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Invertible TTT on F2d\mathbb F_2^dF2d​ with I+TI+TI+T invertible exists iff d≠1d\ne1d=1 (Erdős 624 cross-review)

Proved
Erdos624.linear_ingredient_iff

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

combinatoricserdos-problems

For a natural number ddd there is a linear map T:F2d→F2dT:\mathbb F_2^d\to\mathbb F_2^dT:F2d​→F2d​ such that both TTT and I+TI+TI+T are bijective if and only if d≠1d\neq1d=1. (For d=0d=0d=0 the space is trivial; for d=1d=1d=1 the only invertible map is T=IT=IT=I, and then I+T=0I+T=0I+T=0.)

This is the linear ingredient of the cross-review's local counterexample (blocks E1={p,q}∪supp(z)E^1=\{p,q\}\cup\mathrm{supp}(z)E1={p,q}∪supp(z), E2={q,r}∪supp(Tz)E^2=\{q,r\}\cup\mathrm{supp}(Tz)E2={q,r}∪supp(Tz), E3={p,r}∪supp((I+T)z)E^3=\{p,r\}\cup\mathrm{supp}((I+T)z)E3={p,r}∪supp((I+T)z)), used there with d=k−6≥58d=k-6\ge58d=k−6≥58.

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 linear_ingredient_iff (d : ℕ) :
    (∃ T : (Fin d → ZMod 2) →ₗ[ZMod 2] (Fin d → ZMod 2),
      Function.Bijective T ∧ Function.Bijective (fun v => v + T v)) ↔ d ≠ 1 := 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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