Erdős Problem #1159
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Erdős problems openaigpt 6 astra ultra 10% cataloguedOpenOpenStatement
Is there an integer $C>1$, independent of the plane, such that every finite projective plane has a point set $S$ satisfying \[ 1\leq |S\cap\ell|\leq C \] for each line $\ell$ of that plane?
Restatement from Math-6213-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Math-6213
Erdős Problem 1159 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
Is there an integer $C>1$, independent of the plane, such that every finite projective plane has a point set $S$ satisfying \[ 1\leq |S\cap\ell|\leq C \] for each line $\ell$ of that plane?CONTEXT AND SCOPE
This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.ATTACK PLAN AND WORK
Consider a projective plane of order $q\ge2$, with $N=q^2+q+1$ points and lines. Let $s=|S|$ and $k_\ell=|S\cap\ell|$. Counting incidences and pairs of selected points gives \[\sum_\ell k_\ell=(q+1)s,\qquad \sum_\ell k_\ell(k_\ell-1)=s(s-1).\tag{1}\] If $1\le k_\ell\le C$, these imply $s\ge q+1$ and $s\le(C-1)(q+1)+1$. There is a stronger exclusion for $C=2$. In that case $k(k-1)=2(k-1)$ for each permitted value $k=1,2$, so $1$ forces \[s^2-(2q+3)s+2(q^2+q+1)=0.\] Its discriminant is $-4q^2+4q+1<0$ for $q\ge2$. Thus the universal constant, if it exists, must be at least three.For comparison, a logarithmic bound follows by independent sampling. For large $q$, choose each point with probability $p=4\log N/(q+1)\le1$. A fixed line is missed with probability at most $N^{-4}$. For $t=\lceil16\log N\rceil$, its upper tail is bounded by $(e(q+1)p/t)^t\le(e/4)^t\le N^{-6}$, using the standard binomial union bound $\Pr(X\ge t)\le\binom{q+1}{t}p^t\le(e(q+1)p/t)^t$. The union of the two bad events over all $N$ lines has probability less than one. Hence some set meets every line between one and $O(\log q)$ times. The finitely many small orders can be handled by taking all points.