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We prove a weighted version of the Szemerédi regularity lemma, which is sufficient for the grid application. For disjoint nonempty vertex sets $U,W$ in a graph, write $d(U,W)=e(U,W)/(|U||W|)$ for their edge density of a bipartite graph. Call the pair $\epsilon$-regular if every $U^{\prime }\subseteq U$, $W^{\prime }\subseteq W$ with $|U^{\prime }|\ge \epsilon |U|$ and $|W^{\prime }|\ge \epsilon |W|$ satisfies $|d(U^{\prime },W^{\prime })-d(U,W)|\le \epsilon$.

The weighted Szemerédi regularity lemma states that, for every $0<\epsilon <1$, there are integers $M,v_0$ such that every graph on $v\ge v_0$ vertices has a set partition into at most $M$ nonempty cells $V_1,\dots ,V_k$, each of size at most $\epsilon v$, for whichThe sum is ordered. The cells need not have equal sizes; their weights replace the usual unweighted count of bad pairs.

To prove it, start with $k_0=\lceil 2/\epsilon \rceil$ nearly equal cells, taking $v\ge k_0$. Each has size at most $v/k_0+1\le \epsilon v$. Define the regularity energyHere even diagonal rectangles use the mean adjacency indicator, counting ordered pairs of vertices. Thus $0\le q\le 1$. Under a refinement, the old rectangle density is the weighted mean of the new ones, and the refinement variance identity for regularity energy givesIn particular refinement cannot decrease energy.

If the irregular-pair weight exceeds $\epsilon v^2$, choose witness subsets $U^{\prime }\subseteq U$, $W^{\prime }\subseteq W$ for every irregular unordered pair, and split each cell simultaneously by all its witnesses. There are at most $k-1$ such subsets in any one cell, so the new number of cells is at most $k{2}^k$. In an irregular rectangle, its witness has relative area at least ${\epsilon }^2$ and density discrepancy greater than $\epsilon$. Since the witness rectangle is now a union of refined rectangles, the Cauchy-Schwarz inequality applied to their weighted density discrepancies bounds the variance contribution from below bySumming over both orientations of all irregular pairs raises $q$ by more than ${\epsilon }^5$. Since $q\le 1$, after at most $\lceil {\epsilon }^{-5}\rceil$ rounds the required weight bound must hold. The finite recurrence $k\mapsto k{2}^k$, starting at $k_0$, supplies $M(\epsilon )$; all cells remain small because we only refine. This proves the stated weighted Szemerédi regularity lemma.

We next derive the triangle removal lemma from this version. First, three disjoint classes $U,W,Z$ whose pairs are $\epsilon$-regular pairs of density at least $d$ have many triangles if $\epsilon \le d/2$ and $\epsilon \le 1/4$. Fewer than $\epsilon |U|$ vertices of $U$ have fewer than $(d-\epsilon )|W|$ neighbours in $W$: otherwise that set of vertices, paired with all of $W$, would violate regularity. The same holds for $Z$. For each of the at least $(1-2\epsilon )|U|$ remaining vertices, both neighbour sets have relative size at least $d-\epsilon \ge \epsilon$. Regularity of $(W,Z)$ then supplies at least $(d-\epsilon )$ times the product of those two neighbour-set sizes edges. Thus the regular triangle counting lemma, also for unequal class sizes, givesFor a desired deletion tolerance $0<\eta <1$, take $\epsilon =\eta /100$, $d=\eta /4$, and a weighted regularity partition with at most $M=M(\epsilon )$ cells. Put $\tau =\epsilon /M$. Delete edges inside cells, between irregular pairs, between regular pairs of density less than $d$, and incident to cells smaller than $\tau v$. The first two deletions each cost at most $\epsilon v^2/2$. The sparse-pair deletion costs at most $d{v}^2/2$. For the last deletion, small-cell deletion in weighted regularity removes at most $M\tau v=\epsilon v$ vertices and at most $\epsilon v^2$ incident edges. The total is bounded byA triangle surviving these deletions would lie in three distinct cells, all of size at least $\tau v$, with all their pairs regular and of density at least $d$. The original graph would then have at least $d^3{\tau }^3{v}^3/16$ triangles. Therefore, witha graph on sufficiently many vertices with at most $\kappa v^3$ triangles can be made triangle-free by deleting fewer than $\eta v^2$ edges. This establishes the triangle removal lemma with every parameter depending only on $\eta$.

Now apply a tripartite graph encoding of a grid. Take separate vertex classes $X=[N]$, $Y=[N]$, $Z=\{2,\dots ,2N\}$. For each grid point $(x,y)$ in the given set, put in the three edgeswhere their endpoints belong respectively to $X\times Y$, $X\times Z$, and $Y\times Z$. Each point supplies a triangle. These $|A|$ triangles are edge-disjoint triangles, since any one of the three edge types uniquely determines the point that supplied it.

Conversely, any triangle with vertices $x\in X$, $y\in Y$, $z\in Z$ forces the three pointsto belong to $A$. Put $d=z-x-y$. For $d\ne 0$ these form a corner in an integer grid; for $d=0$ they all arise from the same grid point. Thus if there is no corner, the graph has exactly $|A|\le N^2$ triangles, yet destroying all triangles requires at least $|A|\ge \delta N^2$ edge deletions.

We may suppose $0<\delta \le 1$, since larger densities make the request vacuous. Use the triangle removal lemma with $\eta =\delta /32$. Our graph has $v=4N-1$ vertices, so, for sufficiently large $N$, its bound $N^2$ on the number of triangles is at most $\kappa v^3$. The lemma would delete fewer thanedges to destroy all triangles. This contradicts the lower bound $\delta N^2$ from the edge-disjoint family. Hence every sufficiently large positive-density square grid contains a nondegenerate corner, proving the requested corners theorem and in particular the existence of a suitable $N$.

First derive the triangle removal lemma from the Szemerédi regularity lemma. Given an allowed deletion fraction $0<\eta <1$, take $d=\eta /10$, $m_0=\lceil 10/\eta \rceil$, and $0<\epsilon \le \min (d/4,\eta /100,1/8)$. Apply part (i), with upper class bound $M$. Delete edges incident with $V_0$, inside each nonexceptional class, between irregular pairs, and between regular pairs of subset density less than $d$. Their total number is at mostThese four terms are bounded respectively by $\epsilon n^2$, $n^2/(2m_0)$, $\epsilon n^2$, and $d{n}^2/2$.

If a triangle in a graph remains, its three vertices lie in three distinct equal classes, all three regular pairs with subset density at least $d$. The permitted regular triangle counting lemma gives at least $(d^3/4)t^3$ triangles in a graph in the original graph, where $t$ is the class size. This lower bound can also be checked directly: all but $2\epsilon t$ vertices of the first class have at least $(d-\epsilon )t$ neighbors in each of the other two. Those neighborhoods are large enough to use the regular pair condition on the remaining pair, giving at least $(d-\epsilon {)}^3{t}^2$ edges between them. Hence the total is at least $(1-2\epsilon )(d-\epsilon {)}^3{t}^3\ge (d^3/4)t^3$.

Since $t\ge n/(2M)$, a surviving triangle in a graph forces at least $d^3{n}^3/(32M^3)$ original triangles in a graph. Thus a graph with fewer than $\xi n^3$ triangles in a graph, where $\xi =d^3/(64M^3)>0$, can be made triangle-free by deleting fewer than $\eta n^2$ edges, for all sufficiently large $n$. This proves the removal lemma needed here.

Now use the triangle-removal proof of Roth theorem. Fix a subset density lower bound $\delta >0$ and suppose $A\subseteq [N]$, $|A|\ge \delta N$, has no nonconstant three-term arithmetic progression. Set $m=2N+1$ and $G={\mathbb{Z}}_m$. Use three disjoint copies $X,Y,Z$ of $G$ and put edges according toDivision by two is valid because $m$ is odd. A triangle in a graph corresponds to $a=y-x$, $b=z-y$, $c=(z-x)/2$ with $a,b,c\in A$ and $a+b=2c$. Since $A\subseteq [N]$ and $|a+b-2c|<m$, this congruence is an integer equality. Progression-freeness therefore forces $a=b=c$.

Every triangle in a graph is consequently of the form $(x,x+a,x+2a)$ with $x\in G,a\in A$, so there are exactly $m|A|$ triangles in a graph. These triangles in a graph are edge-disjoint: an $XY$ edge determines $(x,a)$; a $YZ$ edge determines $a$ and then $x$; and an $XZ$ edge determines $a=(z-x)/2$ and $x$. Eliminating them all requires at least $m|A|$ edge deletions.

The graph has $3m$ vertices but only $m|A|\le mN=O(m^2)$ triangles in a graph. Take $\eta =\delta /100$. For large enough $N$ its triangle in a graph count is below $\xi (3m{)}^3$, so removal requires fewer than $\eta (3m{)}^2=9\delta m^2/100$ deletions. Yeta contradiction. Therefore every fixed positive-density subset of a sufficiently long interval contains a nonconstant three-term arithmetic progression.

This proves the qualitative Roth theorem on three-term arithmetic progressions. The quantitative bound from part 2(i) came from the explicit Fourier iteration; the regularity-removal argument gives a density-dependent threshold without that double-logarithmic estimate.

We derive the triangle removal lemma from the permitted Szemerédi regularity lemma, then turn a corner in an integer grid into a triangle in a graph.

The triangle removal lemma says that for every $\eta >0$ there are $\rho >0$ and $N_0$ such that a graph on $N\ge N_0$ vertices with fewer than $\rho N^3$ triangles in a graph can be made triangle-free by deleting at most $\eta N^2$ edges. Here triangles in a graph are unordered triples of distinct vertices. We prove this consequence with the necessary quantitative dependence.

We may assume $0<\eta <1$. Put $d=\eta /4$, choose an integer $m_0\ge 4/\eta$, and chooseApply the Szemerédi regularity lemma to obtain an exceptional class $V_0$ of size at most $\epsilon N$ and equal-sized classes $V_1,\dots ,V_k$ of size $L$, where $m_0\le k\le M$, and at most $\epsilon k^2$ pairs are not regular pairs of vertex sets. The constant $M$ depends only on these chosen parameters. Delete all edges incident with $V_0$, all edges within a class, all edges between irregular pairs, and all edges between pairs whose edge density of a bipartite graph is less than $d$. The respective costs are at mostTheir sum is at most $\eta N^2/2$, hence certainly at most $\eta N^2$.

If a triangle in a graph survives, it lies in three distinct classes whose three pairs are $\epsilon$-regular pairs of vertex sets of edge density of a bipartite graph at least $d$ in the original graph. We now check the needed regular triangle counting lemma. In the first class all but at most $2\epsilon L$ vertices have at least $(d-\epsilon )L$ neighbours in each of the other two classes; otherwise the definition of a regular pair of vertex sets would be violated. For each such vertex $v$, its two neighbour sets each have size at least $\epsilon L$, so their mutual edge density of a bipartite graph is at least $d-\epsilon$. It follows that the original three classes contain at leasttriangles in a graph. The elementary lower bound holds because $\epsilon \le 1/4$ and $\epsilon \le d/4$. Also $L=(N-|V_0|)/k\ge N/(2M)$. Thus a surviving triangle in a graph implies at least $d^3{N}^3/(64M^3)$ original triangles in a graph. Set $\rho =d^3/(128M^3)$ and take $N_0$ large enough for the Szemerédi regularity lemma. Fewer than $\rho N^3$ original triangles in a graph force the cleaned graph to be triangle-free, proving the triangle removal lemma.

Now suppose $0<\delta \le 1$ and $A\subseteq [n{]}^2$. Use the tripartite graph encoding of a grid to construct a tripartite graph with disjoint labelled classesFor each $(a,b)\in A$, put in the three edges joining $a\in X$, $b\in Y$, and $a+b\in Z$. These yield $|A|$ canonical triangles in a graph. They are pairwise edge-disjoint triangles: an $XY$ edge determines $(a,b)$ directly, an $XZ$ edge determines $b=z-a$, and a $YZ$ edge determines $a=z-b$. Any deletion making the graph triangle-free must therefore remove at least $|A|$ edges.

On the other hand, an arbitrary triangle in a graph with labels $(x,y,z)$ gives three points of $A$:Writing $d=z-x-y$, these are $(x,y),(x,y+d),(x+d,y)$. If $d\ne 0$, they form the required corner in an integer grid. If there is no such corner in an integer grid, every triangle in a graph is canonical, and the total number is precisely $|A|\le n^2$.

Apply the triangle removal lemma with $\eta =\delta /32$. The constructed graph has $N=4n$ vertices. For sufficiently large $n$, $N\ge N_0$ andIn the absence of a corner in an integer grid, the triangle removal lemma would destroy all triangles in a graph by deleting at mostedges, contradicting the pairwise edge-disjoint triangles. Every sufficiently large grid therefore has the asserted corner in an integer grid in every subset of positive fixed density of a finite subset. For $\delta >1$ there is no eligible subset, so the assertion is vacuous. This proves the corners theorem, including the stipulated nonzero displacement; no positivity of $d$ was required.

A partition with at most $M$ cells has at most $\epsilon n$ vertices in cells smaller than $\epsilon n/M$. Deleting all edges incident to those vertices costs at most $\epsilon n^2$ edges. Every remaining cell has a fixed positive fraction of all vertices, which permits a regular triangle counting lemma to give a cubic count without requiring equal cell sizes.

For every $\eta >0$, some $\rho >0$ and $N_0$ have this property: a graph on $N\ge N_0$ vertices with fewer than $\rho N^3$ unordered triangles in a graph can be made triangle-free by deleting at most $\eta N^2$ edges. A Szemerédi regularity lemma partition deletes exceptional, intraclass, irregular and sparse-pair edges cheaply. A surviving triangle in a graph would force a positive cubic count by the regular triangle counting lemma.