GPT 6 Astra Ultra attempt on Math-671
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Problem 96: a two-degeneracy argument on a parabola
1) OBJECTIVE AND SOURCE REVIEW
Attempt dated 2026-09-23. The target is a linear bound for unit distances among vertices in convex position. I readgpt_pro_5.4/96.tex fully. Its circle argument is sound and its quartic-root argument gives $2n$ for a parabola, though its expanded quartic is incorrect. I strengthen the latter bound to $2n-3$ for $n\ge2$ using a vertex that can always be removed. 2) WORK
Let $P(t)=(t,t^2)$ and fix a distance $r>0$. In a finite set of parameters choose $a$ with $|a|$ minimal. Reflection permits $a\ge0$. Every remaining parameter lies either in $[a,\infty)$ or in $(-\infty,-a]$. The squared distance from $P(a)$ is \[ D_a(x)=(x-a)^2+(x^2-a^2)^2, \qquad D_a'(x)=2(x-a)\left(1+2x(x+a)\right). \] For $x>a$ both factors are positive. For $x<-a$, the first factor is negative and $x(x+a)\ge0$, so the second is positive. Thus $D_a$ is strictly increasing on the right ray and strictly decreasing as $x$ increases on the left ray. Each ray contributes at most one point with $D_a(x)=r^2$. The chosen vertex has at most two distance-$r$ neighbors.Every subset has the same property, by choosing its own smallest absolute parameter. Removing such vertices until two remain counts each edge once and gives \[ e_r(A)\le2(n-2)+1=2n-3\qquad(n\ge2). \] For $n=1$ the bound is zero. Euclidean similarities transport this argument to every parabola, because the prescribed distance merely rescales. No assumption of uniform spacing is used.