Reused writeup on Math-5799 (gpt pro 5.2)

Erdős Problem 657 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/657

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

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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_5.2/657.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/657.tex.

Is it true that if $A\subset \mathbb{R}^2$ is a set of $n$ points such that every subset of $3$ points determines $3$ distinct distances (i.e. $A$ has no isosceles triangles) then $A$ must determine at least $f(n)n$ distinct distances, for some $f(n)\to \infty$? In [Er73] Erdős attributes this problem (more generally in $\mathbb{R}^k$) to himself and Davies. In [Er97e] he does not mention Davis, but says this problem was investigated by himself, Füredi, Ruzsa, and Pach. In [Er73] Erdős says it is not even known in $\mathbb{R}$ whether $f(n)\to \infty$. Sarosh Adenwalla has observed that this is equivalent to minimising the number of distinct differences in a set $A\subset \mathbb{R}$ of size $n$ without three-term arithmetic progressions. Dumitrescu [Du08] proved that, in these terms,\[(\log n)^c \leq f(n) \leq 2^{O(\sqrt{\log n})}\]for some constant $c>0$. Hunter observed in the comments that a result of Ruzsa coupled with standard tools of additive combinatorics (with details given by Alfaiz and Tang) allow recent progress on the size of subsets without three-term arithmetic progression (see [BlSi23] which improves slightly on the bounds due to Kelley and Meka [KeMe23]) yield\[2^{c(\log n)^{1/9}}\leq f(n)\]for some constant $c>0$. Straus has observed that if $2^k\geq n$ then there exist $n$ points in $\mathbb{R}^k$ which contain no isosceles triangle and determine at most $n-1$ distances. See also [135] . References [BlSi23] T. F. Bloom and O. Sisask, An improvement to the Kelley-Meka bounds on three-term arithmetic progressions . arXiv:2309.02353 $2023$. [Du08] Dumitrescu, Adrian, On distinct distances and {$\lambda$}-free point sets . Discrete Math. $2008$, 6533--6538. [Er73] Erdős, P., Problems and results on combinatorial number theory . A survey of combinatorial theory (Proc. Internat. Sympos., Colorado State Univ., Fort Collins, Colo., 1971) $1973$, 117-138. [Er97e] Erdős, Paul, Some of my favourite unsolved problems . Math. Japon. $1997$, 527-537. [KeMe23] Kelley, Z. and Meka, R., Strong Bounds for 3-Progressions . arXiv:2302.05537 $2023$.

1) FORMAL RESTATEMENT Let $n\ge 1$ and let $A\subset\mathbb{R}^2$ with $|A|=n$. Assume that for every triple of distinct points $a,b,c\in A$ the three pairwise distances $|a-b|,|a-c|,|b-c|$ are all distinct (equivalently, $A$ contains no isosceles triangle). Let \[ D(A):=\left|\left\{|x-y|\ :\ x,y\in A,\ x\neq y\right\}\right| \] be the number of distinct (nonzero) distances determined by $A$. Question: does there exist a function $f:\mathbb{N}\to\mathbb{R}_{>0}$ with $f(n)\to\infty$ such that every such $A$ satisfies $D(A)\ge f(n)\,n$?

Edge cases: for $n\le 2$ the hypothesis is vacuous and $D(A)\in\{0,1\}$.

2) QUICK LITERATURE/CONTEXT CHECK The problem statement records nontrivial lower bounds coming from additive-combinatorial reformulations (Dumitrescu; later improvements in the comments on erdosproblems.com). I do not rely on any external results beyond what is explicitly written in the problem statement.

3) ATTACK PLAN Proof-track ideas: - Use the "no isosceles" constraint to bound multiplicities of repeated distances, then apply double counting to force many distinct distances. - Translate to 1D difference-set questions (as in the statement) and try to import bounds for 3-term-AP-free sets.

Disproof-track ideas: - Attempt to construct point sets with no isosceles triangles but only $O(n)$ distinct distances, e.g. by arranging repeated distances along a large matching of equal-length segments.

At present I only reach linear lower bounds from basic multiplicity control.

4) WORK

Lemma 1 (distinct distances from a fixed apex). If $A\subset\mathbb{R}^2$ has no isosceles triangle and $p\in A$, then the $n-1$ distances $\{|p-q|: q\in A\setminus\{p\}\}$ are all distinct. In particular $D(A)\ge n-1$ for $n\ge 2$.

Proof. Fix $p\in A$. Suppose $q,r\in A\setminus\{p\}$ are distinct and satisfy $|p-q|=|p-r|$. Then the triple $\{p,q,r\}$ forms an isosceles triangle with apex $p$, contradicting the hypothesis. Thus the $n-1$ distances from $p$ to the other points are pairwise distinct. Since $D(A)$ counts distinct distances appearing among all pairs, it is at least the number of distinct distances appearing among the pairs $\{p,q\}$ with $q\neq p$, namely $n-1$. $\square$

Lemma 2 (fixed-length edges form a matching). Let $A\subset\mathbb{R}^2$ have no isosceles triangle, and fix a distance value $d>0$. Consider the graph on vertex set $A$ whose edges are the unordered pairs $\{x,y\}$ with $|x-y|=d$. Then this graph has maximum degree $1$; equivalently, the edges of length $d$ form a matching. In particular, the multiplicity of the distance value $d$ among all $\binom n2$ pairs is at most $\lfloor n/2\rfloor$.

Proof. If a vertex $x\in A$ were incident to two distinct edges $\{x,y\}$ and $\{x,z\}$ of length $d$, then $|x-y|=|x-z|=d$ with $y\neq z$, so $\{x,y,z\}$ would be an isosceles triangle with apex $x$, contradiction. Hence every vertex has degree at most $1$, so the edges form a matching and there are at most $\lfloor n/2\rfloor$ such edges. $\square$

Proposition 3 (double counting). If $A$ has no isosceles triangle and $n\ge 2$, then \[ D(A)\ \ge\ \frac{\binom n2}{\lfloor n/2\rfloor}\ \ge\ n-1. \]

Proof. Write $\binom n2=\sum_{d} m_d$, where the sum runs over distinct distance values $d$ and $m_d$ is the number of pairs at distance $d$. By Lemma 2, $m_d\le \lfloor n/2\rfloor$ for every $d$, so $\binom n2\le D(A)\,\lfloor n/2\rfloor$, giving the first inequality. The second inequality is a direct check: if $n=2m$ then $\binom n2/(n/2)=2m-1=n-1$; if $n=2m+1$ then $\binom n2/\lfloor n/2\rfloor=2m+1=n$. $\square$

FAST REALITY CHECK (brute force on the integer grid $[-2,2]^2$). I searched all $n$-subsets of $[-2,2]^2$ and minimized the number of distinct squared distances subject to the no-isosceles constraint. Results (distinct distances, then one witness set): - $n=3$: $D=3$, e.g. $\{(-2,-2), (-2,-1), (-2,1)\}$. - $n=4$: $D=3$, e.g. $\{(-2,-2), (-2,-1), (0,-2), (0,-1)\}$. - $n=5$: $D=6$, e.g. $\{(-2,-1), (-1,1), (0,2), (1,-2), (2,0)\}$. - $n=6$: $D=7$, e.g. $\{(-2,-2), (-2,-1), (-2,1), (-2,2), (0,-2), (0,2)\}$. These are only sanity checks inside a small search window, not proofs of the true minima.

5) VERIFICATION - Lemma 1: the only step is that $|p-q|=|p-r|$ produces an isosceles triangle with apex $p$, which is exactly the forbidden configuration. - Lemma 2: I checked that forbidding isosceles triangles rules out two equal-length edges sharing a common endpoint; the conclusion that the equal-length edges form a matching follows. - Proposition 3: the decomposition $\binom n2=\sum_d m_d$ and the inequality $m_d\le\lfloor n/2\rfloor$ are correct, so the bound is valid.

6) FINAL: UNRESOLVED (i) Strongest proved partial result: $D(A)\ge n-1$ for every $n\ge 2$, and moreover each fixed distance value occurs at most $\lfloor n/2\rfloor$ times (Lemma 2), hence $D(A)\ge \binom n2/\lfloor n/2\rfloor$. (ii) First gap (crisp): prove any superlinear bound $D(A)\ge n\cdot g(n)$ with $g(n)\to\infty$ (even $g(n)=\log\log n$) for all no-isosceles $A\subset\mathbb{R}^2$. (iii) Top 3 next moves (concrete): 1. Prove an improved multiplicity bound $m_d=o(n)$ for many $d$, e.g. show equal-length matchings cannot cover a positive fraction of all pairs once higher-order constraints are used. 2. Work in the 1D reformulation (distinct differences without 3-term AP) and push the strongest known lower bounds for $|A-A|$ in that setting back to planar configurations. 3. Computation: search for structured large examples with small $D(A)$ to guess extremizers and then attempt a stability theorem around those extremizers. (iv) Minimal counterexample structure (if the conjecture is true): a family $A_n$ with $D(A_n)=O(n)$ would need distance multiplicities close to the matching upper bound $\lfloor n/2\rfloor$ for many distance values, i.e. most pairs partitioned into near-perfect matchings by distance classes.

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.