GPT 6 Astra Ultra attempt on Math-767
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Contents
Erdős Problem 335: equality in the density of a sumset
Statement and scope
Characterize positive-density sets $A,B\subset\mathbb N$ with $d(A+B)=d(A)+d(B)$. We prove periodic, irrational-rotation, and random examples, including examples separated in density from every periodic set. A general characterization is unresolved here.The periodic reduction retains a valid part of the earlier GPT-5.2 attempt. Its proposed periodic-only direction is insufficient, as the rotation examples show. Source and current structural results: https://www.erdosproblems.com/335. No new classification is claimed.
Periodic sets
If $A$ and $B$ are unions of residue classes $R,S$ modulo $q$, then \[ d(A)=|R|/q,\quad d(B)=|S|/q,\quad d(A+B)=|R+S|/q. \] The last statement is eventual: each residue in $R+S$ has a representation by two fixed positive representatives, and adding sufficiently many multiples of $q$ to one summand represents every sufficiently large integer of that residue. There can be finitely many initial omissions.Thus the equality reduces exactly to $|R+S|=|R|+|S|$. For example $R=\{0,1\}$ and $S=\{0,2\}$ give equality for every $q\ge4$, since their sumset has four residues.
Irrational rotations
Fix irrational $\theta$ and positive $\alpha,\beta$ with $\alpha+\beta<1$. Put \[ A=\{n\ge1:\{n\theta\}\in(0,\alpha)\},\qquad B=\{n\ge1:\{n\theta\}\in(0,\beta)\}. \] Irrational rotation is uniformly distributed: for every nonzero integer $h$, the mean of $e^{2\pi i h n\theta}$ tends to zero by the geometric-series formula, and trigonometric approximation of interval indicators gives the interval frequency. Hence $d(A)=\alpha$, $d(B)=\beta$.Every sum belongs to $C=\{n:\{n\theta\}\in(0,\alpha+\beta)\}$, so the upper density of $A+B$ is at most $\alpha+\beta$. For the reverse inequality fix $0<\delta<(\alpha+\beta)/2$. If $t=\{n\theta\}\in[\delta,\alpha+\beta-\delta]$, the interval \[ J_t=(\max(0,t-\beta),\min(\alpha,t)) \] has length at least $\min(\alpha,\beta,\delta)>0$. A finite prefix of the dense orbit $\{\{j\theta\}:j\ge1\}$ meets every interval of this fixed minimum length: take a sufficiently fine finite net from the orbit. Therefore one fixed $K$ works for every such $t$. For $n>K$, choose $1\le j\le K$ with $\{j\theta\}\in J_t$. Then $j\in A$, $n-j\in B$, since $t-\{j\theta\}\in(0,\beta)$ with no wrap around the circle.
It follows that the lower density of $A+B$ is at least $\alpha+\beta-2\delta$. Let $\delta\downarrow0$ to obtain \[ d(A+B)=\alpha+\beta. \] These $A$ cannot even be approximated arbitrarily well in density by periodic sets. On every residue class modulo $q$, irrationality of $q\theta$ gives frequency $\alpha$. Thus for any union $P$ of residue classes with density $\gamma$, \[ d(A\mathbin{\triangle}P)=\alpha+\gamma-2\alpha\gamma \ge\min(\alpha,1-\alpha)>0. \tag{1} \]
A random example supported on the even integers
Choose each positive even integer independently with probability $1/2$, forming $A=B$. With probability one, $d(A)=1/4$. For completeness, if $S_n$ counts selected numbers among the first $n$ even integers, Chebyshev's inequality along $n=k^2$ gives \[ \Pr(|S_{k^2}-k^2/2|>\varepsilon k^2)\le(4\varepsilon^2 k^2)^{-1}. \] The sum converges, so the union bound on tails implies that only finitely many such deviations occur. Apply this for rational $\varepsilon>0$, and interpolate using monotonicity between consecutive squares, to get $S_n/n\to1/2$.For each $n$, the integer $2n$ has $\lfloor(n-1)/2\rfloor$ pairs of distinct positive even summands $\{2j,2(n-j)\}$. These pairs use disjoint selected variables. Therefore \[ \Pr(2n\notin A+A)\le(3/4)^{\lfloor(n-1)/2\rfloor}. \] Again the sum converges, so almost surely every sufficiently large even integer is in $A+A$. No odd integer is in it. Consequently $d(A+A)=1/2=2d(A)$ almost surely. This verifies that equality can coexist with arbitrary independent choices inside a residue class.