Given integers $k,n\ge 1$, I shall write $\Bbb{Z}_k^n := (\Bbb{Z}/k\Bbb{Z})^n$.
Fix $k\ge 3$. Let $r_k(\Bbb{Z}_k^n)$ denote the cardinality of the largest $A\subset \Bbb{Z}_k^n$, such that $A$ does not contain any $k$-term progression $P\subset A$ with $|P| = k$.
It is clear that $r_k(\Bbb{Z}_k^1)=k-1$. Inductively, by taking product sets, we deduce that $r_k(\Bbb{Z}_k^n) \ge (k-1)^n$ for all $n\ge 0$.
Is it known that there exists $c_k>0$ such that $r_k(\Bbb{Z}_k^n) \gg (k-1+c_k)^n$ (equivalently, by considering product sets, that there exists some $n_k$ such that $r_k(\Bbb{Z}_k^{n_k}) > (k-1)^{n_k}$)?
This is known to be true for $k=3$ (see e.g., the construction of Edel or the recent work of Tyrrell), and presumably one could check this for a large number of small $k$ via computer search. But I am wondering if there is a proof that this $c_k>0$ always exists, or at the very least it would be nice if I could be directed to some literature on what is known if this is a hard problem.
Best Answer
I think that this is known. A good source to check recent things would be https://arxiv.org/abs/2211.02588.