GPT 6 Astra Ultra attempt on Math-6239

Erdős Problem 1185: a checked attempt

GPT_6_Astra_Ultra

24 September 2026

Statement and construction

Given $\delta>0$ and $k\ge3$, must there be an $m=m(\delta,k)$ such that, for all sufficiently large $N$, every $A,B\subseteq\{1,\ldots,N\}$ with $|A|\ge\delta N$ and $|B|\ge m$ has a nonconstant $k$-term arithmetic progression in $A$ with difference in $B-B$?

The assertion is false already for $k=3$, and hence for every $k\ge3$. We construct an infinite set $B$ and a set $A$ of natural density $1/50$ such that no nonconstant three-term progression in $A$ has difference in $B-B$. Passing to finite intervals then disproves the exact quantifiers in the question.

The known negative result is attributed to Furstenberg's Recurrence in ergodic theory and combinatorial number theory at https://www.erdosproblems.com/1185. The following quadratic-phase construction supplies a direct proof. Its only distribution input, quadratic equidistribution, is justified below.

The elementary distribution fact

Fix an irrational real $\theta$, for example $\sqrt2$. The pairs \[ (\{n\theta\},\{n^2\theta\})\quad(n\ge1) \] are equidistributed in the unit square. By the Weyl criterion it suffices to show that, for every nonzero integer pair $(h_1,h_2)$, \[ \frac1N\sum_{n\le N}e^{2\pi i\theta(h_1n+h_2n^2)}\longrightarrow0. \tag{1} \] If $h_2=0$, this is a bounded geometric sum divided by $N$. Otherwise, for every fixed positive shift $s$, the correlation of the summands at $n+s$ and $n$ is a linear exponential sum with irrational frequency $2h_2s\theta$, and its normalized limit is zero. The van der Corput inequality, obtained by averaging $H$ shifted sums and applying Cauchy--Schwarz, bounds the limsup of the squared absolute value in $1$ by $O(1/H)$. Let $H\to\infty$. This proves $1$, and the one-coordinate sequence $\{n^2\theta\}$ is equidistributed as well.

Choose an infinite set with controlled squared differences

Construct $b_1<b_2<\cdots$ recursively so that \[ \left|\{\theta b_i^2\}-\frac18\right|<\frac1{1000}, \qquad 0<\{\theta b_i\}<\frac1{100\max\{1,b_{i-1}\}}. \tag{2} \] For $i=1$ put $b_0=1$. At each step the required pair lies in a fixed nonempty open rectangle in the unit square. Equidistribution supplies arbitrarily large choices, so the recursion continues indefinitely.

For $i>j$, write $u_i=\{\theta b_i^2\}$ and $v_i=\{\theta b_i\}$. Modulo an integer, \[ \theta(b_i-b_j)^2\equiv u_i+u_j-2b_jv_i. \] The representative on the right lies strictly between \[ \frac14-\frac2{1000}-\frac2{100}=0.228 \quad\text{and}\quad \frac14+\frac2{1000}=0.252. \tag{3} \] In particular, for every nonzero $d\in B-B$, where $B=\{b_i:i\ge1\}$, \[ \|2\theta d^2\|_{\mathbb R/\mathbb Z}>0.45. \tag{4} \] The sign of $d$ is irrelevant because it is squared.

A positive-density set excludes those differences

Let \[ A=\{n\ge1:\|\theta n^2\|_{\mathbb R/\mathbb Z}<1/100\}. \] By $1$, its natural density is $2/100=1/50$. If $a,a+d,a+2d$ all belonged to $A$, the second-difference identity \[ \theta(a+2d)^2-2\theta(a+d)^2+\theta a^2=2\theta d^2 \] and the triangle inequality for distance to an integer would give \[ \|2\theta d^2\|_{\mathbb R/\mathbb Z}<\frac4{100}. \] This contradicts $4$ whenever $d\in(B-B)\setminus\{0\}$.

Return to the finite formulation

Fix $\delta=1/100$ and let $m$ be any positive integer. For every sufficiently large $N$, the set $A_N=A\cap[1,N]$ has size at least $\delta N$, while $B_N=\{b_1,\ldots,b_m\}$ lies in $[1,N]$ and has size $m$. The preceding argument excludes a three-term progression in $A_N$ with nonzero difference in $B_N-B_N$. It also excludes any longer such progression by taking its first three terms. Thus for this single fixed $\delta$ no choice of $m$ works, even when $N$ is allowed to be arbitrarily large.

Assessment

The density, infinite recursion, second-difference obstruction, and finite-quantifier passage are all established. The argument uses standard Weyl and van der Corput criteria, with the required exponential-sum cancellation supplied explicitly.

Completion rate estimate: 100%.