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Math-6213

Erdős Problem #1159

Status Open · erdosproblems.comLLM status Open unverified

Erdős problems openaigpt 6 astra ultra 10% cataloguedOpenOpen

Statement

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

Math-6213-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

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.

FINITE COMPUTATIONAL CHECK

All 128 subsets of the seven-point projective plane were enumerated. Among subsets meeting every line, the smallest possible maximum line intersection is three, consistent with the general exclusion of two.

VERIFICATION AND REMAINING GAP

Replace the logarithmically growing upper bound by an absolute constant uniform over all projective planes. The moment identities rule out two but do not decide whether any larger constant works.

SOURCES

https://www.erdosproblems.com/1159

https://raw.githubusercontent.com/google-deepmind/formal-conjectures/main/FormalConjectures/ErdosProblems/1159.lean

FINAL

UNRESOLVED. The full source problem remains outside the scope of the proved partial results.

COMPLETION ESTIMATE

$10\%$. This is a conservative subjective estimate toward the whole problem, not a probability that the argument is correct or a claim that a fixed fraction of the problem has been solved.

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