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Jan 2, 2023 at 23:02 history bounty ended CommunityBot
Dec 29, 2022 at 15:30 history edited YCor CC BY-SA 4.0
fixed minor errors and added some material
Dec 27, 2022 at 15:55 comment added YCor @user369335 ah, thanks! indeed I misread your table, from which indeed $(31,10)$ is excluded.
Dec 27, 2022 at 15:42 comment added user369335 For $p=31$, I have checked the solutions and my computational result shows that $|I|=0,31,1,30,6,25,15,16$. $(p,|I|)=(31,10)$ is impossible in my opinion. @YCor
Dec 27, 2022 at 13:49 comment added YCor Note that this explains the case $(p,|I|)=(31,6)$ but not the case $(p,|I|)=(31,10)$.
Dec 27, 2022 at 10:47 comment added YCor I got it. For a prime power $m$, Singer finds an element $T$ of order $p=m^2+m+1$ in $\mathrm{PGL}_3(\mathbf{F}_m)$. So $\langle T\rangle$ acts simply transitively on $\mathrm{P}^2(\mathbf{F}_m)$. Fix $x_0\in \mathrm{P}^2(\mathbf{F}_m)$. Define $I$ as the set of $i\in\mathbf{Z}/p\mathbf{Z}$ such that $x_0,Tx_0,T^ix_0$ are aligned. Then $I$ is the desired subset. Example: $m=5$, $p=31$, $T=\begin{pmatrix}0&0&1\\1&0&0\\0&1&1\end{pmatrix}$, $x_0=[1:0:0]$, $I=\{0,1,3,8,12,18\}$.
Dec 27, 2022 at 8:08 comment added YCor @user369335 thanks! it indeed says there is a solution with $c=1$ (and cardinal $m+1$) modulo $p=m^2+m+1$ for every prime power $m$. This indeed realizes the following values of $(p,q)$: $(7,3)$, $(13,4)$, $(31,6)$, $(73,9)$, $(307,18)$, etc. I'll need to look closer into it to understand the construction of the desired subset.
Dec 27, 2022 at 7:35 comment added user369335 For $p=31=2^5-1$ and $p=73$, I have found something related in this paper -- "A theorem in finite projective geometry and some applications to number theory" by Singer. ams.org/journals/tran/1938-043-03/S0002-9947-1938-1501951-4/…
Dec 27, 2022 at 2:23 vote accept user369335
Dec 26, 2022 at 22:17 history edited YCor CC BY-SA 4.0
added potential values
Dec 26, 2022 at 22:06 history answered YCor CC BY-SA 4.0