Independent notes on Erdős Problem 132

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Status and provenance

Neither assertion has been proved or disproved here for arbitrary planar point sets. In particular, the results below are not a resolution of the problem. They give finite-order cases, extensions of the convex case, quantitative asymptotic results for several families, and explicit restrictions on a hypothetical counterexample.

Only the supplied problem page and papers were consulted. No web search was made, and no earlier research notes were used. The proofs developed during this investigation are recorded in the files linked below.

Throughout, $A\subset\mathbb R^2$ consists of $n\ge5$ distinct points,

$$m_A(r)=\#\{\{x,y\}\subset A:|x-y|=r\},\qquad R(A)=\#\{r:1\le m_A(r)\le n\}.$$

All multiplicities count unordered pairs. Write $d(A)$ for the number of distinct positive distances and list them in decreasing order as

$$d_1>d_2>\cdots>d_{d(A)}.$$

1. Summary of established results

1. Small orders: $R(A)\ge2$ for every $5\le n\le10$. The proof uses the supplied few-distance classifications, with explicit coordinates and checked multiplicity tables for all four nine-point four-distance types. The complete finite-order proof is not retained.

2. Almost convex sets: $R(A)\ge2$ if the convex hull has $n-1$ vertices. Together with the supplied convex-set theorem, this covers hull size at least $n-1$. Also, $R(A)\ge2$ if the hull is cocircular and has $n-2$ vertices. The separate proofs are not retained. The latter statement needs cocircularity; it has not been extended to all hulls with $n-2$ vertices.

3. Convex layers: Let $L_i$ be the successive convex layers, removing only hull vertices at each step, and put $s_k=|L_1\cup\cdots\cup L_k|$. Then

$$\boxed{m_A(d_k)\le(6k-5)s_k.} \tag{1}$$

Thus, if every layer has size at most $H$,

$$\boxed{R(A)\ge \max\left\{1,\left\lfloor\sqrt{\frac n{6H}}\right\rfloor\right\}.} \tag{2}$$

In particular, maximal layer size $o(n)$ implies $R(A)\to\infty$. These statements have elementary proofs, given below; the separate layer write-up is not retained.

4. Diameter and separation: If $\Delta$ is the diameter and $\delta$ is the minimum separation, then

$$\boxed{R(A)\ge \max\left\{1,\left\lfloor \sqrt{\frac{\delta n}{6\pi\Delta}} \right\rfloor\right\}.} \tag{3}$$

Hence $\Delta/\delta=o(n)$ implies divergence. In particular, $\Delta/\delta=O(\sqrt n)$ gives $R(A)=\Omega(n^{1/4})$.

5. Regular-polygon subsets: If $m$ points of $A$ belong to the vertices of a regular $M$-gon, then

$$\boxed{R(A)\ge m-\frac n2-27M^{2/3}-18.} \tag{4}$$

Thus a fixed majority lying in a regular polygon of order $O(n)$ forces linearly many rare distances, even when all remaining points are arbitrary. The proof uses an elementary bound for lattice points on a strictly convex graph, not a genericity assumption. The full incidence and transfer proof is not retained.

The circular-set write-up, linked in Section 5 below, strengthens the last conclusion to a fixed majority on an arbitrary circle.

2. Counting and the finite-order argument

The diameter theorem gives $R(A)\ge1$. A general pair count gives

$$\binom n2\ge (d(A)-R(A))(n+1)+R(A),$$

and therefore

$$R(A)\ge \left\lceil\frac{(n+1)d(A)-\binom n2}{n}\right\rceil. \tag{5}$$

In particular, if $R(A)=1$, every distance except the diameter has multiplicity at least $n+1$, so

$$(d(A)-1)(n+1)+1\le\binom n2, \qquad d(A)\le\lfloor n/2\rfloor. \tag{6}$$

For $n=2k$, the case $d(A)=k$ forces equality:

$$m_A(d_1)=1,\qquad m_A(r)=2k+1\quad(r\ne d_1). \tag{7}$$

Erdős--Fishburn (1996), Theorem 1, supplies $g(2)=5$, $g(3)=7$, and $g(4)=9$, together with the relevant maximum-set classifications. Under (6), for $5\le n\le10$ there must be exactly $\lfloor n/2\rfloor$ distances. The odd orders reduce to:

Every row has at least two multiplicities at most its number of points.

For even $n=6,8,10$, delete either endpoint of the unique diameter in (7). Each deletion is an odd-order set in the preceding table. Every nonregular candidate has a multiplicity exceeding $n+1$, so both deletion sets must be regular polygons. They share at least three vertices. These vertices identify their common circumcircle, and one common vertex then identifies the entire regular polygon of the specified order. Thus the two deletion sets coincide, contrary to the endpoints being distinct.

The coordinate and displacement tables that made the nine-point check independent of missing source figures are not retained. In particular, the exceptional representative is

$$\{a\omega^j:a\in\{1,1+\sqrt3,-2-\sqrt3\},\ j=0,1,2\}, \qquad \omega=e^{2\pi i/3}.$$

Its squared distances $3,6+3\sqrt3,12+6\sqrt3,21+12\sqrt3$ have multiplicities $6,18,6,6$, respectively.

This finite argument stops at $n=10$. A hypothetical eleven-point counterexample would have exactly five distances and diameter multiplicity at most seven. The supplied uniqueness theorem for twelve-point five-distance sets does not classify eleven-point five-distance sets.

3. A proof of the layer bound

3.1. Distance rank controls layer depth

If $p\in L_i$, $q\in L_j$, and $|p-q|=d_k$, then

$$i+j\le k+1. \tag{8}$$

Indeed, if $i>1$, express $p$ as a nontrivial convex combination of vertices of $L_{i-1}$. Strict convexity of $x\mapsto|x-q|^2$ gives a vertex $p'\in L_{i-1}$ with $|p'-q|>|p-q|$. Repeat through all preceding layers, then move $q$ outward in the same way. This creates $i+j-2$ strict increases among distances of $A$. At most $k-1$ distinct distances exceed $d_k$, proving (8).

In particular, every $d_k$-edge lies in the first $k$ layers. This argument also handles nonvertices on hull edges and collinear layers.

3.2. A rank-$k$ distance has only linearly many pairs

We prove for every $N$-point planar set $B$, and its $k$-th largest distance $r$, that

$$m_B(r)\le(6k-5)N. \tag{9}$$

Let

$$C=\{x\in B:|x-p|\le r\text{ for every }p\in B\}, \qquad E=B\setminus C.$$

Fix $q\in B$, and put $S=\{x\in B:|x-q|=r\}$. For each $p\ne q$, the radius-$r$ disk centered at $p$ cuts the radius-$r$ circle centered at $q$ in a closed arc $I_p$ of angular length strictly less than $\pi$, a point, or the empty set. At most $2(k-1)$ points of $S$ lie outside $I_p$: their distances to $p$ exceed $r$, and each of the $k-1$ possible larger-distance circles centered at $p$ meets the circle centered at $q$ in at most two points.

A family of closed arcs of length less than $\pi$ has these elementary properties:

• A nonempty intersection is determined by at most two arcs. Cut at the antipode of a common point and use the largest left endpoint and smallest right endpoint of the resulting intervals.
• An empty intersection is witnessed by at most three arcs. Fix one arc $I_0$. Its intersection with each other short arc is an interval in $I_0$, possibly empty. If none is empty, two of these intervals are already disjoint by the interval intersection property.

Consequently, with $J=\bigcap_{p\ne q}I_p$, either at most $4(k-1)$ points of $S$ lie outside $J$, or $J$ is empty and $|S|\le6(k-1)$. Since $S\cap J=S\cap C$, in both cases

$$\#\{x\in E:|x-q|=r\}\le6(k-1).$$

Sum over $q$. Every $r$-edge not entirely in $C$ is counted at least once, so there are at most $6(k-1)N$ such edges. Edges entirely in $C$, if any, are diameter edges of $C$, hence number at most $|C|\le N$. This proves (9).

3.3. Localization and consequences

By (8), the first $k$ distances and every occurrence of $d_k$ are present in $L_1\cup\cdots\cup L_k$. Applying (9) there proves (1). If

$$(6k-5)s_k\le n, \tag{10}$$

all the first $k$ distances are rare. For $k\ge2$, (10) implies $s_k<n$; a point in a deeper layer and (8) guarantee that these distances exist. For $k=1$, the diameter suffices.

If each layer has at most $H$ points, take $k=\lfloor\sqrt{n/(6H)}\rfloor$. Then $(6k-5)s_k\le6k^2H\le n$, proving (2).

For a nondegenerate layer with $h$ vertices and perimeter $P$, $h\delta\le P\le\pi\Delta$. The last inequality follows by integrating the widths of a convex polygon; a proof was included in the layer write-up, which is not retained. A degenerate layer has at most two vertices, also at most $\pi\Delta/\delta$. Thus $H\le\pi\Delta/\delta$, proving (3).

An exact remaining restriction is worth recording. If $R(A)\le K$, then

$$s_{K+1}>\frac{n}{6K+1}. \tag{11}$$

Otherwise (10) would give $K+1$ rare distances. Therefore some one of the first $K+1$ convex layers has more than

$$\frac{n}{(K+1)(6K+1)}$$

vertices. This is a necessary condition, not a contradiction.

4. Why a regular-polygon subset survives arbitrary added points

Here is the mechanism behind (4); the complete calculations are not retained.

On a circle containing a regular $M$-gon, every chord-distance graph has degree at most two. For a subset $B$ of $m$ vertices, consequently $m_B(r)\le m$ and $B$ determines at least $\lfloor m/2\rfloor$ distances.

For a point $q$ not among the ambient polygon's vertices and not its center, at most

$$J(M)=27M^{2/3}+17$$

vertices $v$ can have $|q-v|$ equal to any chord length of that polygon. To see this, normalize the circumradius to one and write $q=\rho(\cos\phi,\sin\phi)$. For $\rho\notin\{0,1\}$, the incidence equation is

$$\cos(2\pi k/M)=\frac{1-\rho^2}{2} +\rho\cos(2\pi j/M-\phi).$$

Its inverse-cosine graph, with integer coordinates ((j,k)), splits into at most nine strictly convex or concave pieces after at most eight exceptional points are removed. The second derivative has a nonzero quadratic numerator in $\cos(2\pi j/M-\phi)$, so this splitting is uniform in $q$. Each piece lies in a rectangle of width $M$, height $M/2$. Successive lattice-point differences have distinct slopes and total $\ell^1$-length at most $2M$; counting integer vectors of length at most $M^{1/3}$ gives at most $3M^{2/3}+1$ points per piece. If $\rho=1$, equality would make $q$ an ambient polygon vertex, which was excluded. The center affects only the circumradius class.

Take $B=A\cap P_M$, where $P_M$ is the ambient vertex set, and put $t=n-m$. Discard at most the one center-related distance. All other chord classes together receive at most

$$tJ(M)+\binom t2$$

new pairs. A chord class that becomes rich must receive at least $t+1$ new pairs, since its old multiplicity was at most $m$. Hence

$$R(A)\ge\left\lfloor\frac m2\right\rfloor-1 -\frac{tJ(M)+\binom t2}{t+1}, \tag{12}$$

which implies (4). This calculation addresses the subset-to-whole-set transfer explicitly; merely finding a rare distance inside $B$ would not suffice.

5. Circular sets and the additive argument

The regular-polygon method extends to a majority on an arbitrary circle. The self-contained proof and its additive lemma were written in companion files that are not retained; the mechanism is given in Section 8 below. It establishes

$$\boxed{R(A)\ge h-\frac n2-10n^{6/7}} \tag{13}$$

whenever a circle contains $h$ points of $A$. In particular, $h\ge(1/2+\varepsilon)n$ gives $R(A)\ge\varepsilon n-10n^{6/7}$.

The argument is structural, not a deformation of the points. If the angular set $S\subset\mathbb R/\mathbb Z$ has too many differences, (5) already forces many rare distances. Otherwise an elementary minimal-expansion-set argument produces a finite subgroup of order $M\ge2|S|-|S-S|$. Both $S$ and (S-S) lie in boundedly many cosets when the circular majority is fixed. Each coset is a shifted regular angular grid, so the same convex-graph incidence bound and pair-budget transfer apply. Keeping track of the coset counts yields the error $O(n^{6/7})$ in (13).

No theorem on small sumsets is being imported without proof: the minimal-expansion-set argument, including the finite-group boundary case, is proved in the linked files.

6. Source use and exact unresolved gaps

The following supplied passages were read before their results were used:

• Erdős (1946), Section 4 and Theorem 3: the diameter multiplicity bound.
• Erdős--Fishburn (1995), Theorem 2 and Section 5/Theorem 5: the original five- and six-point cases and the exact unordered-pair convention.
• Erdős--Fishburn (1996), Theorem 1 and the accompanying few-distance classifications: the finite-order proof.
• Fishburn (1995), Theorems 1--3: the convex minimum-distance count and equality classifications used for hull size $n-1$.
• Shinohara (2008), Theorems 1.1 and 1.2$b$: confirmation of the nine-point classification and uniqueness of the twelve-point five-distance set.
• Clemen--Dumitrescu--Liu (2025), Sections 1.1 and 2: the convex and first-two-layer conclusions and their proofs. Its Theorem 1.3 in the supplied source, rather than the inconsistent digest transcription, uses

$$\min\{\tfrac32(h_1+h_2),\tfrac43h_1+2h_2,2h_1+h_2\}\le n.$$

• Bhowmick (2024), Section 2: rich-distance constructions. These do not give configurations with only one, or only boundedly many, rare distances.

The two unrestricted questions remain beyond the arguments above. The specific obstacles are:

• A sequence with bounded $R(A)$ can still have a linear-sized convex layer, as required by (11), without having a majority on any circle. No reduction established here forces that layer to be regular or cocircular.
• Fishburn's exact equality classifications cannot be extended, without a proof, to all convex sets whose distance count is a bounded amount above the minimum. Even the asymptotic assertion for arbitrary convex sets is not proved here.
• Rare distances in a selected subset can gain many pairs involving the remaining points. The transfer estimate (12) controls this only for its stated circular structures; it is not valid for an arbitrary convex subset.
• The second-largest distance need not itself be rare. Its source bounds, or the much weaker general rank bound (9), cannot replace a search for another distance at the exact threshold $n$.

Thus there is neither a general counterexample nor a complete general proof in this write-up. The proved partial results and their assumptions are the claims to retain.

7. Additional calculations and the almost-convex proofs

7.1. Exact small examples

The restriction $n\ge5$ is essential. The four points

$$(0,0),\quad(1,0),\quad(1/2,\sqrt3/2),\quad(1/2,-\sqrt3/2)$$

have five unit pairs and one pair at distance $\sqrt3$. Thus $R=1$ at $n=4$, not in the requested range.

There is an eight-point example with $R=2$. Put $a=1+\sqrt3$ and take

$$\{(\pm1,\pm1)\}\cup\{(0,\pm a),(\pm a,0)\}.$$

Its squared distances and multiplicities are

$$\begin{array}{c|rrrr} r^2&4&8&8+4\sqrt3&16+8\sqrt3\\\hline m(r)&12&2&12&2. \end{array}$$

The first class comprises four square sides and eight pairs to nearer outer points; the second comprises two square diagonals; the third comprises four adjacent outer pairs and eight pairs to farther square corners; the last comprises two opposite outer pairs. The counts sum to 28.

The unique twelve-point five-distance type has squared-distance multiplicities

$$\begin{array}{c|rrrrr} r^2&1&3&4&7&9\\\hline m(r)&24&15&12&12&3. \end{array}$$

An explicit triangular-lattice representative and complete displacement counts were recorded in a companion file that is not retained. Removing its point ((0,-2)), in the axial coordinates used there, subtracts degrees ((6,3,2,0,0)). The resulting eleven-point set has multiplicities ((18,12,10,12,3)), hence $R=2$. This is an example, not a proof of the universal assertion for eleven points.

The triangular-lattice nine-point representatives used in Section 2 are, with squared norm $i^2+ij+j^2$,

$$\begin{aligned} P={}&\{(0,0),(1,0),(0,1),(-1,1),(-1,0),(0,-1),(1,-1),(1,1),(2,-1)\},\\ Q={}&\{(0,1),(0,2),(1,1),(1,2),(1,3),(2,0),(2,1),(2,2),(3,0)\}. \end{aligned}$$

At squared lengths $1,3,4,7$, their multiplicities are respectively ((16,9,7,4)) and ((15,9,6,6)). These counts, together with the explicit three-triangle configuration in Section 2, establish four nonsimilar nine-point four-distance sets including $R_9$. The supplied classification allows at most four types, so no identification from absent pictures is needed.

7.2. Why one nonvertex can be handled

Let $H$ now denote the hull vertex set, $h=|H|$, and suppose $h=n-1$. A nonvertex cannot be a diameter endpoint: its expression as a nontrivial convex combination of hull vertices, and strict convexity of squared distance, would give a strictly longer pair.

If $H$ is cocircular with center $O$, and its one remaining point $p$ is not $O$, every distance from $p$ is attained by at most two hull vertices. None is the diameter. Consequently

$$d(A)\ge\left\lceil\frac h2\right\rceil+1 >\lfloor n/2\rfloor,$$

contradicting (6) under $R=1$. If $p=O$, adding $p$ changes just the circumradius class. For $n\ge7$, the hull has at least three distance classes, each of multiplicity at most $h$, so at least two stay rare.

It remains to justify cocircularity, rather than assume it. For odd $n=2k+1\ge9$, (6) and the convex distinct-distance lower bound force $H$, a convex $2k$-gon, to have exactly $k$ distances. Fishburn's even equality classification makes $H$ cocircular.

For even $n=2k\ge10$, a convex $H$ of size $2k-1$ has either $k-1$ or $k$ distances. In the former case it is regular by the odd equality classification. In the latter case $A$ has $k$ distances, so (7) gives a unique diameter pair $a,b\in H$. Each of $H\setminus\{a\}$ and $H\setminus\{b\}$ is a convex $(2k-2)$-gon with exactly $k-1$ distances. Both are cocircular by Fishburn's even equality classification. They share at least three noncollinear points, so their circles coincide and $H$ is cocircular. For $n=8$, Fishburn's classification of convex heptagons with four distances supplies the same conclusion. Smaller cases follow from the finite-order proof.

7.3. Two nonvertices require a separate argument

Suppose the hull is cocircular and has $h=n-2$ vertices, with remaining points $p,q$. At least one is not the circle center. For odd $n$, its distances to the hull already give $d(A)\ge\lceil h/2\rceil+1>\lfloor n/2\rfloor$, excluding $R=1$.

For even $n=2k\ge8$, assuming $R=1$ instead forces $d(A)=k$, a unique diameter, and exactly $k-1$ nonmaximal classes of multiplicity $n+1$. If one of $p,q$ is the center, a nonmaximal class different from the circumradius and (|p-q|) receives at most $h+2=n$ pairs, a contradiction. There are at least three nonmaximal classes, so such a choice exists.

If neither point is the center, each sees every nonmaximal distance exactly twice among the $h=2k-2$ hull vertices. Reflection in the line joining the center to $p$ therefore preserves the hull and fixes no hull vertex; likewise for $q$. Both reflections must swap the endpoints of the unique diameter, so their axes coincide. In coordinates with this axis horizontal, write $p=(t,0)$, $q=(s,0)$, and the hull as $(x_i,\pm y_i)$, where $x_i^2+y_i^2=\rho^2$. The numbers $x_i$ are distinct. Equality of the two sets of squared distances says that $X=\{x_i\}$ is preserved by

$$x\longmapsto \frac tsx+\frac{s^2-t^2}{2s}.$$

The diameter of a finite real set with at least two elements is preserved only if $|t/s|=1$. Since $p\ne q$, $s=-t$, and this map is $x\mapsto-x$. Thus the hull is centrally symmetric and has several antipodal diameter pairs, contradicting uniqueness.

Without cocircularity, this two-nonvertex argument is unavailable. The unhandled convex hulls have one more distance than the classified minimum; no classification of that regime for all orders was proved here.

8. Detailed mechanism of the circular-majority theorem

The following records the delicate steps of (13), so the result is not merely a reference to a proposed structural lemma.

8.1. The additive lemma and its finite-group issue

Let $S\subset\mathbb R/\mathbb Z$ be finite, $a=|S|$, $b=|S-S|<2a-1$. Translate $S$ to contain zero and work in $G=\langle S\rangle$. If $S-S=G$, the group is finite and can itself be used below. Otherwise, for nonempty finite $X$ with $X+S\ne G$, minimize

$$f(X)=|X+S|-|X|.$$

Its minimum $\kappa$ exists and satisfies $\kappa\le b-a<a-1$, by taking $X=-S$. Choose a minimizer $U$ of least size and translate it to contain zero.

The inequality

$$f(X\cap Y)+f(X\cup Y)\le f(X)+f(Y)$$

follows directly from intersections and unions of sumsets. To apply it to two intersecting translates of $U$, their union must still be admissible. This is automatic when $G$ is infinite. When $G$ is finite, the exterior $G\setminus(U+S)$ is a minimizer for expansion by (-S), and reflection makes it a minimizer for expansion by $S$. This follows from the inclusions

$$\bigl(G\setminus(X+S)\bigr)-S\subseteq G\setminus X$$

and equality of the two minimum expansion values. Minimality of (|U|) then gives $|G|\ge2|U|+\kappa$.

If the neighborhoods of intersecting translates $U,V$ covered $G$, their intersection would have at most $\kappa$ elements. But it contains a translate of $S$, of size $a>\kappa$, a contradiction. Thus the union is admissible. Submodularity now makes $U\cap V$ a minimizer, so minimal cardinality forces $U=V$. In particular $U+u=U$ for each $u\in U$; hence $U$ is a finite subgroup $H$.

Put $M=|H|$, and let $k$ be the number of $H$-cosets met by $S$. Then $\kappa=(k-1)M$, and

$$M\ge2a-b.$$

The total number $u=kM-a$ of missing elements in these cosets satisfies $u<M$. Two different occupied fibers have sizes summing to more than $M$, so their differences fill the corresponding entire $H$-coset. At least one fiber has more than $M/2$ elements, so its self-difference fills $H$. Therefore (S-S) is a union of complete $H$-cosets. Since

$$b\ge a+\kappa=(2k-1)M-u>(2k-2)M, \qquad b<2a\le2kM,$$

divisibility by $M$ forces

$$b=(2k-1)M,\qquad u<M/2. \tag{14}$$

The case $S-S=G$ satisfies the same conclusion with $k=1$.

This proves the needed small-difference structure without invoking an unprovided additive-combinatorics theorem. A finite subgroup of $\mathbb R/\mathbb Z$ of order $M$ is the regular angular grid $\{j/M:0\le j<M\}$.

8.2. Shifted angular grids and the incidence constant

Let $B=A\cap C$, $|B|=h$, and normalize $C$ to the unit circle. Suppose the angular set $S$ lies in $k$ cosets of an order-$M$ subgroup and (S-S) lies in $l$ cosets of that subgroup. An off-circle, noncentral point $q$ has at most

$$F=kl(27M^{2/3}+17) \tag{15}$$

incidences into $B$ whose lengths occur within $B$.

For a source coset $\alpha+H$ and a target difference coset $\beta+H$, choose principal angular differences in ([0,1/2]). This is valid because (S-S) is symmetric under negation. The incidence equation has the integer-coordinate graph

$$v=\frac{M}{2\pi}\arccos\!\left( \frac{1-\rho^2}{2}+\rho\cos(2\pi(\alpha+j/M)-\phi) \right)-M\beta.$$

The horizontal interval has length less than $M$; the vertical interval has length at most $M/2$. Thus the proof of $J(M)$ in Section 4 applies to each pair of cosets. Using principal angles is important: it avoids an unnecessary extra factor for the branches of cosine.

Since $B$ includes every point of $A$ on $C$, no off-circle point falls into the degenerate case $\rho=1$. The center, if present, affects only one distance class, which is discarded. The old chord classes then receive at most $tF+\binom t2$ new unordered pairs, where $t=n-h$. Consequently, exactly as in (12),

$$R(A)\ge h-\frac n2-F-1. \tag{16}$$

8.3. The exponent (6/7)

Write $s=R(A)$ and $y=h-n/2-s$. The pair count (5) implies

$$|S-S|\le2d(A)+1\le n+2s.$$

If $y>1/2$, the additive lemma applies and gives

$$M\ge2y,\qquad l=|S-S|/M=2k-1,\qquad k\le l.$$

Also $|S-S|<2n$. Thus (15) gives

$$F\le44\left(\frac{|S-S|}{M}\right)^2M^{2/3} \le \frac{176n^2}{(2y)^{4/3}}.$$

By (16), $y\le F+1$. For $y\ge2$, therefore,

$$y^{7/3}\le\frac{352}{2^{4/3}}n^2 <10^{7/3}n^2,$$

which gives $y<10n^{6/7}$. The cases $y<2$ are immediate. This is the claimed bound (13). No asymptotic estimate is used to disguise a missing transfer inequality.

9. Other approaches tried, and what failed

Counting alone

The inequality (5) is useful when the number of distances exceeds roughly $n/2$. It does not prove the problem for arbitrary sets: the supplied lattice constructions have far fewer than $n/2$ distinct distances for large orders. A lower bound tending to infinity on the total number of distances does not force the number of rare ones to tend to infinity.

Taking the next-largest distance

The supplied second-largest-distance bound is $3n/2$, not $n$, and examples can exceed $n$. Clemen--Dumitrescu--Liu, Proposition 1.5 and its proof, were read: the minimum distance and the second-largest distance can both have multiplicity $9n/8+o(n)$. Selecting just those two candidates therefore cannot yield a universal proof. The rank estimate (9) succeeds only after sufficiently strong support localization; on a convex set it gives $(6k-5)n$, which is too large as soon as $k\ge2$.

An earlier, weaker support bound

Before the disk-core argument, the two-circle intersection property gave a fully elementary bound for a distance graph on $s$ vertices:

$$\sum_v\binom{\deg(v)}2\le2\binom s2.$$

If it has $e$ edges, Cauchy--Schwarz gives

$$\frac{2e^2}{s}-e\le s(s-1), \quad e\le U(s):=\frac{s}{4}\bigl(1+\sqrt{8s-7}\bigr)\le s^{3/2}.$$

Combined with layer localization, $s_k\le n^{2/3}$ makes the first $k$ distances rare. If all layers have at most $H$ vertices, this gives $R(A)\ge\lfloor n^{2/3}/H\rfloor$. It remains useful for very small $H$, but only proves divergence when $H=o(n^{2/3})$.

The source-quoted unit-distance bound $O(s^{4/3})$, explicitly stated in Clemen--Dumitrescu--Liu Section 1.4, similarly gives $R(A)\ge\lfloor c n^{3/4}/H\rfloor$ for an absolute $c>0$. Its proof was not supplied there and was not independently recreated. The stronger qualitative conclusion $H=o(n)\Rightarrow R(A)\to\infty$ instead follows from the self-contained disk-core proof, without this quoted incidence theorem.

A much less efficient fixed-rank proof

An independent alternative was developed before the disk-core proof: pairwise disjoint equal-length segments force a longer endpoint distance. Color each pair of such segments by an endpoint choice and one of the $k-1$ larger lengths. A monochromatic five-segment configuration would give three common points of two distinct circles, so a finite Ramsey bound limits the number of disjoint segments. A vertical-line decomposition then bounds the edges by a constant depending on $k$ times the vertex count.

This gives a valid but very inefficient constant, for example

$$C_k=3\big((4(k-1))^{16(k-1)+2}-1\big)^4\quad(k\ge2).$$

The proof is not retained. It is superseded for the present application by $C_k=6k-5$, not used as an unproved intermediate step.

Subsets and deleting diameter endpoints

Deleting endpoints of a unique diameter works at $n=6,8,10$ because the resulting odd-order few-distance sets are classified and their multiplicities can be compared with the original threshold. It does not give an induction for arbitrary $n$. At twelve points, the corresponding deletions have eleven points and five distances, a class not classified by the supplied sources. Similarly, a diameter-independent subset may have rare distances that gain many pairs when the deleted points are restored. No unrestricted transfer bound was established.

Near-minimal convex distance counts

The convex theorem and the one-nonvertex extension use exact minimum distance counts. For hull size $n-2$, the remaining case is generally one above that minimum. Assuming these hulls are automatically cocircular would be a gap. Fishburn's introduction explicitly distinguishes exact equality from the broader near-minimum classification problem. A stability theorem strong enough to deal with bounded excess was considered as a possible route, but no such theorem was proved here.

Algebraic and deformation considerations

Equal-distance constraints can be written as equations between squared lengths; merely noting their algebraic form supplied no useful inequality at the exact threshold $n$. An affine or generic perturbation can split one rich distance into many rare distances, so proving a statement after such a perturbation would not prove it for the original configuration. No generic-deformation extension, field-embedding argument, or unseen number-field note was used in any claim of this record.

10. Proof status and checks

The layer-rank lemma, short-arc argument, exact pair budgets, circle-grid incidence estimate, and additive-atom lemma all have proofs in this record or the linked detailed files. Particular checks were made of the following delicate points:

• A layer step uses strict convexity even when a point is on a hull edge, and strictly increasing lengths cannot introduce coincident endpoints.
• The disk-center $p=q$ is omitted from the circle-intersection count; for every other center the circles really are distinct.
• Short arcs have length strictly less than a semicircle; empty and singleton arcs are allowed. The finite-group union-admissibility step in the additive lemma was proved rather than assumed.
• All new-pair budgets use $t+1$, because richness means at least $n+1$ unordered pairs, not at least $n$.
• The circumcenter is an explicit exception in the incidence bound. All same-circle points are included in $B$ in the arbitrary-circle theorem.
• The exceptional nine-point model was counted using signed radii; its three noncorresponding cross-orbits contribute $6+6+6=18$ pairs.
• The finite-order classifications are invoked only within their stated orders and number-of-distances hypotheses. Source figures absent from the supplied files were not treated as though they had been inspected.

The important remaining uncertainty is mathematical, not numerical: no argument excludes all configurations with a linear-sized convex layer but without a large circular subset. The proofs do not establish either universal assertion, and do not construct a counterexample in the requested range.

The detailed companion records (small orders, near-convex sets, layers, regular polygons, circular majorities, and the additive atom lemma) and the alternate rank proofs are not retained. No general resolution is claimed.