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Math-6141 · mathematics · Mathematics

Erdős Problem #1087

Status Open · erdosproblems.comLLM status Open unverified

Statement

Let $P\subset\mathbb{R}^2$ be a set of $n$ points. Call a 4-element subset $Q\subset P$ degenerate if among the $\binom{4}{2}=6$ pairwise distances $\{\|x-y\|: x,y\in Q, x\ne y\}$ there exist two distinct unordered pairs $\{x,y\}\ne\{u,v\}$ with $\|x-y\|=\|u-v\|$. Let \[ f(n):=\max\{\,\#\{Q\subset P: |Q|=4,\ Q\text{ degenerate}\}: P\subset\mathbb{R}^2,\ |P|=n\,\}. \] Equivalently, $f(n)$ is the smallest number such that every $n$-point set has at most $f(n)$ degenerate 4-subsets. The problem asks for estimates of $f(n)$, in particular whether $f(n)\le n^{3+o(1)}$.

Restatement from Math-6141-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Math-6141-A1 · No claim recorded · LLM-generated · gpt pro 5.2 ·

Reused writeup on Math-6141 (gpt pro 5.2)

Erdős Problem 1087 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1087

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1087.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1087.tex.

Erdős Problem 1087 ("degenerate" 4-sets with a repeated distance)

FORMAL RESTATEMENT

Let $P\subset\mathbb{R}^2$ be a set of $n$ points. Call a 4-element subset $Q\subset P$ degenerate if among the $\binom{4}{2}=6$ pairwise distances $\{\|x-y\|: x,y\in Q, x\ne y\}$ there exist two distinct unordered pairs $\{x,y\}\ne\{u,v\}$ with $\|x-y\|=\|u-v\|$. Let \[ f(n):=\max\{\,\#\{Q\subset P: |Q|=4,\ Q\text{ degenerate}\}: P\subset\mathbb{R}^2,\ |P|=n\,\}. \] Equivalently, $f(n)$ is the smallest number such that every $n$-point set has at most $f(n)$ degenerate 4-subsets. The problem asks for estimates of $f(n)$, in particular whether $f(n)\le n^{3+o(1)}$.

QUICK LITERATURE/CONTEXT CHECK

The problem statement records bounds of Erdős--Purdy: $n^3\log n \ll f(n) \ll n^{7/2}$. I do not use further external results.

ATTACK PLAN

  • Proof track (partial). Provide a clean combinatorial upper bound on the number of degenerate 4-sets in terms of distance multiplicities (counts of equal-length pairs). Provide an explicit construction (regular $n$-gon) yielding $\Omega(n^3)$ degenerate 4-sets.
  • Disproof track. Not applicable.

WORK

Fast reality check.
For $P$ the vertices of a regular $n$-gon, brute-force computation shows: for $n=6,7,8,9,10,11$, every 4-subset is degenerate (so the count equals $\binom{n}{4}$); for $n=12$, there are $447$ degenerate 4-subsets out of $\binom{12}{4}=495$.
Lemma (Degenerate 4-sets controlled by distance multiplicities): For a finite set $P\subset\mathbb{R}^2$, and each distance value $r>0$, let $a_r$ be the number of unordered pairs $\{x,y\}\subset P$ with $\|x-y\|=r$. Then the number $D(P)$ of degenerate 4-subsets of $P$ satisfies \[ D(P)\le \sum_{r} \binom{a_r}{2}. \]
Proof: For each distance value $r$, form the graph $G_r$ on vertex set $P$ whose edges are exactly the pairs at distance $r$. Then $|E(G_r)|=a_r$. If $Q\subset P$ is a degenerate 4-set, by definition there exist two distinct unordered pairs $e_1,e_2\subset Q$ with $|e_1|=|e_2|=2$ and $\mathrm{dist}(e_1)=\mathrm{dist}(e_2)=r$ for some $r$. In particular, $e_1$ and $e_2$ are two distinct edges of $G_r$. Thus each degenerate 4-set contributes at least one unordered pair of edges from some $G_r$. For each fixed $r$, the total number of unordered edge-pairs in $G_r$ is $\binom{a_r}{2}$, and the number of degenerate 4-sets that can be "certified" by distance $r$ is at most that. Summing over $r$ gives $D(P)\le \sum_r \binom{a_r}{2}$. ∎
Lemma (Regular $n$-gon gives $\Omega(n^3)$ degenerate 4-sets): Let $P$ be the vertex set of a regular $n$-gon in $\mathbb{R}^2$, with $n\ge 6$. Then \[ f(n)\ge D(P)\ge c\,n^3 \] for an absolute constant $c>0$ (e.g. $c=1/48$ works for all sufficiently large $n$).
Proof: In a regular $n$-gon, for each step size $k\in\{1,2,\dots,\lfloor n/2\rfloor\}$, all chords connecting vertices with cyclic separation $k$ have the same length; denote this length by $\ell_k$. For each fixed $k$ (with $1\le k< n/2$), there are exactly $n$ edges of length $\ell_k$: namely $\{i,i+k\}$ for $i\in\mathbb{Z}/n\mathbb{Z}$. Among these $n$ edges, at most $2n$ unordered pairs of edges share a vertex (each edge shares a vertex with at most $2$ other edges of the same step size at each endpoint), so the number of unordered disjoint edge pairs of length $\ell_k$ is at least \[ \binom{n}{2}-2n = \frac{n(n-1)}{2}-2n \ge \frac{n^2}{3} \] for all $n\ge 6$. Each disjoint pair of edges $\{a,b\}$ and $\{c,d\}$ of the same length determines a 4-set $\{a,b,c,d\}$ which is degenerate (it contains two equal distances).

Let $T$ be the number of pairs $(k,\{e_1,e_2\})$ where $1\le k\le \lfloor n/3\rfloor$ and $e_1,e_2$ are disjoint edges of length $\ell_k$. By the bound above, for each such $k$ there are at least $n^2/3$ choices of $\{e_1,e_2\}$, so \[ T \ge \left\lfloor\frac{n}{3}\right\rfloor\cdot \frac{n^2}{3} \ge \frac{n^3}{12}-\frac{n^2}{9}. \] On the other hand, a fixed 4-set $Q$ can contribute to at most $3$ such pairs $(k,\{e_1,e_2\})$, because a 4-set has exactly $3$ ways to partition its vertices into two disjoint unordered pairs. Therefore $T\le 3D(P)$, hence \[ D(P)\ge \frac{T}{3} \ge \frac{n^3}{36}-\frac{n^2}{27}. \] Taking $c=1/48$ works for all sufficiently large $n$, and in any case this proves $D(P)=\Omega(n^3)$. Since $f(n)$ is the maximum possible number of degenerate 4-sets over all $n$-point sets, $f(n)\ge D(P)$. ∎

VERIFICATION

  • In Lemma 1, the injection "degenerate 4-set $\to$ some distance class edge-pair" may have multiplicities (a 4-set can be certified by multiple distances), but that only strengthens the inequality in the correct direction.
  • In Lemma 2, the disjoint-edge-pair count is a lower bound; even if the constant $2n$ is loose, it suffices to obtain $\Omega(n^3)$.
  • Computation for regular $n$-gons: for $n\le 11$ all $\binom{n}{4}$ quadruples are degenerate; for $n=12$ the exact number of degenerate quadruples is $447$ (out of $495$).

FINAL

UNRESOLVED.
  1. (i) Strongest proved partial result here: explicit construction shows $f(n)\ge c n^3$. Also, for any point set $P$, the number of degenerate 4-sets is at most $\sum_r \binom{a_r}{2}$ where $a_r$ are multiplicities of each realized distance.
  2. (ii) First gap: prove a universal upper bound $f(n)\le n^{3+o(1)}$ (or construct configurations with $\gg n^{3+\varepsilon}$ degenerate 4-sets).
  3. (iii) Top 3 next moves: $1$ bound $\sum_r \binom{a_r}{2}$ sharply using incidence geometry to control distance multiplicities $a_r$; $2$ look for configurations (e.g. integer lattices) where many distances repeat and estimate the resulting degenerate-4 count computationally; $3$ attempt to count non-degenerate 4-sets (all 6 distances distinct) from below to deduce upper bounds on $f(n)$.
  4. (iv) Minimal counterexample structure (to $f(n)\le n^{3+o(1)}$): sets with many distance classes each containing $\Theta(n)$ pairs, and with many of those pairs disjoint so that $\sum_r \binom{a_r}{2}$ is $\gg n^{3+\varepsilon}$.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.

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Math-6141-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-6141

Erdős Problem 1087 ("degenerate" 4-sets with a repeated distance)

FORMAL RESTATEMENT

Let $P\subset\mathbb{R}^2$ be a set of $n$ points. Call a 4-element subset $Q\subset P$ degenerate if among the $\binom{4}{2}=6$ pairwise distances $\{\|x-y\|: x,y\in Q, x\ne y\}$ there exist two distinct unordered pairs $\{x,y\}\ne\{u,v\}$ with $\|x-y\|=\|u-v\|$. Let \[ f(n):=\max\{\,\#\{Q\subset P: |Q|=4,\ Q\text{ degenerate}\}: P\subset\mathbb{R}^2,\ |P|=n\,\}. \] Equivalently, $f(n)$ is the smallest number such that every $n$-point set has at most $f(n)$ degenerate 4-subsets. The problem asks for estimates of $f(n)$, in particular whether $f(n)\le n^{3+o(1)}$.

QUICK LITERATURE/CONTEXT CHECK

The problem statement records bounds of Erdős--Purdy: $n^3\log n \ll f(n) \ll n^{7/2}$. I do not use further external results.

ATTACK PLAN

  • Proof track (partial). Provide a clean combinatorial upper bound on the number of degenerate 4-sets in terms of distance multiplicities (counts of equal-length pairs). Provide an explicit construction (regular $n$-gon) yielding $\Omega(n^3)$ degenerate 4-sets.
  • Disproof track. Not applicable.

WORK

Fast reality check.
For $P$ the vertices of a regular $n$-gon, brute-force computation shows: for $n=6,7,8,9,10,11$, every 4-subset is degenerate (so the count equals $\binom{n}{4}$); for $n=12$, there are $447$ degenerate 4-subsets out of $\binom{12}{4}=495$.
Lemma (Degenerate 4-sets controlled by distance multiplicities): For a finite set $P\subset\mathbb{R}^2$, and each distance value $r>0$, let $a_r$ be the number of unordered pairs $\{x,y\}\subset P$ with $\|x-y\|=r$. Then the number $D(P)$ of degenerate 4-subsets of $P$ satisfies \[ D(P)\le \sum_{r} \binom{a_r}{2}. \]
Proof: For each distance value $r$, form the graph $G_r$ on vertex set $P$ whose edges are exactly the pairs at distance $r$. Then $|E(G_r)|=a_r$. If $Q\subset P$ is a degenerate 4-set, by definition there exist two distinct unordered pairs $e_1,e_2\subset Q$ with $|e_1|=|e_2|=2$ and $\mathrm{dist}(e_1)=\mathrm{dist}(e_2)=r$ for some $r$. In particular, $e_1$ and $e_2$ are two distinct edges of $G_r$. Thus each degenerate 4-set contributes at least one unordered pair of edges from some $G_r$. For each fixed $r$, the total number of unordered edge-pairs in $G_r$ is $\binom{a_r}{2}$, and the number of degenerate 4-sets that can be "certified" by distance $r$ is at most that. Summing over $r$ gives $D(P)\le \sum_r \binom{a_r}{2}$. ∎
Lemma (Regular $n$-gon gives $\Omega(n^3)$ degenerate 4-sets): Let $P$ be the vertex set of a regular $n$-gon in $\mathbb{R}^2$, with $n\ge 6$. Then \[ f(n)\ge D(P)\ge c\,n^3 \] for an absolute constant $c>0$ (e.g. $c=1/48$ works for all sufficiently large $n$).
Proof: In a regular $n$-gon, for each step size $k\in\{1,2,\dots,\lfloor n/2\rfloor\}$, all chords connecting vertices with cyclic separation $k$ have the same length; denote this length by $\ell_k$. For each fixed $k$ (with $1\le k< n/2$), there are exactly $n$ edges of length $\ell_k$: namely $\{i,i+k\}$ for $i\in\mathbb{Z}/n\mathbb{Z}$. Among these $n$ edges, at most $2n$ unordered pairs of edges share a vertex (each edge shares a vertex with at most $2$ other edges of the same step size at each endpoint), so the number of unordered disjoint edge pairs of length $\ell_k$ is at least \[ \binom{n}{2}-2n = \frac{n(n-1)}{2}-2n \ge \frac{n^2}{3} \] for all $n\ge 6$. Each disjoint pair of edges $\{a,b\}$ and $\{c,d\}$ of the same length determines a 4-set $\{a,b,c,d\}$ which is degenerate (it contains two equal distances).

Let $T$ be the number of pairs $(k,\{e_1,e_2\})$ where $1\le k\le \lfloor n/3\rfloor$ and $e_1,e_2$ are disjoint edges of length $\ell_k$. By the bound above, for each such $k$ there are at least $n^2/3$ choices of $\{e_1,e_2\}$, so \[ T \ge \left\lfloor\frac{n}{3}\right\rfloor\cdot \frac{n^2}{3} \ge \frac{n^3}{12}-\frac{n^2}{9}. \] On the other hand, a fixed 4-set $Q$ can contribute to at most $3$ such pairs $(k,\{e_1,e_2\})$, because a 4-set has exactly $3$ ways to partition its vertices into two disjoint unordered pairs. Therefore $T\le 3D(P)$, hence \[ D(P)\ge \frac{T}{3} \ge \frac{n^3}{36}-\frac{n^2}{27}. \] Taking $c=1/48$ works for all sufficiently large $n$, and in any case this proves $D(P)=\Omega(n^3)$. Since $f(n)$ is the maximum possible number of degenerate 4-sets over all $n$-point sets, $f(n)\ge D(P)$. ∎

VERIFICATION

  • In Lemma 1, the injection "degenerate 4-set $\to$ some distance class edge-pair" may have multiplicities (a 4-set can be certified by multiple distances), but that only strengthens the inequality in the correct direction.
  • In Lemma 2, the disjoint-edge-pair count is a lower bound; even if the constant $2n$ is loose, it suffices to obtain $\Omega(n^3)$.
  • Computation for regular $n$-gons: for $n\le 11$ all $\binom{n}{4}$ quadruples are degenerate; for $n=12$ the exact number of degenerate quadruples is $447$ (out of $495$).

FINAL

UNRESOLVED.
  1. (i) Strongest proved partial result here: explicit construction shows $f(n)\ge c n^3$. Also, for any point set $P$, the number of degenerate 4-sets is at most $\sum_r \binom{a_r}{2}$ where $a_r$ are multiplicities of each realized distance.
  2. (ii) First gap: prove a universal upper bound $f(n)\le n^{3+o(1)}$ (or construct configurations with $\gg n^{3+\varepsilon}$ degenerate 4-sets).
  3. (iii) Top 3 next moves: $1$ bound $\sum_r \binom{a_r}{2}$ sharply using incidence geometry to control distance multiplicities $a_r$; $2$ look for configurations (e.g. integer lattices) where many distances repeat and estimate the resulting degenerate-4 count computationally; $3$ attempt to count non-degenerate 4-sets (all 6 distances distinct) from below to deduce upper bounds on $f(n)$.
  4. (iv) Minimal counterexample structure (to $f(n)\le n^{3+o(1)}$): sets with many distance classes each containing $\Theta(n)$ pairs, and with many of those pairs disjoint so that $\sum_r \binom{a_r}{2}$ is $\gg n^{3+\varepsilon}$.

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