High-energy Roth theorem Created 2026-10-07 Updated 2026-10-14
For fixed , a sufficiently large finite set of integers with additive energy at least has a nonconstant three-term arithmetic progression. The Balog-Szemerédi-Gowers theorem and Ruzsa modelling lemma reduce the problem to the Roth theorem on three-term arithmetic progressions in a dense cyclic model.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 129 4 c Solution 2026-09-28
The Dense Bogolyubov-Ruzsa lemma says that, for every , if has density at least in a cyclic group of prime order, then contains a proper generalized arithmetic progression of rank and size .
Now suppose and . The Ruzsa modelling lemma, taken at a sufficiently high fixed Freiman order, supplies with and a Freiman isomorphism from to a subset , where is prime, , and . Apply the dense Bogolyubov-Ruzsa lemma to and transfer the resulting progression back through the Freiman model. We obtain a proper progressionof rank and size .
The Plünnecke-Ruzsa inequality givesThere are pairs with and , distributed among the sums in . Some therefore has at least representations . Equivalently,Set . Translation and negation preserve properness and rank. Moreover and another use of Plünnecke gives . Thusas required.
For every , some has this property: any nonempty finite with has an arithmetic progression of length at least in . The small-difference-set form of the Balog-Szemerédi-Gowers theorem, Ruzsa modelling lemma, cyclic Bogolyubov lemma and nonwrapping progression in a cyclic Bohr set prove it. An order-eight Freiman s-isomorphism suffices to lift the progression because its consecutive second-difference equations expand into equalities of eight-term sums.
Ruzsa modelling lemma Created 2026-09-28 Updated 2026-11-05
For a nonempty finite set with bounded doubling constant and a fixed positive integer , the Ruzsa modelling lemma finds a subset of size at least that is Freiman s-isomorphic to a dense subset of a finite cyclic group. The model order is at most when . The cyclic Freiman model of a small-doubling integer set gives an explicit order-eight version.
The integer, or appropriate torsion-free, setting matters. If , then for every . A Freiman 2-isomorphism into a cyclic group would force . All image differences would lie in its subgroup of elements killed by two, which has at most two elements. Thus no subset with more than two points can be Freiman 2-isomorphic to a cyclic model. General abelian groups need models retaining their torsion, rather than an unrestricted cyclic-model assertion.