Let $q$ be a Sophie Germain prime number, i.e. $2q+1=p$ is prime. Consider the set $\{1,2,3,\ldots,p-1\}$. Then what is the maximum size of a subset of this set, such that the subset contains no two elements, $a,b$, such that:
$b\equiv \{a,2a,3a,4a,5a\}\pmod {p-1}$
$a\equiv \{b,2b,3b,4b,5b\}\pmod {p-1}$.
For example, for $q=11$ we have $p=23$, and we can pick the 5 element subset $\{2,5,11,16,21\}\subset \{1,2,3,4,\ldots,22\}$ that satisfies the property.
In general, how big can we make "5"?
I haven't been able to find any UPPER bounds, but I know from experimenting a lot that it cannot be a linear (or higher) function of $p$. So it has to be like $\sqrt{2p}$ or something.
EDIT: I added the word "upper" before "bound", as I am looking mainly for upper bounds
Every new number rules out the nine numbers $a,2a,3a,4a,5a,a/2,a/3,a/4,a/5$. So there should be at least $\lfloor (p+8)/9\rfloor$ numbers in the set. Perhaps there is overlap, and you can find more.