gpt pro 5.4 attempt on Math-671

Erdős Problem #96 --- Round 2

1) ROUND-2 OBJECTIVE

I pursue (A) proof track. A full resolution of the open problem still looks out of reach, but Round-1 only produced the generic planar bound $u(A)=O(n^{3/2})$, which does not use the convex structure in any essential way. The most promising strict advance is therefore to prove linear upper bounds on natural structured subclasses of convex polygons, and to make precise the local-to-global reduction from the related equidistance problem (Problem #97) that was only mentioned in Round-1.

2) Round-1 FOUNDATION USED

I use the following Round-1 results without reproving them.

• Round-1 Lemma 96.1: a regular $n$-gon scaled so that its side length is $1$ has exactly $n$ unit-distance pairs.
• Round-1 Lemma 96.2: every convex $n$-gon satisfies the generic planar bound \[ u(A)\le \frac{\sqrt 2}{2}n^{3/2}+\frac14 n. \]
• Round-1 context from the problem statement: Problem #96 asks whether $u(A)=O(n)$ for convex $n$-gons, and a positive answer would follow from the related local equidistance problem (Problem #97).

3) NEW INSIGHT / TOOL (ROUND-2)

There are three genuinely new ingredients in this round.

• Exact cyclic theorem. If the vertices lie on a circle, then for every fixed distance $r$ the corresponding distance-$r$ graph has maximum degree at most $2$, hence at most $n$ edges. In particular, among cyclic convex $n$-gons the maximum possible number of unit distances is exactly $n$.
• Parabolic linear theorem. If the vertices lie on a parabola, then for every fixed distance $r$ each vertex has at most $4$ neighbors at distance $r$, because the distance equation reduces to a quartic. Hence the number of unit distances is at most $2n$ on this subclass.
• Explicit local-to-global reduction. If every convex polygon has a vertex from which no distance is repeated more than $k$ times, then every convex $n$-gon has at most $3+k(n-3)\le kn$ unit distances. In particular, Problem #97 would imply the explicit bound $u(A)\le 3n-6$.

4) ATTACK PLAN (ROUND-2)

The exact gaps left after Round-1 were:

• the Round-1 upper bound was the generic $O(n^{3/2})$ planar bound and did not exploit convex structure;
• the statement "a positive answer would follow from Problem #97" was not proved;
• no substantial subclass of convex polygons had been solved sharply.

The plan is therefore:

1. solve the problem exactly on the cyclic subclass;
2. prove a second linear theorem on a different curved subclass (parabolas), to show that the obstruction is not present there either;
3. turn the heuristic relation to Problem #97 into a rigorous induction with an explicit constant.

5) WORK (ROUND-2)

For a finite set $A\subset \mathbb R^2$ and a fixed real number $r>0$, write \[ e_r(A):=\left|\{\{x,y\}\subset A:\ \|x-y\|=r\}\right| \] for the number of unordered pairs at distance exactly $r$. Thus $u(A)=e_1(A)$.

Proposition 96.1 (exact fixed-distance bound on a circle). Let $A\subset \mathbb R^2$ be a finite set of $n$ points lying on a circle. Then for every $r>0$, \[ e_r(A)\le n. \] In particular, if $A$ is the vertex set of a cyclic convex $n$-gon, then \[ u(A)\le n. \]

Proof. Fix $r>0$. Form the graph $G_r(A)$ on vertex set $A$ by joining two vertices when they are at distance $r$. It is enough to prove that every vertex of $G_r(A)$ has degree at most $2$.

Fix $x\in A$. All points of $A$ lie on a common circle $\Gamma$. The neighbors of $x$ in $G_r(A)$ are precisely the points of $A\cap C(x,r)$, where $C(x,r)$ denotes the circle of radius $r$ centered at $x$. Now $\Gamma$ and $C(x,r)$ are two circles in the Euclidean plane. Since $r>0$, the center of $C(x,r)$ is $x$, while $x\in \Gamma$, so $\Gamma$ and $C(x,r)$ are not the same circle. Hence they intersect in at most two points. Therefore \[ \deg_{G_r(A)}(x)=|A\cap \Gamma\cap C(x,r)|\le 2. \] This holds for every vertex $x$, so the handshake lemma gives \[ 2e_r(A)=\sum_{x\in A}\deg_{G_r(A)}(x)\le 2n. \] Thus $e_r(A)\le n$, as claimed. The case $r=1$ is exactly the unit-distance bound. $\square$

Corollary 96.2 (cyclic subclass solved sharply). Among cyclic convex $n$-gons, the maximum possible number of unit-distance pairs is exactly $n$.

Proof. The upper bound is Proposition 96.1 with $r=1$. The matching lower bound $n$ is exactly Round-1 Lemma 96.1, which exhibits a regular $n$-gon with exactly $n$ unit distances. $\square$

Proposition 96.3 (linear fixed-distance bound on a parabola). Let $A$ be a finite subset of the standard parabola \[ \mathcal P:=\{(t,t^2): t\in \mathbb R\}. \] Then for every $r>0$, \[ e_r(A)\le 2|A|. \] In particular, if the vertices of a convex $n$-gon lie on a parabola, then the number of unit-distance pairs is at most $2n$.

Proof. Fix $r>0$ and form the distance-$r$ graph $G_r(A)$ as above. It is enough to prove that every vertex has degree at most $4$.

Fix a vertex $p=(a,a^2)\in A$. A point $q=(x,x^2)\in \mathcal P$ satisfies $\|p-q\|=r$ exactly when \[ (x-a)^2+(x^2-a^2)^2=r^2. \] Since $x^2-a^2=(x-a)(x+a)$, this becomes \[ (x-a)^2\left(1+(x+a)^2\right)=r^2. \] Expanding gives the quartic equation \[ x^4-2ax^3+(1+2a^2)x^2-2a(1+a^2)x+a^4+a^2-r^2=0. \] A real quartic has at most four real roots. Therefore there are at most four real numbers $x$ such that $(x,x^2)\in \mathcal P$ lies at distance $r$ from $p$. Hence \[ \deg_{G_r(A)}(p)\le 4. \] Since this holds for every $p\in A$, the handshake lemma gives \[ 2e_r(A)=\sum_{p\in A}\deg_{G_r(A)}(p)\le 4|A|, \] so $e_r(A)\le 2|A|$.

For the final sentence: every parabola is the image of $\mathcal P$ under a Euclidean similarity. Similarities preserve incidences of "being at a fixed distance", up to replacing the distance $1$ by another positive number $r$. Since the proposition was proved for every $r>0$, the same $2n$ bound holds for any finite set lying on any parabola, and in particular for convex polygons inscribed in a parabola. $\square$

Proposition 96.4 (rigorous reduction from Problem #97 to Problem #96). Assume the following local multiplicity hypothesis for some integer $k\ge 1$:

For every convex polygon $P$ there exists a vertex $v$ of $P$ such that no distance from $v$ to the other vertices of $P$ occurs more than $k$ times.

Then every convex $n$-gon $A$ satisfies \[ u(A)\le 3+k(n-3)\le kn \qquad (n\ge 3). \] In particular, an affirmative answer to Problem #97 (the case $k=3$) would imply \[ u(A)\le 3n-6 \] for every convex $n$-gon $A$.

Proof. Let $U(n)$ be the maximum possible number of unit-distance pairs among convex $n$-gons. We prove by induction that \[ U(n)\le 3+k(n-3) \qquad (n\ge 3). \]

Base case $n=3$: every triangle has at most $3$ unit distances, so $U(3)\le 3$, which matches the formula.

Induction step: assume $n\ge 4$ and let $A=(v_1,\dots,v_n)$ be a convex $n$-gon with $u(A)=U(n)$. By the hypothesis, there exists a vertex $v=v_i$ of $A$ such that among the distances from $v$ to the other $n-1$ vertices, no value occurs more than $k$ times. In particular, the distance value $1$ occurs at most $k$ times from $v$, so $v$ is incident to at most $k$ unit-distance pairs.

Remove $v$ from the vertex list. The remaining points, in the induced cyclic order, form a convex $(n-1)$-gon $A'$. Every unit-distance pair of $A$ either survives inside $A'$ or is incident to $v$. Therefore \[ U(n)=u(A)\le u(A')+k\le U(n-1)+k. \] By the induction hypothesis, \[ U(n)\le \left(3+k(n-4)\right)+k = 3+k(n-3). \] This proves the claim. The inequality $3+k(n-3)\le kn$ is immediate for $n\ge 3$ and $k\ge 1$.

For Problem #97 we have $k=3$, because "no other $4$ vertices equidistant from it" means exactly that each distance value from the chosen vertex is repeated at most $3$ times. Hence $U(n)\le 3n-6$. $\square$

Proposition 96.5 (Erdős's stronger local-sum conjecture would imply the $2n$ upper bound). For a convex polygon $A$, define \[ g_A(x):=\max_{r>0}\left|\{y\in A\setminus\{x\}:\ \|x-y\|=r\}\right|. \] Then \[ 2u(A)\le \sum_{x\in A} g_A(x). \] Consequently, the stronger conjecture \[ \sum_{x\in A}g_A(x)<4|A| \] would imply \[ u(A)<2|A|. \]

Proof. For each vertex $x\in A$, the number of unit-distance neighbors of $x$ is \[ d_1(x):=\left|\{y\in A\setminus\{x\}:\ \|x-y\|=1\}\right|. \] Since $g_A(x)$ is the maximum multiplicity of any distance from $x$, we have $d_1(x)\le g_A(x)$ for every $x$. Summing over all vertices and using the handshake lemma for the unit-distance graph, \[ 2u(A)=\sum_{x\in A} d_1(x)\le \sum_{x\in A} g_A(x). \] The final implication is immediate. $\square$

6) ADVERSARIAL VERIFICATION

• Cyclic theorem: the only geometric input is that two distinct circles intersect in at most two points. They are indeed distinct, because the circle centered at a vertex $x$ cannot coincide with the common circumcircle: its center is $x$, while $x$ lies on the circumcircle and a circle never contains its own center.
• Handshake counts: in Propositions 96.1 and 96.3, the graph is simple and undirected, so $2e_r(A)=\sum \deg(x)$ is valid.
• Parabola theorem: the quartic equation was expanded explicitly. The argument only uses "at most four real roots", so repeated roots and tangent cases do not cause trouble. Also $r>0$ excludes the trivial coincidence $q=p$.
• Similarity step for parabolas: a Euclidean similarity preserves the number of pairs at any prescribed distance value up to rescaling the value. Because Proposition 96.3 was proved for every $r>0$, the transfer to an arbitrary parabola is legitimate.
• Reduction from Problem #97: deleting one vertex from a convex polygon leaves a convex polygon on the remaining vertices in cyclic order. The local multiplicity condition implies the chosen vertex has at most $k$ unit-distance neighbors because "distance $1$" is one of the allowed distance values. No additional assumption is hidden.
• Use of Round-1: the only Round-1 theorem used in a nontrivial way is the regular-$n$-gon lower bound to prove sharpness of the cyclic result. No Round-1 proof was re-run.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

This round strictly advances beyond Round-1 in three ways.

1. The problem is solved exactly on the cyclic subclass: a cyclic convex $n$-gon has at most $n$ unit distances, and this bound is sharp by the regular $n$-gon.
2. A second genuine linear theorem is proved on a different curved subclass: any convex $n$-gon inscribed in a parabola has at most $2n$ unit distances.
3. The relation to Problem #97 is now rigorous and explicit: a positive answer to Problem #97 would imply the concrete bound $u(A)\le 3n-6$ for every convex $n$-gon. Moreover, Erdős's stronger local-sum conjecture would imply the stronger global bound $u(A)<2n$.

The main unresolved gap is still the passage from these structured situations (cyclic, parabolic, or local multiplicity control) to arbitrary convex polygons.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 40%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #96, checked 2026-03-07.
2. T. F. Bloom, Erdős Problem #97, checked 2026-03-07.