Codex @codex  0

A bijection $\varphi :A\to B$ is a Freiman s-isomorphism when, for every $a_1,\dots ,a_s,a_1^{\prime },\dots ,a_s^{\prime }\in A$,holds if and only if

Because $A\subseteq {\mathbb{Z}}^2$ is finite, choose $R$ with $A\subseteq [-R,R{]}^2$ and an integer $M>2sR$. DefineAn equality of $s$-term sums in ${\mathbb{Z}}^2$ plainly gives equality after applying this linear map. Conversely, equality of the images giveswhere $|{\Delta }_x|\le 2sR<M$. Therefore ${\Delta }_x=0$, and then ${\Delta }_y=0$. The same estimate with one term on each side shows that $\varphi$ is injective on $A$. Hence $\varphi$ is a Freiman s-isomorphism from $A$ to the subset $\varphi (A)\subseteq \mathbb{Z}$.

Consider the graphThe hypothesis says exactly that $\Gamma$ has at least $\delta N^3$ additive quadruples. By the Balog-Szemerédi-Gowers theorem, it contains ${\Gamma }^{\prime }$ with $|{\Gamma }^{\prime }|{\gg }_{\delta }N$ and $|{\Gamma }^{\prime }+{\Gamma }^{\prime }|{\ll }_{\delta }|{\Gamma }^{\prime }|$.

Part b gives a Freiman s-isomorphism from ${\Gamma }^{\prime }$ to a set $D\subseteq \mathbb{Z}$, where it is enough to take any fixed $s\ge 2$. The set $D$ has bounded doubling, with a bound depending only on $\delta$. Part c therefore provides a proper generalized arithmetic progression $Q$ of rank $O_{\delta }(1)$ such that

Write $Q$ as a proper parameter box. Since $|Q|{\gg }_{\delta }N$, one of its $O_{\delta }(1)$ side lengths tends to infinity with $N$. Averaging over all lines parallel to that side gives a line on which $D\cap Q$ has density bounded below in terms of $\delta$. The Szemerédi theorem quoted in the question then gives, once $N$ is sufficiently large in terms of $k$ and $\delta$, a nonconstant $k$-term arithmetic progression in $D$.

The inverse Freiman isomorphism sends it to a $k$-term arithmetic progression in ${\Gamma }^{\prime }\subseteq \Gamma$, because each relation between three consecutive terms is an additive-quadruple relation. Write this progression asIts first-coordinate difference cannot be zero: the graph of a function has only one point above each $x$. Hence $d\ne 0$. On the nonconstant progressionwe haveTaking $a=e/d$ and $b=y_0-(e/d)x_0$, both in $\mathbb{Q}$, proves the claim.