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Let $\mathcal{F} = \{A_1, \cdots, A_n\}$ be a family of sets with the following conditions.

  1. For all $i$, $|A_i| = p^k+1$ where $p$ is prime and $k$ is an integer
  2. If $i \neq j$, $|A_i \cap A_j| = 1$
  3. $\bigcap_{i=1}^n A_i = \varnothing$

I should find the largest $n$, with such $\mathcal{F}$ exists, where $p$, $k$ are fixed.

Considering the conditions about $p^k$ and the intersection, I think this is the problem using the finite field. I might construct the family by thinking about "lines" on finite field: namely, let $A_i$'s correspond to the sets of tuples $(x,y) \in GF(p^k)^2$ such that $y = ax+b$.

This gives me a vague idea about the answer: $n = p^{2k} + O(p^k) + \cdots$? However I cannot specify those foggy thoughts to actual proof.

Thanks in advance for any form of help, hint, or solution.

mathhello
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    The projective plane $\Bbb{P}^2K$, $K=\Bbb{F}_{q}, q=p^k$ has $q^2+q+1$ lines, each pair intersecting at exactly one point. The projective space is more "homogeneous" than its affine brother, where we have exceptional cases like parallel lines and such. – Jyrki Lahtonen Aug 24 '22 at 04:50
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    The third condition (together with the second) implies that there cannot be more than $q+1$ "lines" intersecting at a given point. Do you see why? I think that should lead to a proof that $q^2+q+1$ is the maximum. Without the third condition there obviously won't be any upper bound at all. I'm fairly sure there exists a sleek counting argument involving the incidence data leading to the maximum :-) – Jyrki Lahtonen Aug 24 '22 at 05:04
  • Just in case this contest is current, I'm leaving it with the extended hints above. Enjoy! – Jyrki Lahtonen Aug 24 '22 at 05:12
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    @JyrkiLahtonen, $+1$ for the phrase "affine brother" – MAS Aug 24 '22 at 05:16

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