Erdoes #1207 - working report - Erdős Problem a Day

Erdős problem 1207 — wave w038

Date: 2026-07-29 (UTC)

Result

[b] The conjectural part of the live problem has just been proved, although the live page has not yet incorporated the result. Lee, Pohoata, and Zhu, The Minkowski grid has robustly many repeated distances, arXiv:2607.05374v1 (submitted 2026-07-06), prove that

P_2(n)\ll n^{1-\delta}

for an absolute $\delta>0$. I downloaded the v1 source, checked that its definition $f_{\rm iso}(n)$ is exactly the live page's $P_2(n)$, audited the complete sieve and parameter argument, and independently checked its finite counting inequalities and exponent algebra. The qualification [b] is important: the proof uses the published tower theorem of Hajir--Maire--Ramakrishna and Chebotarev as black boxes, and the July paper itself is currently an eight-page v1 preprint.

This answers “is it true that $P_2(n)<n^{1-c}$?” affirmatively for all sufficiently large $n$. It does not determine the asymptotic order of $P_d(n)$, so the broader “estimate $P_d(n)$” part remains open.

Claim labels

[a] elementary-rigorous, with the proof given here;
[b] rigorous modulo the explicitly named theorem(s);
[c] plausible or search-based, not proved;
[d] computational-only or a direct observation of external metadata.

No novelty is claimed for the July 2026 proof. The useful work in this run is the live-page/status check, primary-source discovery and verification, a from-scratch audit that isolates the black-box input, an explicit exponent extraction, and a standalone finite checker.

Step 0: mandatory live-page audit before mathematics

[d] I fetched both the live problem page and its discussion thread through the Bright Data browser. On 2026-07-29 the page showed:

status OPEN;
0 claimed proofs for this problem;
Currently working on this problem: None;
Interested in collaborating: None;
one comment;
last problem-page edit: 2026-04-08.

Thus the task's mandatory collision/claimed-proof stop rule did not trigger. The forum comment is not registered as a claimed proof, and the site expressly labels comments unverified.

The exact live statement, copied verbatim from the site's LaTeX view, is:

Let $P_d(n)$ be such that in any set of $n$ points in $\mathbb{R}^d$ there exist at least $P_d(n)$ many points which do not contain an isosceles triangle. Estimate $P_d(n)$ - in particular, is it true that\[P_2(n)<n^{1-c}\]for some constant $c>0$?

[d] The listed known-results text says:

Erdős attributes the question to Riddell and records

$P_d(n)>n^{\epsilon_d}$, with $\epsilon_d\to0$ as $d\to\infty$. The page gives the sharper displayed consequence \[ P_d(n)\ge f_d(n)\ge n^{1/(3d-3)-o(1)} \] from the distinct-distance-subset problem #1208.

Erdős suggests lattice points in a small-radius sphere as a possible worst

configuration.

The $d=1$ case is the problem of finding a three-term-progression-free

subset.

The page attributes $P_2(n)\gg n^{0.432}$ to the Pach--Tardos

isosceles-triangle bound plus random deletion.

It says the regular-polygon $O(n^{1/2})$ claim in Brass--Moser--Pach,

§5.3, appears incorrect. What the regular polygon actually gives is $P_2(n)\ll r_3(n)$, because its isosceles triples are cyclic three-term progressions.

It points to #1208 and #657 for the stronger all-distances-distinct

variant.

The sole comment, by Kenta Kitamura at 23:43 on 2026-07-08, reads:

I would like to mention the recent preprint of Lee--Pohoata--Zhu, “The Minkowski grid has robustly many repeated distances”. It appears to settle the conjectural part of Problem #1207. The authors explicitly say in the abstract that their first result “confirms a conjecture of Erdős from 1980”, and after Corollary 3 they again write that “the first result confirms a conjecture by Erdős from 1980”.

Definition alignment: there is no hidden mismatch

For an $n$-point set $P\subset\mathbb R^2$, define

\alpha_{\rm iso}(P)= \max\{|A|:A\subseteq P,\ A\text{ has no isosceles triple}\}.

Then the largest integer guaranteed in every $n$-point set is

P_2(n)=\min_{|P|=n}\alpha_{\rm iso}(P). \tag{1}

[a] The live problem counts equally spaced collinear triples: otherwise its stated $d=1$ equivalence with three-term arithmetic progressions would make no sense. Lee--Pohoata--Zhu explicitly adopt the same convention (“degenerate isosceles triangles ... are also forbidden”) and define

f_{\rm iso}(n)=\min_{|P|=n}\alpha_{\rm iso}(P).

Consequently $f_{\rm iso}(n)=P_2(n)$ exactly, not merely up to a comparison.

Primary-source literature check

[d] Erdős's source. The cited paper really is P. Erdős,

A survey of problems in combinatorial number theory, Annals of Discrete Mathematics 6 (1980), 89--115. On p. 110 it defines $P(n,k)$, attributes it to Riddell, records the power lower bound, asks for a sublinear power in the plane, and suggests lattice points in a sphere. Primary PDF

[d] Pach--Tardos. J. Pach and G. Tardos,

Isosceles triangles determined by a planar point set, Graphs and Combinatorics 18 (2002), 769--779, DOI 10.1007/s003730200063, exists. The author's PDF states Theorem 1 with isosceles-triple exponent \[ \alpha=\frac{11e-3}{5e-1}=2.136464617\ldots . \] Direct one-stage random deletion gives $(3-\alpha)/2=2e/(5e-1)=0.431767691\ldots$, with the usual arbitrarily small exponent loss. This calculation is included in the checker as a diagnostic; it is not used below, and I do not rely on a three-decimal rendering of the historical bound. Author PDF

[b] June 2026 precursor. Croot, Mao, Pohoata, Sheffer, and Yip,

A combinatorial large sieve for Sidon sets, distances, and norm forms, arXiv:2606.17487v2, proves for the $N\times N$ square grid that an isosceles-free subset has size \[ O\!\left(N^2\exp\!\left[-c\frac{\log N}{\log\log N}\right]\right). \] Its concluding remarks define exactly the min--max quantity in (1) and announce, without supplying the separate construction there, that vertical amplification should give $P_2(n)\ll n^{1-c}$.

[b] July 2026 proof. Sungchul Lee, Cosmin Pohoata, and Daniel G. Zhu,

The Minkowski grid has robustly many repeated distances, arXiv:2607.05374v1, exists with the stated authors and submission time 2026-07-06 17:52:52 UTC. The downloaded v1 source archive has SHA-256 ede1c4ce70713446ec7b6f41c8babc6557d1b3b219879852c3260bf07896ee34. Theorem 2, Corollary 3, and their proofs say exactly what the forum comment reports.

[b] Deep input. Farshid Hajir, Christian Maire, and Ravi Ramakrishna,

Cutting towers of number fields, Annales Mathématiques du Québec 45 (2021), 321--345, DOI 10.1007/s40316-021-00156-8, proves the tower/splitting result invoked by the July preprint. Its abstract and main theorem explicitly produce infinite asymptotically good extensions in which infinitely many primes split completely. arXiv:1901.04354

[c] Searches by the exact problem wording, $f_{\rm iso}$, the July paper's title and identifier, and isosceles-free planar subsets found no primary-source retraction, correction, competing claimed proof, or later improvement. This is a documented search miss, not a proof of absence.

Elementary reduction from robust distance multiplicity to $P_2(n)$

For a finite $A\subset\mathbb R^2$, let

\mu(A)=\max_{\lambda>0} \#\{(a,b)\in A^2:a\ne b,\ |a-b|=\lambda\}, \tag{2}

where ordered pairs are counted.

[a] Lemma. If $A$ contains no isosceles triangle, including no equally spaced collinear triple, then

\mu(A)\le |A|. \tag{3}

Proof. For a fixed $\lambda$, make a graph on $A$, joining two points at distance $\lambda$. Two edges sharing a vertex give three distinct points with two equal distances from that vertex, hence an isosceles triple. Thus this graph is a matching. It has at most $|A|/2$ unordered edges and at most $|A|$ ordered pairs. Maximize over $\lambda$. $\square$

[b] Lee--Pohoata--Zhu Theorem 2. There is $\delta_0>0$ such that, for every $n$, some $n$-point $P\subset\mathbb R^2$ satisfies, for every $A\subseteq P$ with $|A|\ge2$,

\mu(A)\gg \frac{|A|^2}{n^{1-\delta_0}}. \tag{4}

Their proof gives an exact constant-one version of (4) after a fixed threshold, with a possibly smaller $\delta_0$. Combining that version with (3), every isosceles-free $A\subseteq P$ satisfies

\frac{|A|^2}{n^{1-\delta_0}}\le\mu(A)\le |A|, \qquad\text{so}\qquad |A|\le n^{1-\delta_0}. \tag{5}

By (1),

P_2(n)\le n^{1-\delta_0} < n^{1-\delta_0/2}\quad(n>1). \tag{6}

Thus the live question has the strict requested form with $c=\delta_0/2>0$ for all sufficiently large $n$.

[a] Dimension monotonicity. A planar configuration can be embedded isometrically into $\mathbb R^d$. Since $P_d(n)$ is a minimum over all $d$-dimensional configurations,

P_d(n)\le P_2(n)\qquad(d\ge2). \tag{7}

Combining the live page's lower bound with (6) gives the currently verified but very wide window, for each fixed $d\ge2$,

n^{1/(3d-3)-o(1)} \ \le\ P_d(n)\ \le\ P_2(n)\ \le\ n^{1-\delta_0}. \tag{8}

The middle inequality is equality when $d=2$. Formula (8) explains why the particular conjecture is settled while the requested “estimate” is not.

Audit of the July proof

1. The isolated non-elementary input

[b] Proposition 4 of the July preprint, deduced from Hajir--Maire--Ramakrishna Theorem 4 and Chebotarev, supplies totally real fields

\mathbb Q=K_0\subset K_1\subset K_2\subset\cdots

with

[K_i:\mathbb Q]=2^i,\qquad \operatorname{rd}(K_i)\le D,

and an infinite set $\mathcal P$ of rational primes congruent to $1\bmod4$ that split completely in every $K_i$. The July paper explains the $1\bmod4$ refinement by applying Chebotarev to the compositum with $\mathbb Q(i)$.

This proposition is the exact black box a from-scratch finite checker cannot certify. Everything below is conditional only on this tower and the standard Minkowski-box lattice estimate quoted as Lemma 5 in the paper.

2. The finite sieve, reconstructed

Fix $K=K_i$, $d=[K:\mathbb Q]$, distinct $p_1,\ldots,p_k\in\mathcal P$, and

Q=\prod_{i=1}^k p_i.

Let

B_K(X)=\{\alpha\in\mathcal O_K: |\sigma_j(\alpha)|\le X\text{ for all real embeddings }\sigma_j\}.

For $A\subseteq B_K(X)^2$, put

\[ q(a,b)=(a_1-b_1)^2+(a_2-b_2)^2. \]

[a] Under any real embedding $\sigma:K\hookrightarrow\mathbb R$, injectivity of $\sigma$ makes equality of two $q$-values equivalent to equality of the corresponding squared Euclidean distances. Thus the algebraic multiplicity used in the sieve is exactly (2) after embedding.

For every prime ideal $\mathfrak p_{i,j}\mid p_i$ and every choice of sign $\epsilon_{i,j}\in\{\pm1\}$, the factorization of $x^2+y^2$ modulo $\mathfrak p_{i,j}$ defines an additive subgroup $L_\epsilon\subset\mathcal O_K^2$. Complete splitting and the Chinese remainder theorem give

|\mathcal O_K^2/L_\epsilon|=Q^d. \tag{9}

If $S_\epsilon$ is the set of ordered distinct pairs $(a,b)\in A^2$ with $a-b\in L_\epsilon$, Cauchy--Schwarz over the $Q^d$ cosets gives, whenever $|A|\ge2Q^d$,

|S_\epsilon| \ge \frac{|A|^2}{Q^d}-|A| \ge \frac{|A|^2}{2Q^d}. \tag{10}

There are $2^{kd}$ sign vectors. Every pair in their union has $Q\mid q(a,b)$. If $w(r)$ counts the prime ideals $\mathfrak p_{i,j}$ for which $\mathfrak p_{i,j}^2\mid r$, a pair is counted for at most $2^{w(q(a,b))}$ sign vectors. Since $q(a,b)\in B_K(8X^2)$,

\frac{2^{kd}|A|^2}{2Q^d} \le \mu(A)\cdot \sum_{r\in Q\mathcal O_K\cap B_K(8X^2)}2^{w(r)}. \tag{11}

For $r$ divisible by $Q$, $2^{w(r)}$ counts the ideal divisors common to $r/Q$ and $Q$. Therefore the last sum is exactly

\sum_{\mathfrak a\mid Q} |Q\mathfrak a\cap B_K(8X^2)|.

The elementary separation estimate in Lemma 5 yields, for $X\ge Q$,

\begin{aligned} \sum_{\mathfrak a\mid Q} |Q\mathfrak a\cap B_K(8X^2)| &\le \left(\frac{17X^2}{Q}\right)^d \sum_{\mathfrak a\mid Q}\frac1{N(\mathfrak a)}\\ &= \left( \frac{17X^2}{Q}\prod_{i=1}^k\left(1+\frac1{p_i}\right) \right)^d. \tag{12} \end{aligned}

Substituting (12) into (11) gives

\mu(A)\ge \frac{|A|^2}{2} \left( \frac1{17X^2} \prod_{i=1}^k\frac{2p_i}{p_i+1} \right)^d \ge |A|^2 \left( \frac1{34X^2} \prod_{i=1}^k\frac{2p_i}{p_i+1} \right)^d. \tag{13}

[a] The last absorption is valid for every $d\ge1$ because $\tfrac12\,17^{-d}\ge34^{-d}$. The standalone checker reconstructs every set, multiplicity, overlap weight, image value, and inequality in (9)--(13) for two nontrivial $K=\mathbb Q$ instances.

3. Parameter choice and an explicit symbolic exponent

Choose enough common split primes that

\frac1{34}\prod_{i=1}^k\frac{2p_i}{p_i+1}\ge2D, \tag{14}

and retain the resulting fixed $Q$. For $n\ge100Q^2$, the intervals

[(10Q)^{2d},(10Q)^{4d}),\qquad d=1,2,4,8,\ldots,

tile the entire range $[(10Q)^2,\infty)$. Hence there is a unique power of two $d$ with

(10Q)^{2d}\le n<(10Q)^{4d}. \tag{15}

Set $K=K_i$ of degree $d$ and $X=n^{1/(2d)}\sqrt D$. The Minkowski-box lower estimate gives $|B_K(X)|\ge\sqrt n$, so $B_K(X)^2$ contains an $n$-point set $P$. Equations (13)--(14) imply, for $|A|\ge(2Q)^d$,

\mu(A)\ge |A|^2\,\frac{2^d}{n}. \tag{16}

The paper defines $\delta_1,\delta_2>0$ by

(100Q^2)^{1/2-\delta_1}=2Q,\qquad (10000Q^4)^{\delta_2}=2.

[a] Solving these exactly gives

\delta_1=\frac{\log5}{2\log(10Q)},\qquad \delta_2=\frac{\log2}{4\log(10Q)}. \tag{17}

For large $A$, (15)--(17) turn (16) into $\mu(A)\ge |A|^2/n^{1-\delta_2}$. For $2\le|A|<n^{1/2-\delta_1}$, the trivial $\mu(A)\ge1$ gives the same shape with exponent $2\delta_1$. Thus one may take

\delta_0=\min(2\delta_1,\delta_2) =\delta_2 =\frac{\log2}{4\log(10Q)}>0, \tag{18}

and the strict exponent in the live question may be taken as

c=\frac{\delta_0}{2} =\frac{\log2}{8\log(10Q)}>0. \tag{19}

This is symbolic rather than numerical because the proof does not list the chosen common split primes, so it does not give a numerical $Q$. Positivity, not optimization, is all the Erdős question requires.

Crucially, (15) covers every sufficiently large $n$; the proof is not merely a construction on a sparse subsequence. This is the uniformity step that actually closes the conjectural part.

Standalone checker and exact finite data

The checker is erdos1207_wavew038_reverify.py. Run:

python runs/erdos1207_wavew038_reverify.py

It uses only the Python standard library and completes in about $0.32$ seconds on this VM.

Exact regular-polygon computation

[a] On the regular $n$-gon, chord length depends only on cyclic separation. For distinct vertices $a,b,c$,

|a-b|=|a-c| \iff b-a\equiv -(c-a)\pmod n \iff b+c\equiv2a\pmod n.

Thus its isosceles triples are exactly nontrivial cyclic three-term progressions. The checker constructs the two hyperedge sets independently (once from chord classes, once from modular progressions) and asserts equality.

[d] Two separate exhaustive maximum-independent-set implementations give:

| $n$ | maximum isosceles-free subset of regular $n$-gon | $r_3(n)$ | |---:|---:|---:| | 3 | 2 | 2 | | 4 | 2 | 3 | | 5 | 2 | 4 | | 6 | 4 | 4 | | 7 | 3 | 4 | | 8 | 4 | 4 | | 9 | 4 | 5 | | 10 | 4 | 5 | | 11 | 4 | 6 | | 12 | 4 | 6 | | 13 | 4 | 7 | | 14 | 6 | 8 | | 15 | 4 | 8 | | 16 | 6 | 8 |

Here $r_3(n)$ is independently recomputed for the interval $\{0,\ldots,n-1\}$. The cyclic value is always at most $r_3(n)$, as the live page asserts. These are exact values for this explicit configuration, not exact values of the min--max function $P_2(n)$.

Multiplicity and sieve checks

[d] The checker exhausts all 2,040 subsets of regular $n$-gons for $3\le n\le10$, verifies directly that isosceles-free is equivalent to every fixed-distance graph being a matching, and checks $\mu(A)\le|A|$.

For the $K=\mathbb Q$, $p=Q=X=5$ specialization of the sieve lemma it checks both the full 121-point grid $[-5,5]^2$ and a deterministic irregular 83-point subset. Representative exact output is:

full:      mu=720, sum |S_epsilon|=5616,
           weighted-pair sum=6504, weighted-image sum=98,
           lemma RHS=14641/510
irregular: mu=304, sum |S_epsilon|=2596,
           weighted-pair sum=3002, weighted-image sum=98,
           lemma RHS=6889/510

The code separately checks the Cauchy--Schwarz lower bound, sign-overlap bound, divisor identity, lattice-count upper bound, the $1/34$ constant, the power-of-two interval endpoints, (17)--(19), and finishes with ALL CHECKS PASSED.

What remains, and the exact wall

[b] The particular planar power-saving question is answered by the

July preprint, modulo its named tower input.

[c] The asymptotic order of $P_2(n)$, and still more of $P_d(n)$,

remains far from determined: (8) has a large exponent gap.

[c] The construction is existential in its number-field layer. A

numerical coordinate generator would require defining polynomials for an unbounded tower $K_i$, an explicit common list of completely split $1\bmod4$ primes, and effective bounds for $D$ and their sizes. Those data are not supplied by the July paper. This is why the finite checker audits the sieve rather than pretending to instantiate the asymptotic point sets.

[d] No computation exceeding a few CPU-minutes was run. Heavy search

is not the missing ingredient here; the remaining broad problem needs new upper/lower-bound theory, while numerical realization needs effective algebraic-number-field data.

FOUND: Lee--Pohoata--Zhu, arXiv:2607.05374v1, source-audited modulo the Hajir--Maire--Ramakrishna/Chebotarev tower input, proves $P_2(n)<n^{1-c}$ for some $c>0$ and all sufficiently large $n$; the full estimation of $P_d(n)$ remains open.

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