The <Ruzsa triangle inequality> applied to $A-A$ also bounds $|A+A|$ in terms of $C|A|$, so $A$ has bounded doubling. The <Freiman-Ruzsa theorem> places $A$ inside a proper coset progression $P$ whose rank and ratio $|P|/|A|$ are bounded only in terms of $C$. Hence $A$ has positive density bounded in terms of $C$ inside a bounded-rank progression.

For sufficiently large $|A|$, the <Szemerédi theorem in a bounded-rank coset progression> gives a nontrivial three-term <arithmetic progression> in $A$. This proves the assertion.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 129 / 3 / iii / Solution

For every $C$ there is $N(C)$ such that a set $A\subseteq\mathbb Z$ with $|A|\geq N(C)$ and $|A-A|\leq C|A|$ contains a nontrivial three-term <arithmetic progression>. The <Freiman-Ruzsa theorem> places $A$ densely in a bounded-rank coset progression, where the <Szemerédi theorem in a bounded-rank coset progression> applies.

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 Incoming links: Small difference set forces a three-term arithmetic progression