Erdős Problem #668

Erdős Problem 668 --- collection record

PROVENANCE AND STATUS

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_5.2/668.tex.

REUSED SOURCE 1

FORMAL RESTATEMENT. For $n\ge 1$, let \[ u(n) := \max\{\#\{\{i,j\}: 1\le i<j\le n,\ \|x_i-x_j\|=1\}\ :\ x_1,\dots,x_n\in\mathbb{R}^2\ \text{distinct}\} \] be the maximum possible number of unit distances among $n$ points in the Euclidean plane. Let $\mathcal{M}(n)$ denote the set of congruence classes (under Euclidean isometries of $\mathbb{R}^2$) of $n$-point configurations achieving $\nu(n)$. The problem asks:

1. Does $|\mathcal{M}(n)|\to\infty$ as $n\to\infty$?
2. Is it true that $|\mathcal{M}(n)|>1$ for all sufficiently large $n$?

Ambiguity. The text asks "always $>1$ for $n>3$" but immediately notes $|\mathcal{M}(4)|=1$. So the literal statement "$|\mathcal{M}(n)|>1$ for all $n>3$" is already false (with counterexample $n=4$). A minimal corrected version would ask "for all $n>4$" or "for all sufficiently large $n$".

QUICK LITERATURE/CONTEXT CHECK. The supplied statement reports computational evidence (up to $n\le 21$) suggesting $|\mathcal{M}(n)|=1$ for many small $n$, though only checked up to graph isomorphism rather than congruence. Per project rules, I do not use external sources beyond what is written in the prompt.

ATTACK PLAN.

• Reality check for very small $n$: determine $\nu(n)$ and $|\mathcal{M}(n)|$ for $n=2,3,4$ by direct geometric reasoning.
• Main questions: no general method is available here; I will provide only rigorous small-$n$ structure and then state the main problem as unresolved with concrete next steps.

WORK.

Fast reality check (tiny $n$). For $n=2$, $\nu(2)=1$ and $|\mathcal{M}(2)|=1$ (two points at distance $1$). For $n=3$, $\nu(3)=3$ and $|\mathcal{M}(3)|=1$ (equilateral triangle of side $1$). For $n=4$, we prove below that $\nu(4)=5$ and $|\mathcal{M}(4)|=1$.

Lemma 668.1 (Upper bound $\nu(4)\le 5$). For any $4$ distinct points in $\mathbb{R}^2$, the number of unit distances among them is at most $5$.

Proof. There are $\binom{4}{2}=6$ unordered pairs of points. Suppose for contradiction that all $6$ pairs are at distance $1$. Pick any three of the points, call them $A,B,C$. Then $|AB|=|BC|=|CA|=1$, so $\triangle ABC$ is an equilateral triangle of side length $1$. Now consider the fourth point $D$. The conditions $|DA|=|DB|=1$ mean that $D$ lies in the intersection of the circles of radius $1$ centered at $A$ and $B$. In the plane, two circles of equal radius with centers at distance $1$ intersect in exactly two points: these are precisely the two vertices of equilateral triangles on base $AB$. One of these intersection points is $C$; let the other be $C'$ (the reflection of $C$ across the line $AB$). Thus $D\in\{C,C'\}$. Since the four points are distinct, $D\ne C$, so $D=C'$. But then the distance between $C$ and $C'$ is $|CC'|=\sqrt{3}$ (this is a standard computation from equilateral triangles, and can be verified by coordinates, e.g. $A=(0,0)$, $B=(1,0)$, $C=(1/2,\sqrt3/2)$, $C'=(1/2,-\sqrt3/2)$). In particular $|CC'|\ne 1$, contradicting the assumption that all pairs are at distance $1$. Therefore at least one pair is not a unit distance, so there are at most $5$ unit distances. $\square$

Lemma 668.2 (Characterization and uniqueness at $n=4$). One has $\nu(4)=5$. Moreover, any configuration of $4$ points in $\mathbb{R}^2$ with exactly $5$ unit distances is congruent to two unit equilateral triangles joined along a common edge. In particular $|\mathcal{M}(4)|=1$.

Proof. Existence: Let $A=(0,0)$ and $B=(1,0)$. Let $C=(1/2,\sqrt3/2)$ and $D=(1/2,-\sqrt3/2)$. Then $\triangle ABC$ and $\triangle ABD$ are equilateral with side length $1$. Therefore the five distances $|AB|,|AC|,|BC|,|AD|,|BD|$ equal $1$. The remaining distance is $|CD|=\sqrt3\ne 1$. So this configuration has exactly $5$ unit distances. By Lemma 668.1, this is maximal, hence $\nu(4)=5$.

Uniqueness: Let $\{P_1,P_2,P_3,P_4\}$ be any configuration with $5$ unit distances. By Lemma 668.1, not all $6$ pairwise distances can be $1$, so there is exactly one unordered pair (say $\{P_3,P_4\}$) with $|P_3P_4|\ne 1$, and every other pair has distance $1$. In the unit-distance graph (vertices $P_i$, edges between unit-distance pairs), this means the graph is $K_4$ minus a single missing edge. The two vertices incident to the missing edge have degree $2$, and the other two have degree $3$. Renaming if necessary, we may assume $P_1$ and $P_2$ are the degree-$3$ vertices. Then \[ |P_1P_2|=|P_1P_3|=|P_2P_3|=1 \quad\text{and}\quad |P_1P_2|=|P_1P_4|=|P_2P_4|=1. \] In particular, both triples $(P_1,P_2,P_3)$ and $(P_1,P_2,P_4)$ form equilateral triangles of side $1$ with the same base $P_1P_2$. As in Lemma 668.1, for fixed $P_1,P_2$ with $|P_1P_2|=1$, there are exactly two points in the plane at distance $1$ from both $P_1$ and $P_2$. Therefore $\{P_3,P_4\}$ must equal this pair of equilateral-triangle vertices on the two sides of the line $P_1P_2$. Consequently the configuration is congruent to the explicit model above (two equilateral triangles glued on a common edge), and there is only one congruence class. $\square$

VERIFICATION.

• The proof of Lemma 668.1 uses only basic circle intersection facts and the explicit computation $|CC'|=\sqrt3$.
• Lemma 668.2 checks both existence and uniqueness, and the uniqueness argument uses only that the unit-distance graph must be $K_4$ minus one edge.
• Boundary cases $n=2,3$ are consistent with standard geometry and were sanity-checked.

FINAL. UNRESOLVED.

(i) Strongest proved partial results.

• For $n=3$, $\nu(3)=3$ and the maximizer is unique up to congruence (equilateral triangle).
• For $n=4$, $\nu(4)=5$ and the maximizer is unique up to congruence (two equilateral triangles joined by an edge), hence $|\mathcal{M}(4)|=1$ (Lemmas 668.1--668.2).
• Therefore the literal subquestion "$|\mathcal{M}(n)|>1$ for all $n>3$" is false (counterexample $n=4$).

(ii) First gap (crisp statement). Determine whether $|\mathcal{M}(n)|\to\infty$ as $n\to\infty$. Even the weaker question "does there exist some $n_0$ such that $|\mathcal{M}(n)|>1$ for all $n\ge n_0$?" is open here.

(iii) Top 3 next moves.

1. Rigorous small-$n$ certification: for $5\le n\le 10$, attempt a full congruence-level classification of maximizers by combining exact distance-geometry constraints with computer-assisted solving (not just graph isomorphism).
2. Stability analysis: prove that near-extremal unit distance graphs must be close (in a precise sense) to a specific geometric template; then classify all exact extremizers as perturbations of that template.
3. Search for multiplicity mechanisms: identify a construction that, for infinitely many $n$, admits a continuous or discrete family of noncongruent realizations while keeping the unit-distance count maximal.

(iv) What a minimal counterexample would likely look like.

• If $|\mathcal{M}(n)|\to\infty$ is false, there should exist an infinite sequence $n_t\to\infty$ such that the extremal unit-distance graph(s) have essentially unique geometric realization up to congruence.
• If instead the "eventually $>1$" claim is false, one would need infinitely many $n$ with a unique maximizer (as already happens at $n=3,4$).

COMPLETION ESTIMATE

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

FORMAL RESTATEMENT. For $n\ge 1$, let \[ u(n) := \max\{\#\{\{i,j\}: 1\le i<j\le n,\ \|x_i-x_j\|=1\}\ :\ x_1,\dots,x_n\in\mathbb{R}^2\ \text{distinct}\} \] be the maximum possible number of unit distances among $n$ points in the Euclidean plane. Let $\mathcal{M}(n)$ denote the set of congruence classes (under Euclidean isometries of $\mathbb{R}^2$) of $n$-point configurations achieving $\nu(n)$. The problem asks:

1. Does $|\mathcal{M}(n)|\to\infty$ as $n\to\infty$?
2. Is it true that $|\mathcal{M}(n)|>1$ for all sufficiently large $n$?

Ambiguity. The text asks "always $>1$ for $n>3$" but immediately notes $|\mathcal{M}(4)|=1$. So the literal statement "$|\mathcal{M}(n)|>1$ for all $n>3$" is already false (with counterexample $n=4$). A minimal corrected version would ask "for all $n>4$" or "for all sufficiently large $n$".

QUICK LITERATURE/CONTEXT CHECK. The supplied statement reports computational evidence (up to $n\le 21$) suggesting $|\mathcal{M}(n)|=1$ for many small $n$, though only checked up to graph isomorphism rather than congruence. Per project rules, I do not use external sources beyond what is written in the prompt.

ATTACK PLAN.

• Reality check for very small $n$: determine $\nu(n)$ and $|\mathcal{M}(n)|$ for $n=2,3,4$ by direct geometric reasoning.
• Main questions: no general method is available here; I will provide only rigorous small-$n$ structure and then state the main problem as unresolved with concrete next steps.

WORK.

Fast reality check (tiny $n$). For $n=2$, $\nu(2)=1$ and $|\mathcal{M}(2)|=1$ (two points at distance $1$). For $n=3$, $\nu(3)=3$ and $|\mathcal{M}(3)|=1$ (equilateral triangle of side $1$). For $n=4$, we prove below that $\nu(4)=5$ and $|\mathcal{M}(4)|=1$.

Lemma 668.1 (Upper bound $\nu(4)\le 5$). For any $4$ distinct points in $\mathbb{R}^2$, the number of unit distances among them is at most $5$.

Proof. There are $\binom{4}{2}=6$ unordered pairs of points. Suppose for contradiction that all $6$ pairs are at distance $1$. Pick any three of the points, call them $A,B,C$. Then $|AB|=|BC|=|CA|=1$, so $\triangle ABC$ is an equilateral triangle of side length $1$. Now consider the fourth point $D$. The conditions $|DA|=|DB|=1$ mean that $D$ lies in the intersection of the circles of radius $1$ centered at $A$ and $B$. In the plane, two circles of equal radius with centers at distance $1$ intersect in exactly two points: these are precisely the two vertices of equilateral triangles on base $AB$. One of these intersection points is $C$; let the other be $C'$ (the reflection of $C$ across the line $AB$). Thus $D\in\{C,C'\}$. Since the four points are distinct, $D\ne C$, so $D=C'$. But then the distance between $C$ and $C'$ is $|CC'|=\sqrt{3}$ (this is a standard computation from equilateral triangles, and can be verified by coordinates, e.g. $A=(0,0)$, $B=(1,0)$, $C=(1/2,\sqrt3/2)$, $C'=(1/2,-\sqrt3/2)$). In particular $|CC'|\ne 1$, contradicting the assumption that all pairs are at distance $1$. Therefore at least one pair is not a unit distance, so there are at most $5$ unit distances. $\square$

Lemma 668.2 (Characterization and uniqueness at $n=4$). One has $\nu(4)=5$. Moreover, any configuration of $4$ points in $\mathbb{R}^2$ with exactly $5$ unit distances is congruent to two unit equilateral triangles joined along a common edge. In particular $|\mathcal{M}(4)|=1$.

Proof. Existence: Let $A=(0,0)$ and $B=(1,0)$. Let $C=(1/2,\sqrt3/2)$ and $D=(1/2,-\sqrt3/2)$. Then $\triangle ABC$ and $\triangle ABD$ are equilateral with side length $1$. Therefore the five distances $|AB|,|AC|,|BC|,|AD|,|BD|$ equal $1$. The remaining distance is $|CD|=\sqrt3\ne 1$. So this configuration has exactly $5$ unit distances. By Lemma 668.1, this is maximal, hence $\nu(4)=5$.

Uniqueness: Let $\{P_1,P_2,P_3,P_4\}$ be any configuration with $5$ unit distances. By Lemma 668.1, not all $6$ pairwise distances can be $1$, so there is exactly one unordered pair (say $\{P_3,P_4\}$) with $|P_3P_4|\ne 1$, and every other pair has distance $1$. In the unit-distance graph (vertices $P_i$, edges between unit-distance pairs), this means the graph is $K_4$ minus a single missing edge. The two vertices incident to the missing edge have degree $2$, and the other two have degree $3$. Renaming if necessary, we may assume $P_1$ and $P_2$ are the degree-$3$ vertices. Then \[ |P_1P_2|=|P_1P_3|=|P_2P_3|=1 \quad\text{and}\quad |P_1P_2|=|P_1P_4|=|P_2P_4|=1. \] In particular, both triples $(P_1,P_2,P_3)$ and $(P_1,P_2,P_4)$ form equilateral triangles of side $1$ with the same base $P_1P_2$. As in Lemma 668.1, for fixed $P_1,P_2$ with $|P_1P_2|=1$, there are exactly two points in the plane at distance $1$ from both $P_1$ and $P_2$. Therefore $\{P_3,P_4\}$ must equal this pair of equilateral-triangle vertices on the two sides of the line $P_1P_2$. Consequently the configuration is congruent to the explicit model above (two equilateral triangles glued on a common edge), and there is only one congruence class. $\square$

VERIFICATION.

• The proof of Lemma 668.1 uses only basic circle intersection facts and the explicit computation $|CC'|=\sqrt3$.
• Lemma 668.2 checks both existence and uniqueness, and the uniqueness argument uses only that the unit-distance graph must be $K_4$ minus one edge.
• Boundary cases $n=2,3$ are consistent with standard geometry and were sanity-checked.

FINAL. UNRESOLVED.

(i) Strongest proved partial results.

• For $n=3$, $\nu(3)=3$ and the maximizer is unique up to congruence (equilateral triangle).
• For $n=4$, $\nu(4)=5$ and the maximizer is unique up to congruence (two equilateral triangles joined by an edge), hence $|\mathcal{M}(4)|=1$ (Lemmas 668.1--668.2).
• Therefore the literal subquestion "$|\mathcal{M}(n)|>1$ for all $n>3$" is false (counterexample $n=4$).

(ii) First gap (crisp statement). Determine whether $|\mathcal{M}(n)|\to\infty$ as $n\to\infty$. Even the weaker question "does there exist some $n_0$ such that $|\mathcal{M}(n)|>1$ for all $n\ge n_0$?" is open here.

(iii) Top 3 next moves.

1. Rigorous small-$n$ certification: for $5\le n\le 10$, attempt a full congruence-level classification of maximizers by combining exact distance-geometry constraints with computer-assisted solving (not just graph isomorphism).
2. Stability analysis: prove that near-extremal unit distance graphs must be close (in a precise sense) to a specific geometric template; then classify all exact extremizers as perturbations of that template.
3. Search for multiplicity mechanisms: identify a construction that, for infinitely many $n$, admits a continuous or discrete family of noncongruent realizations while keeping the unit-distance count maximal.

(iv) What a minimal counterexample would likely look like.

• If $|\mathcal{M}(n)|\to\infty$ is false, there should exist an infinite sequence $n_t\to\infty$ such that the extremal unit-distance graph(s) have essentially unique geometric realization up to congruence.
• If instead the "eventually $>1$" claim is false, one would need infinitely many $n$ with a unique maximizer (as already happens at $n=3,4$).