GPT 6 Astra Ultra attempt on Math-5421
mathematicsErdős problemscontributionManuscript
Contents
Erdős Problem 92: many equidistant neighbors at every point
1) Statement and the source of a counterexample
For a finite planar set $A$, put \[ g(A)=\min_{x\in A}\max_{r>0} |\{y\in A\setminus\{x\}:|x-y|=r\}|, \qquad f(n)=\max_{|A|=n}g(A). \] Is $f(n)\leq n^{o(1)}$, or even $f(n)\leq n^{O(1/\log\log n)}$? Both assertions are false. We give the complete graph-theoretic deduction from the established negative resolution of Problem 90. This is a reconstruction of the known implication, not a new counterexample construction.The explicit external input is the theorem of OpenAI's Planar Point Sets with Many Unit Distances: there is a fixed $\delta>0$ and sets $P_j$ of $N_j\to\infty$ planar points whose unit-distance graphs have at least $N_j^{1+\delta}$ edges. The geometric part of a reconstruction is also given in the companion attempt for Problem 90; its arithmetic existence input is not reproved here.
2) Converting many edges into a minimum-degree bound
Let a finite graph have $N$ vertices and $E>0$ edges. Repeatedly delete a vertex whose current degree is less than $E/N$, where $E/N$ remains the original fixed threshold. If all vertices were deleted, count each edge at the moment its first endpoint is removed. The resulting count would be strictly less than $N(E/N)=E$, a contradiction. A nonempty induced subgraph therefore remains, with minimum degree at least $E/N$.Apply this to the unit-distance graph on $P_j$. The remaining vertices form a planar set $Q_j$ of order $M_j\leq N_j$, with \[ \deg_{Q_j}(x)\geq E_j/N_j\geq N_j^\delta \quad\hbox{for every }x\in Q_j. \] All these neighbors have the same radius $1$ at every vertex, a stronger requirement than allowing a different radius at each point. Since degree is at most $M_j-1$, we also have \[ M_j\geq N_j^\delta+1\longrightarrow\infty. \] Finally, $M_j\leq N_j$ implies \[ f(M_j)\geq g(Q_j)\geq N_j^\delta\geq M_j^\delta. \] Thus the counterexamples occur at unbounded orders, not merely in a bounded core repeatedly extracted from larger sets.
3) Negating both asymptotic assertions
The estimate $f(n)\leq n^{o(1)}$ would imply, for example, $f(n)\leq n^{\delta/2}$ for all sufficiently large $n$, contradicting the preceding subsequence. If instead some constant $C$ gave $f(n)\leq n^{C/\log\log n}$ eventually, choose $M_j$ so large that $C/\log\log M_j<\delta$. The same subsequence contradicts that bound as well. No monotonicity of $f$ is needed.4) An elementary upper-bound consistency check
For an $n$-point set with $g(A)=t$, select at each center a circle containing at least $t$ other points. The $n$ circles are distinct because their centers are distinct. Let $r_p$ count how many selected circles contain $p\in A$, and put $I=\sum_p r_p\geq nt$. Two distinct circles meet at most twice, so double counting gives \[ \sum_{p\in A}\binom{r_p}{2}\leq2\binom n2=n(n-1). \] By Cauchy--Schwarz the left side is at least $(I^2/n-I)/2$. Writing $z=I/n\geq t$ yields $z^2-z\leq2(n-1)$, and hence \[ f(n)\leq\frac{1+\sqrt{8n-7}}2. \] This weak upper bound is compatible with a fixed positive lower exponent. It does not restore either conjectured subpolynomial bound.Outcome: SOLVED, known negative answer reconstructed using the explicit unit-distance theorem. The pruning argument and quantifier checks are proved in full. The batch verification exhausts small graphs to check the fixed-threshold pruning procedure; it is not a formal or numerical verification of the number-field theorem.
5) Sources
- OpenAI, Planar Point Sets with Many Unit Distances, Theorem 1.1 and the discussion immediately following it,
https://cdn.openai.com/pdf/74c24085-19b0-4534-9c90-465b8e29ad73/unit-distance-proof.pdf. - T. F. Bloom, Problem 92,
https://www.erdosproblems.com/92, checked 23 September 2026.