Reused writeup on Math-5702 (gpt pro 5.2)

Erdős Problem 510 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/510

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_5.2/510.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/510.tex.

FORMAL RESTATEMENT

For each finite set $A\subset\mathbb{Z}$ with $|A|=N\ge 1$, define \[ S_A(\theta):=\sum_{n\in A}\cos(n\theta),\qquad \theta\in\mathbb{R}. \] Question: Does there exist an absolute constant $c>0$ (independent of $A$ and $N$) such that for every such $A$ there exists some real $\theta$ with \[ S_A(\theta) < -c\,N^{1/2}? \]

Edge cases: If $0\in A$, then $S_A(\theta)$ has a constant $+1$ contribution. The statement asks for a negative value anyway.

QUICK LITERATURE/CONTEXT CHECK

The problem statement records: Ruzsa proved a bound of size $-\exp(O(\sqrt{\log N}))$; and more recent work (Bedert; Jin--Milojević--Tomon--Zhang) gives polynomially large negative values, currently $-cN^{1/7}$. I do not re-derive those results here.

ATTACK PLAN

• Proof-track idea: treat $S_A$ as a real part of an exponential sum $\sum_{n\in A} e^{in\theta}$; use $L^2$ identities and higher-moment bounds to force a substantially negative real part somewhere.
• Disproof-track idea: attempt to build $A$ with "random-like" residue distribution so that $S_A(\theta)$ stays close to $0$ for most $\theta$; then try to show its minimum cannot be as negative as $-c\sqrt{N}$.

The full $-c\sqrt{N}$ bound remains out of reach here.

WORK

Fast reality check (small $N$; brute force on a grid).

I brute-forced small sets $A\subset\{1,2,\dots,8\}$ and numerically approximated $\min_{\theta} S_A(\theta)$ by sampling $\theta$ on an $8192$-point uniform grid on $[0,2\pi)$. For each $N$ I then took the worst case (largest) of these approximate minima over all $A$ of size $N$.

If $0$ is allowed (universe $\{0,1,\dots,8\}$), the worst-case approximate minima shift up by $+1$ in the obvious way for sets containing $0$; for example for $N=4$ one worst-case set was $(0,2,4,6)$ with grid-minimum about $-0.315556$.

These computations are only sanity checks (grid-based, not exact global minima).

Lemma 1 (mean value). For any finite $A\subset\mathbb{Z}$, \[ \frac{1}{2\pi}\int_0^{2\pi} S_A(\theta)\,d\theta = \begin{cases} 1,& 0\in A,\\ 0,& 0\notin A. \end{cases} \]

Proof. For $n\neq 0$, \[ \frac{1}{2\pi}\int_0^{2\pi} \cos(n\theta)\,d\theta = 0 \] because $\cos(n\theta)$ has integral $0$ over a full period. For $n=0$, $\cos(0\cdot\theta)=1$ and the integral equals $1$. Summing over $n\in A$ gives the claim. $\square$

Lemma 2 (a universal constant negative value when $0\notin A$). Assume $A\subset\mathbb{Z}$ is finite, $|A|=N$, and $0\notin A$. Then there exists $\theta\in\mathbb{R}$ such that \[ S_A(\theta)\le -\tfrac12. \]

Proof. Define $g(\theta)=S_A(\theta)$ on $[0,2\pi]$. By Lemma 1, $\int_0^{2\pi} g(\theta)\,d\theta=0$. Let $M:=\max_{\theta\in[0,2\pi]} g(\theta)$ and $m:=\min_{\theta\in[0,2\pi]} g(\theta)$. We will use the variance bound for bounded real-valued functions with mean $0$: \begin{equation}\label{wsref-1-65-71-3a-76-61-72-72-61-6e-67-65} \frac{1}{2\pi}\int_0^{2\pi} g(\theta)^2\,d\theta \le (-m)\,M. \end{equation} To justify (eq:varrange): view $g(\Theta)$ as a random variable on the probability space $([0,2\pi], d\theta/(2\pi))$ with mean $0$ and range $[m,M]$. For any such random variable $X$ with $\mathbb{E}X=0$ and $m\le X\le M$, one has $\mathrm{Var}(X)=\mathbb{E}(X^2)\le (-m)M$; this follows because the maximal variance at fixed endpoints and mean is achieved by a two-point distribution supported on $\{m,M\}$ (a standard convexity argument), and in that extremal case $\mathbb{E}(X^2)=(-m)M$.

Next, compute a lower bound on the $L^2$ norm. Since $0\notin A$, \[ \int_0^{2\pi} \cos(n\theta)\cos(m\theta)\,d\theta = 0\quad\text{for } |n|\neq |m|, \] and \[ \int_0^{2\pi} \cos(n\theta)^2\,d\theta = \pi\quad\text{for }n\neq 0. \] However, if both $n$ and $-n$ lie in $A$, then the corresponding cosine terms are identical and contribute a factor of $2$ in $g$. Writing \[ A_+:=\{k\ge 1: k\in A\text{ or }-k\in A\},\qquad m_k:=|\{n\in A: |n|=k\}|\in\{1,2\}, \] we have $g(\theta)=\sum_{k\in A_+} m_k \cos(k\theta)$ and hence by orthogonality \[ \int_0^{2\pi} g(\theta)^2\,d\theta = \pi\sum_{k\in A_+} m_k^2 \ge \pi\sum_{k\in A_+} m_k = \pi N. \] Therefore \[ \frac{1}{2\pi}\int_0^{2\pi} g(\theta)^2\,d\theta \ge \frac{N}{2}. \] On the other hand, evaluating at $\theta=0$ gives $g(0)=\sum_{n\in A}1=N$, so $M\ge N$. Combining with (eq:varrange) gives \[ \frac{N}{2}\le (-m)\,M \le (-m)\,N, \] hence $-m\ge 1/2$, i.e. $m\le -1/2$. This means there exists $\theta$ with $g(\theta)\le -1/2$. $\square$

Lemma 3 (a large negative value for structured sets). If $A$ consists entirely of odd integers (so $|A|=N$ and $n\equiv 1\pmod 2$ for all $n\in A$), then taking $\theta=\pi$ gives \[ S_A(\pi)=\sum_{n\in A}\cos(n\pi)=\sum_{n\in A}(-1)=-N. \]

Proof. For an odd integer $n$, $n\pi$ is an odd multiple of $\pi$, so $\cos(n\pi)=\cos(\pi)= -1$. Summing yields $-N$. $\square$

VERIFICATION

• Lemma 2 uses an inequality $\mathrm{Var}(X)\le (-m)M$ for mean-zero random variables supported in $[m,M]$. I verified it separately: among distributions on $[m,M]$ with fixed mean, the second moment is maximized by putting all mass at endpoints (convexity of $x\mapsto x^2$), giving $\mathbb{E}(X^2)\le (-m)M$ when mean is $0$.
• Orthogonality step: I reduced the sum to distinct absolute values to avoid double-counting the identical functions $\cos(n\theta)=\cos(-n\theta)$.
• The obtained bound $-1/2$ is far from $-c\sqrt{N}$; it is consistent with the brute-force numerics for small $N$.

FINAL

UNRESOLVED

(i) Strongest proved partial result: If $0\notin A$ there exists $\theta$ with $\sum_{n\in A}\cos(n\theta)\le -\tfrac12$ (Lemma 2). Also, for highly structured sets such as all-odd $A$, one can achieve $-N$ (Lemma 3).

(ii) First gap (crisp): Prove or disprove the existence of an absolute $c>0$ such that for every $A$ with $|A|=N$ there exists $\theta$ with $S_A(\theta)<-c\sqrt{N}$.

(iii) Top 3 next moves:

1. Use higher moments: compute/estimate $\int |\sum_{n\in A} e^{in\theta}|^{2k}$ and relate it to additive energy of $A$; then try to convert a large $L^{2k}$ norm into a large negative real part at some $\theta$.
2. Try explicit "hard" families: difference sets $A=B-B$ with $B$ Sidon (as mentioned in the problem statement) and numerically estimate the true minimum of $S_A$ for moderate sizes to see scaling.
3. Attempt a discretisation-to-continuum reduction: approximate the minimum of $S_A$ by evaluating on a net of $\theta$ values with certified error bounds (e.g., via Lipschitz control using $\max |S_A'(\theta)|$), enabling rigorous small-$N$ extremal computations.

(iv) Minimal counterexample structure (if the $\sqrt{N}$ bound is false): would likely be a set $A$ with additive structure forcing the exponential sum $\sum_{n\in A} e^{in\theta}$ to have small real part oscillations for all $\theta$; candidate families include highly additive sets such as difference sets of Sidon sets or sets with small additive energy.

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.