Reused writeup on Math-6099 (gpt pro 5.2)

Erdős Problem 1045 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1045

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_5.2/1045.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1045.tex.

1) FORMAL RESTATEMENT

Fix an integer $n\ge 2$. Let $z_1,\dots,z_n\in\mathbb C$ satisfy the diameter constraint \[ |z_i-z_j|\le 2\quad\text{for all }i,j. \] Define \[ \Delta(z_1,\dots,z_n):=\prod_{i\ne j}|z_i-z_j|. \] (Equivalently, if $P:=\prod_{1\le i<j\le n}|z_i-z_j|$ is the unordered product, then $\Delta=P^2$.) Determine \[ M_n:=\max\{\Delta(z_1,\dots,z_n): |z_i-z_j|\le 2\ \forall i,j\}, \] and decide whether maximizers can be taken to be the vertices of a regular $n$-gon (scaled so that the diameter equals $2$).

2) QUICK LITERATURE/CONTEXT CHECK

The provided text states several known bounds and constructions, and explicitly notes that for $n=4$ and $n=6$ there are examples showing the regular polygon is not a maximizer. In this writeup I will not rely on any external results; I will give an explicit counterexample for $n=4$ with full verification.

3) ATTACK PLAN

Proof track (small $n$): - For $n=2,3$, determine $M_n$ directly by bounding each distance by $2$. - For regular polygons, compute $\Delta$ via Vandermonde determinants.

Disproof track (polygon optimality): - Search numerically for $n=4$ to find a configuration beating the square; then simplify it to exact algebraic coordinates and verify rigorously.

4) WORK

FAST REALITY CHECK (small $n$)

For $n=2$: - Constraint gives $|z_1-z_2|\le 2$. - Then $\Delta=|z_1-z_2|^2\le 4$, achieved by any pair at distance $2$. So $M_2=4$.

For $n=3$: - There are three unordered distances, each $\le 2$, so $P=\prod_{i<j}|z_i-z_j|\le 2^3=8$, hence $\Delta=P^2\le 64$. - Equality requires all three distances $=2$, which is realized by an equilateral triangle of side $2$. So $M_3=64$.

Lemma 4.1 (Scaling reduction to diameter $2$)

Let $P(z_1,\dots,z_n)=\prod_{i<j}|z_i-z_j|$. If $\operatorname{diam}:=\max_{i<j}|z_i-z_j|<2$, then scaling all points by factor $2/\operatorname{diam}$ produces a new configuration satisfying the constraints with larger product $P$, hence larger $\Delta$. Therefore any maximizer satisfies $\max_{i<j}|z_i-z_j|=2$.

Proof. If we replace each $z_i$ by $z_i' := \lambda z_i$ with $\lambda>0$, then all distances scale by $\lambda$. Thus $P$ scales by $\lambda^{\binom{n}{2}}$ and $\Delta$ scales by $\lambda^{n(n-1)}$. If the diameter is $\operatorname{diam}<2$, choosing $\lambda=2/\operatorname{diam}>1$ keeps all pairwise distances $\le 2$ and strictly increases $P$ and $\Delta$. $\square$

Lemma 4.2 (Regular $n$-gon value)

Let $\zeta= e^{2\pi i/n}$ and $u_k=\zeta^k$ for $k=0,1,\dots,n-1$. Then \[ \prod_{0\le i<j\le n-1} |u_i-u_j| = n^{n/2}, \] so for the regular $n$-gon on the unit circle (circumradius $1$), the ordered product is \[ \Delta_{\mathrm{unit\ circle}} = n^n. \] To enforce diameter $2$: - If $n$ is even, the unit-circle polygon already has diameter $2$, hence $\Delta_{\mathrm{reg}}=n^n$. - If $n$ is odd, the maximal chord length on the unit circle is $2\cos(\pi/(2n))$, so scaling by $\sec(\pi/(2n))$ makes the diameter $2$ and multiplies $\Delta$ by the factor $\sec(\pi/(2n))^{n(n-1)}$. Thus \[ \Delta_{\mathrm{reg}} = \cos\!\left(\frac{\pi}{2n}\right)^{-n(n-1)}\,n^n. \]

Proof. The Vandermonde determinant identity gives \[ \prod_{0\le i<j\le n-1} (u_j-u_i) = \prod_{j=1}^{n-1} (\zeta^j-1)^{n-j}. \] Taking absolute values, using $|\zeta|=1$ and symmetry, one can also use the standard identity \[ \prod_{j=1}^{n-1} |1-\zeta^j| = n, \] which follows from evaluating $x^n-1=(x-1)\prod_{j=1}^{n-1}(x-\zeta^j)$ at $x=1$ and taking limits: \[ n = \lim_{x\to 1} \frac{x^n-1}{x-1} = \prod_{j=1}^{n-1} (1-\zeta^j). \] Taking absolute values yields $\prod_{j=1}^{n-1}|1-\zeta^j|=n$. A standard Vandermonde computation then gives $\prod_{i<j}|u_i-u_j|=n^{n/2}$ (equivalently, the discriminant of $x^n-1$ has absolute value $n^n$). Thus $\Delta=(\prod_{i<j}|u_i-u_j|)^2=n^n$.

For the diameter scaling: if $n$ is even, opposite vertices exist and are distance $2$ on the unit circle. If $n$ is odd, the largest angular separation is $(n-1)\pi/n=\pi-\pi/n$, giving chord length $2\sin((\pi-\pi/n)/2)=2\cos(\pi/(2n))$. Scaling distances by $\sec(\pi/(2n))$ multiplies $\Delta$ by $\sec(\pi/(2n))^{n(n-1)}$. $\square$

Explicit COUNTEREXAMPLE for $n=4$ (regular polygon is not a maximizer)

Consider the four points \[ z_1=-1,\quad z_2=1,\quad z_3=(1-\sqrt 3)+i,\quad z_4=(1-\sqrt 3)-i. \] We verify the constraint $|z_i-z_j|\le 2$ and compute $\Delta$.

Step 1: Pairwise distances.

- $|z_1-z_2|=|-2|=2$. - $|z_3-z_4|=|2i|=2$. - $|z_2-z_3| = |1-(1-\sqrt3)-i| = |\sqrt3-i|=\sqrt{3+1}=2$. Similarly $|z_2-z_4|=2$. - $|z_1-z_3| = |-1-(1-\sqrt3)-i| = |(\sqrt3-2)-i|$. Hence \[ |z_1-z_3|^2 = (\sqrt3-2)^2+1 = (7-4\sqrt3)+1 = 8-4\sqrt3 =4(2-\sqrt3), \] so $|z_1-z_3|=2\sqrt{2-\sqrt3}$. Likewise $|z_1-z_4|=2\sqrt{2-\sqrt3}$.

All distances are $\le 2$ because $2\sqrt{2-\sqrt3}<2$. Thus the constraint is satisfied and diameter is $2$.

Step 2: Product $\Delta$.

The unordered product is \[ P=\prod_{i<j}|z_i-z_j| = (2)^4\cdot\left(2\sqrt{2-\sqrt3}\right)^2 = 16\cdot 4(2-\sqrt3)=64(2-\sqrt3). \] Therefore \[ \Delta = P^2 = 4096(2-\sqrt3)^2. \] For the regular square of diameter $2$ (vertices at $\pm 1,\pm i$), the unordered product is $P_{\square}=(\sqrt2)^4\cdot 2^2=16$, so $\Delta_{\square}=256$.

To compare, note \[ P>16 \iff 64(2-\sqrt3)>16 \iff 2-\sqrt3>\tfrac14 \iff 7>4\sqrt3 \iff 49>48, \] which holds. Hence $\Delta>256$.

Thus the regular $4$-gon (square) is not a maximizer.

(FAST REALITY CHECK via computation) A short numerical optimization with constraints (not reproduced here) found essentially the same configuration. A direct numerical evaluation gives $\Delta\approx 294.0795688>256$.

5) VERIFICATION

- All pairwise distances were computed exactly in radicals and shown $\le 2$. - The strict inequality $P>16$ was reduced to $49>48$, leaving no numerical ambiguity. - Therefore this is a fully rigorous counterexample to the claim that the regular polygon always maximizes $\Delta$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. - Exact maxima for $n=2,3$: $M_2=4$ and $M_3=64$. - Exact value of $\Delta$ for regular $n$-gons scaled to diameter $2$ (Lemma 4.2). - A fully explicit counterexample for $n=4$ showing the regular polygon is not a maximizer.

(ii) First gap (crisp statement). Determine the exact value of $M_n$ (or even its sharp asymptotics) for general $n\ge 4$.

(iii) Top 3 next moves. 1. For fixed small $n$ (e.g. $n=4,5,6$), perform rigorous global optimization (or certified numerics) to conjecture exact maximizers and possibly algebraic coordinates. 2. Prove structural constraints for maximizers (e.g. diameter achieved by multiple pairs, boundary of convex hull, or equal-distance conditions from KKT/Lagrange multipliers). 3. Seek a general upper bound of the form $\Delta\le C^n n^n$ with explicit best constant $C$, and attempt matching constructions.

(iv) Minimal counterexample structure. Any exact maximizer different from the regular polygon should exploit the lack of opposite vertices (for odd $n$) or introduce "cluster + antipode" structure for even $n$ (as in the $n=4$ example), so that many distances are forced to equal the diameter $2$ while only a few become moderately small.

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.