GPT 6 Astra Ultra attempt on Math-6261

Erdős Problem 1207 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

For fixed dimension $d$, define $P_d(n)$ as the greatest integer $m$ such that every $n$-point subset of $\mathbb R^d$ contains an $m$-point subset without an isosceles triple. Estimate $P_d(n)$. In particular, does some absolute $c>0$ satisfy \[P_2(n)<n^{1-c}\] for every sufficiently large $n$?

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.

The discussion identifies the one-dimensional case with avoidance of three-term arithmetic progressions. Thus isosceles triple includes three distinct collinear points when two of their mutual distances coincide; it is not restricted to nondegenerate triangles. The current official remarks credit Lee, Pohoata, and Zhu [LPZ26] with proving $P_2(n)<n^{1-c}$ for some $c>0$: this highlighted subquestion is resolved, while the general estimation problem remains open.

ATTACK PLAN AND WORK

The collinear case admits a direct greedy lower bound. Identify a line with the real numbers and let $S$ be an isosceles-triple-free subset of size $m$. For each pair $a,b\in S$, a new point creates a three-term arithmetic progression only at \[x=(a+b)/2,\qquad x=2a-b,qquad x=2b-a.\tag{1}\] These are exactly the collinear isosceles possibilities for three distinct points. If $S$ is maximal in an $n$-point collinear set, $1$ therefore gives \[n\le m+3\binom m2,\qquad m\ge\frac{1+\sqrt{1+24n}}6.\tag{2}\] The bound follows by solving the resulting quadratic inequality, with integer rounding understood. It proves a square-root lower bound for $P_1(n)$ and for any higher-dimensional input set that happens to lie on a line.

This point-count argument cannot be copied unchanged in the plane. Equal distances from a new point to two old ones constrain it to a whole perpendicular bisector; equal distances from an old point constrain it to a circle. Either curve can contain many points of the original set, so the three-forbidden-points bound per pair is no longer valid. The official page already records a solution of the highlighted planar upper-bound subquestion; this restricted lower-bound argument makes no claim to reprove that result.

VERIFICATION AND REMAINING GAP

Estimate the extremal function for unrestricted configurations in fixed dimension, dealing with concentration on circles and bisectors. The proved one-dimensional bound does not resolve the general estimation problem.

SOURCES

https://www.erdosproblems.com/1207

https://www.erdosproblems.com/history/1207

FINAL

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

COMPLETION ESTIMATE

$5\%$. 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.