4≤2132n3, then n≤2156n\le2^{156}n≤2156.

In the cross-review, s=∣Q2∣s=|\mathcal Q_2|s=∣Q2​∣ and the bound ∣Q2∣4≤2132n3|\mathcal Q_2|^4\le2^{132}n^3∣Q2​∣4≤2132n3 is the a=2a=2a=2 case of the assumed packing theorem (P); the conclusion gives the k≥157k\ge157k≥157 exclusion. (P) is not claimed here; it enters only as the numeric hypothesis h2h_2h2​.

Context: Erdős Problem 624 asks whether H(n)−log⁡2n→∞H(n)-\log_2 n\to\inftyH(n)−log2​n→∞, where H(n)H(n)H(n) is the least mmm such that some f:P(X)→Xf:\mathcal P(X)\to Xf:P(X)→X with ∣X∣=n|X|=n∣X∣=n satisfies f[P(Y)]=Xf[\mathcal P(Y)]=Xf[P(Y)]=X for all ∣Y∣≥m|Y|\ge m∣Y∣≥m. This is an auxiliary finite statement from Sol's Round 12 Step 17 cross-review / resilience analysis of the two-track (PAIR) case. It does not settle Erdős Problem 624, PAIR or Step 17.

Preamble
import Mathlib
open Finset
Formal statement
namespace Erdos624

theorem rank_rich_exclusion_arith (n s : ℕ) (hn : 0 < n) (h1 : n ≤ 64 * s)
    (h2 : s ^ 4 ≤ 2 ^ 132 * n ^ 3) : n ≤ 2 ^ 156 := by sorry

end Erdos624
Source
Erdős Problem 624 (https://www.erdosproblems.com/624); Round 12: Sol's 'Step 17: First-Round Cross-Review' and 'Step 17: resilient relations below the top layer'; derived from GPT 6.1 Sol's analysis, relayed by Ethan Cheung; hand-checked and Lean-verified by Grok Bot for Ethan Cheung (Oct 24 2026). Auxiliary result only; does NOT settle PAIR / Step 17 / #624.

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