GPT 6 Astra Ultra attempt on Math-5454
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FORMAL RESTATEMENT
Erdős #139; reconstruction, 2026-09-23. For a fixed integer $k\ge2$, let $r_k(N)$ be the largest size of a subset of $[N]$ containing no $k$-term arithmetic progression with nonzero common difference. Prove $r_k(N)=o(N)$ as $N\to\infty$.QUICK LITERATURE AND CONTEXT CHECK
This is Szemerédi's theorem. We give the standard hypergraph-removal route, verifying the algebraic encoding and the edge-disjoint copies. The hypergraph removal lemma is a substantial external theorem and is not proved here. This is a reconstruction from that explicit input, not an independent elementary proof or a new result.WORK
We use the following form of the hypergraph removal lemma. For fixed $r$ and a fixed $r$-uniform hypergraph $F$, a hypergraph on $v$ vertices with $o(v^{|V(F)|})$ labelled copies of $F$ can be made $F$-free by deleting $o(v^r)$ edges. Equivalently, for every $\epsilon>0$ there is $\delta>0$ such that at most $\delta v^{|V(F)|}$ copies suffice for an $\epsilon v^r$ deletion bound, once $v$ is large enough. This theorem follows from the work of Gowers and independently Nagle--Rödl--Schacht--Skokan; see the primary reference below.For $k=2$, any two distinct elements form a progression, so $r_2(N)=1$. Fix $k\ge3$ and a progression-free $A\subseteq[N]$. By Bertrand's postulate choose a prime $p$ with $2N<p<4N$, taking $N$ large enough that $p>k$. Work over $\mathbb F_p$ and form $k$ vertex classes $V_1,\ldots,V_k$, each a copy of $\mathbb F_p$. A $(k-1)$-tuple omitting class $i$ is an edge exactly when \[ \psi_i(x_1,\ldots,\widehat{x_i},\ldots,x_k) :=\sum_{j\ne i}(j-i)x_j\in A. \tag{1} \] A copy of the complete $(k-1)$-uniform hypergraph on $k$ vertices must use all $k$ classes. Indeed, every pair of its vertices lies in some edge and hence cannot lie in the same class.
For a transversal tuple $(x_1,\ldots,x_k)$ put $u=\sum_jjx_j$ and $d=\sum_jx_j$. Its $k$ edge labels are \[ a_i=\psi_i=u-id\quad(1\le i\le k). \tag{2} \] If all these labels belong to $A$, regard them as integers in $[N]$. Every integer second difference $a_i-2a_{i+1}+a_{i+2}$ is zero modulo $p$ and has absolute value less than $2N<p$. It is therefore exactly zero. Thus the labels form an ordinary integer arithmetic progression. Since $A$ is progression-free, it is constant, so $d=0$ in the field. Conversely, $d=0$ and $u=a\in A$ give a complete copy.
The linear forms $\sum_jx_j$ and $\sum_jjx_j$ are independent over $\mathbb F_p$, so each pair of prescribed values $(0,a)$ has exactly $p^{k-2}$ preimages. The number of complete transversal copies is therefore exactly \[ |A|p^{k-2}\le Np^{k-2}=O(p^{k-1})=o(p^k). \tag{3} \] More importantly, these copies are edge-disjoint. Once a face omitting $i$ is fixed, there is exactly one possible missing coordinate with $\sum_jx_j=0$, namely $x_i=-\sum_{j\ne i}x_j$. There cannot be a second complete copy containing that face, by $2$ and progression freeness. This also shows that every edge in $1$ belongs to exactly one complete copy.
Apply removal with $r=k-1$ and $F$ the complete $(k-1)$-uniform hypergraph on $k$ vertices. There are $kp$ vertices, and $3$ gives $o((kp)^k)$ labelled copies (the fixed labelling factor is harmless). We can destroy all copies by removing $o(p^{k-1})$ edges. But their edge-disjointness requires at least one different deleted edge for each of the $|A|p^{k-2}$ copies. Hence \[ |A|p^{k-2}=o(p^{k-1}),\qquad |A|=o(p)=o(N). \] The constants in the removal statement depend only on fixed $k$, so the conclusion is uniform over $A$ and proves $r_k(N)=o(N)$. $\square$