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](https://arxiv.org/abs/2209.10045)), 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.



Beating trivial bound for $k$-AP-free sets in characteristic $k$
I think that this is known. A good source to check recent things would be
<https://arxiv.org/abs/2211.02588>.


