I found this as an exercise in the context of capping Van der Waerden numbers:
For each arithmetic progression $S \subseteq [n] = \{1,...,n\}$ with $|S| = k$ there are at most $nk$ other such progressions $T$ with $S \cap T \neq \emptyset$. I can't find a solution on how to solve this. Ideas:
If $S = \{a_1,a_1+d_1,a_1+2d_1,\dots\}$ and we want to pick an intersecting $T = \{a_2,a_2+d_2,a_2+2d_2,\dots\}$ then $$a_1+id_1 = a_2+jd_2$$ where $i,j \le k$. So if we pick $a_2 \le n-k$, then $d_2$ is defined by $i,j$ but I think there are more than $k$ valid possibilities. Maybe picking $a_2$ arbitrary was too much? I have a really difficult time finding out how to pick $a$ depending on $d$ (or the other way around) such that intersection happens.
Furthermore:
We know that $d_2 \le \frac{n}{k-1}$ but I couldn't find how that helps. Same with the total number of progressions being capped by $\frac{(n-k)n}{k-1}$. I even got that lowered to $\frac{2n^2 - (n-k+1)(n-k+2)}{(k-1)2}$ which doesn't help.