ERDŐS/DAILY

← back to the ledger

ERDőS #604 · PARTIAL

Erdős problem #604 — wave w002

Access date: 2026-07-28 (UTC). This is a partial result, not a solution of the asymptotic problem.

Claim labels

explicitly named published theorem used as an input.

miss, not a theorem.

without claiming a general proof.

Step 0: authoritative live-page check

I fetched both the live problem page and its discussion thread through the Bright Data browser path, not datacenter curl.

[b/page] Verbatim current statement:

Given \(n\) distinct points \(A\subset\mathbb R^2\) must there be a point \(x\in A\) such that \[ > \#\{d(x,y):y\in A\}\gg n^{1-o(1)}? > \] Or even \(\gg n/\sqrt{\log n}\)?

[b/page] Eligibility and markers on the live page:

“if there exists a single point, then by greedily repeating this you will of course get \(\gg n\) such points.” The thread says the site was updated to address the comment.

Thus the mandatory stop condition did not trigger.

[b/page] Known-results text recorded from the live page:

\[ \sum_{x\in A}d(x)\gg \frac{n^2}{\sqrt{\log n}}, \] where \(d(x)\) is the number of distinct distances from \(x\).

for one good point or for \(\gg n\) good points.

Harborth.

\[ \Delta_{\rm pin}(A)\gg n^{c-o(1)},\qquad c=\frac{48-14e}{55-16e}=0.864137\ldots, \] due to Katz and Tardos.

The literal page set includes \(d(x,x)=0\). The papers conventionally omit zero. This changes every exact finite count below by \(+1\), and changes none of the asymptotics.

Literature audit

[b] Katz–Tardos really proves the quoted pinned bound. Corollary 6 of N. H. Katz and G. Tardos, A new entropy inequality for the Erdős distance problem, Contemporary Mathematics 342 (2004), 119–126, says that for every \(\varepsilon>0\), every \(n\)-point planar set has a point seeing

\[ \Omega\!\left(n^{(48-14e)/(55-16e)-\varepsilon}\right) \]

distinct distances. The exact paper is available from Tardos's publication page, and its DOI is 10.1090/conm/342/06136. The checker independently recomputes the exponent as 0.86413751027277106375....

[b] Later primary sources still identify this as the real-plane record. Lund and Petridis explicitly call \(\Omega(N^{0.864\ldots})\) the best known bound in Bisectors and pinned distances, arXiv:1810.00765, Discrete & Computational Geometry 64 (2020), 995–1012, DOI 10.1007/s00454-019-00122-w. Pham, Senger, and Tran again call Katz–Tardos the “best current record” over \(\mathbb R\) in Remark 1.1 of Distribution of pinned distance trees in the plane \(\mathbb F_p^2\), Discrete Mathematics 346 (2023), 113613, DOI 10.1016/j.disc.2023.113613. Their results concern finite fields and do not improve the present real-plane problem.

[b] The exact-small-case input is a published locally-distance theorem. Nozaki and Shinohara define a locally \(k\)-distance set exactly as a set in which every point sees at most \(k\) nonzero distances. In On a generalization of distance sets, arXiv:0906.0199, JCTA 117 (2010), 810–826, DOI 10.1016/j.jcta.2009.11.001, they record:

Their eight-point result cites Erdős and Fishburn, Distinct distances in finite planar sets, Discrete Mathematics 175 (1997), 97–132, DOI 10.1016/S0012-365X(96)00145-800145-8). The latter paper's abstract explicitly states the sharp minimum \(\sum_x d(x)=24\) for eight points.

[c] Search miss, not a novelty claim. Searches by the exact exponent, “pinned distance”, and “locally four-distance” through 2026 found no later real finite-set exponent and no source for the ten-point construction below. This does not prove that the construction is new.

A clean reformulation

For \(n\ge2\), define the nonzero pinned extremal function

\[ f(n)=\min_{\substack{A\subset\mathbb R^2\\|A|=n}} \max_{x\in A}\#\{\lVert x-y\rVert:y\in A,\ y\ne x\}. \]

Let

\[ L(k)=\max\{|A|:A\subset\mathbb R^2,\ \#\{\lVert x-y\rVert:y\ne x\}\le k\text{ for every }x\in A\}. \]

[a] Exact inverse relation.

\[ f(n)=\min\{k:n\le L(k)\}. \]

Indeed, a subset of a locally \(k\)-distance set is still locally \(k\)-distance, while \(n>L(k-1)\) rules out maximum pin count \(k-1\).

[a] An elementary finiteness bound.

\[ L(k)\le 1+k(2k+1). \]

Fix \(x\). The other points lie on at most \(k\) circles centered at \(x\). If one such circle contains \(m\) points of \(A\), choose a point \(p\) on that circle. A circle centered at \(p\) meets the first circle in at most two points, so \(p\) sees at least \(\lceil(m-1)/2\rceil\) distances within that circle. Hence \(m\le2k+1\), and summing the \(k\) shells proves the bound.

[a] The requested bounds translate into one uniform missing lemma:

to \[ L(k)=O(k\sqrt{\log k}). \]

For the second equivalence, use the elementary quadratic bound above to replace \(\log L(k)\) with \(O(\log k)\). The Katz–Tardos result currently gives only

\[ L(k)\le k^{1/c+o(1)} =k^{1.157223229\ldots+o(1)}. \]

[a] Rich line/circle cases are already linear. If a line or circle contains \(M\) points of \(A\), some point of \(A\) sees at least \(\lceil(M-1)/2\rceil\) distances just to that subset. For a line one can choose an extreme point and get \(M-1\). For a circle, every fixed positive distance has multiplicity at most two. In particular, an all-circular set has at least \(\lfloor n/2\rfloor\) pinned distances at every point, and the regular \(n\)-gon attains equality. Thus the hard regime has no line or circle containing \(\Omega(n/\sqrt{\log n})\) points.

Explicit proper locally four-distance set of size 10

[a] Construction. Put

\[ \rho=\frac{3-\sqrt5}{2}=\varphi^{-2}. \]

Take an outer regular pentagon of circumradius \(1\),

\[ O_j=(\cos(2\pi j/5),\sin(2\pi j/5)),\qquad 0\le j<5, \]

and an inner regular pentagon of circumradius \(\rho\), rotated by \(36^\circ\),

\[ I_j=\rho\bigl(\cos((2j+1)\pi/5),\sin((2j+1)\pi/5)\bigr), \qquad 0\le j<5. \]

Write

\[ A=\frac{5-\sqrt5}{2},\quad B=\frac{5+\sqrt5}{2},\quad C=5-2\sqrt5,\quad E=\frac{15-5\sqrt5}{2},\quad F=\frac{25-11\sqrt5}{2}. \]

These are five distinct positive numbers.

[a] Exact distance calculation. The squared distances from every outer point are exactly

\[ \{C,A,E,B\}, \]

and those from every inner point are exactly

\[ \{F,C,A,E\}. \]

To see the coincidences, same-ring outer squared distances are \(A,B\); same-ring inner squared distances are \(\rho^2A,\rho^2B\); and the three cross-ring angular separations are \(36^\circ,108^\circ,180^\circ\). Direct simplification gives

\[ d^2_{\rm cross}(108^\circ)=A,\qquad d^2_{\rm cross}(36^\circ)=\rho^2B=C, \]
\[ d^2_{\rm cross}(180^\circ)=E,\qquad \rho^2A=F. \]

Consequently all ten points see exactly four nonzero distances, while the whole set determines five distances. It is a proper locally four-distance set, so

\[ L(4)\ge10. \]

The standalone checker performs this calculation in the exact field \(\mathbb Q(\sqrt5)\), with rational coefficient pairs and no numerical tolerance.

Exact pinned values through \(n=10\)

[a] Matching eight-point construction. The known locally three-distance extremizer can be written as the four vertices \((\pm\frac12,\pm\frac12)\) of a unit square together with

\[ \left(\pm\frac{1+\sqrt3}{2},0\right),\qquad \left(0,\pm\frac{1+\sqrt3}{2}\right). \]

These are the apices of four outward equilateral triangles. A square vertex has squared-distance set \(\{1,2,2+\sqrt3\}\); an apex has \(\{1,2+\sqrt3,4+2\sqrt3\}\). Hence every point sees exactly three distances. The checker verifies this exactly in \(\mathbb Q(\sqrt3)\).

[b] Exact table. Use the elementary \(L(1)=3\), the published \(L(2)=5\), the published \(L(3)=8\), the eight-point construction above, and the ten-point construction above:

| \(n\) | \(f(n)\), nonzero convention | literal live-page count \(f(n)+1\) | |---:|---:|---:| | 2 | 1 | 2 | | 3 | 1 | 2 | | 4 | 2 | 3 | | 5 | 2 | 3 | | 6 | 3 | 4 | | 7 | 3 | 4 | | 8 | 3 | 4 | | 9 | 4 | 5 | | 10 | 4 | 5 |

In particular,

\[ \boxed{f(10)=4} \]

is rigorous modulo the named Nozaki–Shinohara theorem \(L(3)=8\). The construction proves \(f(10)\le4\), and \(10>8=L(3)\) proves \(f(10)\ge4\).

[b] Precisely the next unresolved finite question under this reduction is

\[ f(11)\in\{4,5\}, \]

and deciding between the two is equivalent to deciding whether a planar locally four-distance set of size 11 exists. The upper value \(5\) follows from the regular 11-gon; the lower value \(4\) follows from \(L(3)=8\).

Exact square-grid obstruction

Let

\[ G_m=\{0,1,\ldots,m-1\}^2,\qquad n=m^2. \]

[a] Closed exact formula for the maximum pin count:

\[ \Delta_{\rm pin}(G_m) =D_m =\#\{a^2+b^2:0\le a,b\le m-1,\ (a,b)\ne(0,0)\}. \]

Every difference vector between two grid points has \((|a|,|b|)\in[0,m-1]^2\), so every pin's squared-distance set is contained in the displayed set. A corner realizes every displayed pair, proving equality.

[d] Exact recomputed table:

| \(m\) | \(n=m^2\) | \(D_m\) | |---:|---:|---:| | 2 | 4 | 2 | | 3 | 9 | 5 | | 4 | 16 | 9 | | 5 | 25 | 14 | | 10 | 100 | 50 | | 20 | 400 | 179 | | 50 | 2500 | 992 | | 100 | 10000 | 3663 | | 200 | 40000 | 13647 | | 512 | 262144 | 82489 |

For \(2\le m\le12\), the checker also enumerates every pin independently and confirms that its distance set is contained in the corner set and that a corner attains the maximum.

[b] Asymptotic verification modulo the Landau–Ramanujan theorem. If \(B(X)\) counts positive integers at most \(X\) representable as a sum of two squares, then

\[ B((m-1)^2)\le D_m\le B(2(m-1)^2). \]

Landau–Ramanujan gives \(B(X)=\Theta(X/\sqrt{\log X})\), hence

\[ D_m=\Theta\!\left(\frac{m^2}{\sqrt{\log m}}\right) =\Theta\!\left(\frac{n}{\sqrt{\log n}}\right). \]

This proves, rather than merely samples, why the grid has the order stated on the live page.

Verification

Run:

python runs/erdos604_wavew002_reverify.py

The checker uses only the Python standard library. It:

  1. computes all squared distances in the eight-point construction exactly in

\(\mathbb Q(\sqrt3)\);

  1. computes all squared distances in the ten-point construction exactly in

\(\mathbb Q(\sqrt5)\);

  1. verifies the construction half of the exact \(f(n)\) table;
  2. recomputes every grid-table entry and independently enumerates all pins for

\(m\le12\); and

  1. recomputes the Katz–Tardos decimal to 60-digit working precision.

Observed final line:

ALL CHECKS PASSED

Honest boundary of the result

[a] The ten-point construction and the grid formula are exact, but neither is a uniform lower bound for arbitrary \(n\). They do not improve the Katz–Tardos exponent.

[b] The exact missing theorem remains the uniform estimate

\[ L(k)=O(k\sqrt{\log k}) \]

(or \(L(k)\le k^{1+o(1)}\) for the weaker question). Existing global Guth–Katz distance machinery does not supply this pinned estimate; the later primary literature above continues to distinguish the two problems.

[c] An unstructured exact search for the next case is not realistic: already \(K_{11}\) has 55 edges and \(B_{55} =359334085968622831041960188598043661065388726959079837\) unrestricted edge-equality partitions. Local-degree constraints and symmetry would reduce this drastically, so this is not a lower bound on a good algorithm; it explains why a naive semialgebraic enumeration was not run.

PARTIAL: Constructed and exactly checked a 10-point proper locally four-distance set, proving \(f(10)=4\) modulo the published sharp theorem \(L(3)=8\); the asymptotic Erdős problem remains open.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger