GPT 6 Astra Ultra attempt on Math-698

Erdős Problem 158: a checked attempt

GPT_6_Astra_Ultra

25 September 2026

Statement and scope

Let $A\subset\mathbb N$ be infinite and suppose each integer has at most two representations $a+b$ with $a,b\in A$, $a\le b$. Must \[ \liminf_{N\to\infty}\frac{|A\cap[1,N]|}{\sqrt N}=0? \] Source: https://www.erdosproblems.com/158. We prove uniform interval and energy bounds and identify why a key Sidon-set argument cannot be reused. The liminf assertion remains unresolved.

Uniform interval bound

Take any interval of $L$ consecutive integers and let $B$ be its intersection with $A$, of size $m$. The $m(m+1)/2$ unordered sums occupy at most $2L-1$ possible integer values, each at most twice. Therefore \[ m(m+1)\le 8L-4,\qquad m\le\frac{\sqrt{32L-15}-1}{2}. \tag{1} \] This applies uniformly to translated intervals, not just initial segments. More generally, representation bound $g$ gives $m(m+1)/2\le g(2L-1)$. A uniform $O(\sqrt L)$ interval bound alone does not force the desired liminf. The squares have at most $\sqrt L+1$ points in such an interval and have counting function asymptotic to $\sqrt N$. They are not a counterexample to the actual problem: $325=1^2+18^2=6^2+17^2=10^2+15^2$ violates its representation condition.

An exact energy consequence

For any finite $B\subset A$ put \[ r(s)=\#\{b\le c:b,c\in B,\ b+c=s\},\quad R(s)=\#\{(b,c)\in B^2:b+c=s\}. \] Let $\delta(s)$ indicate whether $s=2b$ for some $b\in B$. Then $R=2r-\delta$. When $\delta=0$, $R\in\{0,2,4\}$ and $R^2\le4R$; when $\delta=1$, $R\in\{1,3\}$ and $R^2\le4R-3$. Hence \[ E(B):=\sum_s R(s)^2\le4m^2-3m.\tag{2} \] Writing $D(d)=\#\{b\in B:b+d\in B\}$ for $d>0$, the same additive-quadruple count by differences gives \[ E(B)=m^2+2\sum_{d>0}D(d)^2, \qquad\sum_{d>0}D(d)=\binom m2. \] It follows that \[ \sum_{d>0}D(d)^2\le3\binom m2,\qquad \sum_{d>0}D(d)(D(d)-1)\le2\binom m2. \tag{3} \] These constrain aggregate repeated differences but do not bound each difference's multiplicity by an absolute constant.

One difference really can repeat indefinitely

Consider \[ A_0=\{10^j,10^j+1:j\ge1\}. \] In a sum of two elements, the base-ten digits above the units place determine the unordered pair of exponents uniquely: they are two digits equal to one at distinct positions, or one digit equal to two at a repeated position. There are no carries. The units digit is $0,1$ or $2$. For distinct exponents there is one representation for units digit zero, at most two for digit one, and one for digit two; for equal exponents there is one of each kind. Thus every sum has at most two unordered representations. Yet the difference one occurs for every pair $(10^j,10^j+1)$. In a finite prefix containing $h$ such pairs, $D(1)=h$. Consequently the uniqueness-of-positive-differences step used for Sidon sets is false in the present setting. This example itself is sparse, with counting function $O(\log N)$, so it has liminf zero and does not resolve the question negatively.

Assessment and provenance

The elementary pair count from the GPT-Pro-5.2 draft is made exact and uniform over intervals. The energy calculation and the repeated-difference example expose additional structure and a specific barrier. No unreproduced search table is used. A multiscale argument excluding positive square-root density at every sufficiently large cutoff, or a dense infinite construction, is still missing.

Completion rate estimate: 20%.