gpt pro 5.2 attempt on Math-713

Erdős problem 196

1) FORMAL RESTATEMENT

A permutation of $\mathbb{N}$ is a bijection $\pi:\mathbb{N}\to\mathbb{N}$ viewed as a sequence $(\pi(1),\pi(2),\dots)$. A monotone $k$-term arithmetic progression in $\pi$ means there exist indices $i_1<\dots<i_k$ such that $\pi(i_1),\dots,\pi(i_k)$ form an arithmetic progression and are monotone in that index order. Question: Must every permutation of $\mathbb{N}$ contain a monotone 4-term arithmetic progression?

2) QUICK LITERATURE/CONTEXT CHECK

Problem text states: DEGS77 proved every permutation contains a monotone 3-term AP. The 4-term case is asked and is treated as open in the statement.

3) ATTACK PLAN

Provide two rigorous lemmas: $1$ a clean equivalence formulation in terms of the position function, $2$ a compactness/extension lemma explaining why large finite avoiders do not automatically yield infinite avoiders. Then give small computational evidence (finite avoiders exist for large $n$ but may fail to extend).

4) WORK

Lemma 4.1 (Position-function formulation).

Let $\mathrm{pos}(x)$ be the unique index with $\pi(\mathrm{pos}(x))=x$. Then $\pi$ contains a monotone $k$-AP iff there exist $a,d\ge 1$ such that the sequence $a,a+d,\dots,a+(k-1)d$ has positions $\mathrm{pos}(a),\dots,\mathrm{pos}(a+(k-1)d)$ that are monotone (increasing or decreasing). Proof. If $\pi(i_1),\dots,\pi(i_k)$ are values of a $k$-AP in monotone order along increasing indices, then those values can be written as $a+td$ and their indices are exactly $\mathrm{pos}(a+td)$ which are monotone. Conversely if the positions of a value-AP are monotone, reading $\pi$ at those indices yields a monotone AP subsequence. $\square$

Lemma 4.2 (Kőnig-tree extension principle).

Fix $k\ge 3$. Let $\mathcal{T}$ be the rooted tree whose vertices at depth $n$ are permutations of $\{1,\dots,n\}$ that avoid monotone $k$-APs, with an edge from $\sigma\in S_n$ to $\tau\in S_{n+1}$ if $\tau$ restricts to $\sigma$ on $\{1,\dots,n\}$. If every node has at least one child (i.e. every avoiding permutation extends), then there exists an infinite permutation of $\mathbb{N}$ avoiding monotone $k$-APs. Proof. $\mathcal{T}$ is finitely branching (each node has at most $n+1$ children). If every depth has at least one node and every node has a child, then by Kőnig’s lemma there is an infinite root-to-infinity path, i.e. a coherent sequence of extensions $\sigma_1\prec\sigma_2\prec\cdots$. This defines an infinite permutation of $\mathbb{N}$ whose every finite restriction avoids monotone $k$-APs; hence the whole permutation avoids. $\square$

FAST REALITY CHECK (computed)

Heuristic local-search found explicit finite permutations avoiding monotone 4-APs for $n=20,25,30$. Example for $n=20$ (verified by direct check): \[ (11,2,15,13,18,1,17,6,14,12,8,7,3,10,5,16,20,9,19,4) \] contains no monotone 4-term AP among values in $\{1,\dots,20\}$ (checked by brute enumeration of all 4-APs in $[20]$).

6) FINAL

UNRESOLVED

(i) Strongest proved partial results: clean reformulation via $\mathrm{pos}$ (Lemma 4.1) and an extension/compactness criterion (Lemma 4.2). Finite avoiders exist for sizeable $n$ (computed examples), so any infinite proof must use nontrivial extension obstructions.

(ii) First gap: prove that every infinite permutation must contain a monotone 4-AP, or construct one avoiding them.

(iii) Top 3 next moves: 1. Prove that the extension tree $\mathcal{T}$ for $k=4$ has only finite height (no infinite branch). 2. Try to show a density/regularity property of position function forces monotone 4-AP. 3. Search systematically for partial permutations that cannot be extended beyond some length while avoiding monotone 4-AP.

(iv) Minimal counterexample structure (if 4-AP is unavoidable): existence of some finite $N$ such that every permutation of $\{1,\dots,N\}$ already contains a monotone 4-AP, forcing it inside any infinite permutation via restriction.