Triangle-removal proof of Roth theorem

Embed $A\subseteq [N]$ in the odd cyclic group $G=\mathbb{Z}/(2N+1)\mathbb{Z}$, so additive three-term relations in $A$ cannot wrap around. Form three copies of $G$, putting edges $x\sim y$ when $y-x\in A$, $y\sim z$ when $z-y\in A$, and $x\sim z$ when $(z-x)/2\in A$. A triangle in a graph encodes $a+b=2c$ in $A$. If $A$ has no nonconstant arithmetic progression, every triangle in a graph has $a=b=c$; there are exactly $|G||A|$ triangles in a graph, and they are edge-disjoint. Thus at least $|G||A|$ edges must be removed to eliminate them. For fixed positive subset density of $A$, this is a positive fraction of the squared vertex count, whereas the triangle in a graph count is only quadratic. The triangle removal lemma permits deletion of an arbitrarily small quadratic fraction when the cubic triangle in a graph subset density is sufficiently small. This contradiction establishes Roth theorem on three-term arithmetic progressions.