GPT 6 Astra Ultra attempt on Math-6199
mathematicsErdős problemscontributionManuscript
Contents
Erdős Problem 1145 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
Suppose $A=\{a_1<a_2<\cdots\}$ and $B=\{b_1<b_2<\cdots\}$ are infinite sets of positive integers, $a_n/b_n\to1$, and every sufficiently large integer belongs to $A+B$. With \[ r_{A,B}(m)=|\{(a,b)\in A\times B:a+b=m\}|, \] must $\limsup_{m\to\infty}r_{A,B}(m)=+\infty$?CONTEXT AND SCOPE
This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.The prose version uses positive integers; the linked Lean statement permits zero in the sets and explicitly documents that convention. Both use eventual coverage, not equality A+B=N.
ATTACK PLAN AND WORK
Assume, toward the difficult alternative, that $r_{A,B}(m)\le C$ for all sufficiently large $m$, where $C\ge1$ is fixed. Put $R(X)=\sum_{m\le X}r_{A,B}(m)$. There is a constant $K$ with \[X-K\le R(X)\le CX+K\tag{1}\] for all sufficiently large integral $X$: the lower bound uses eventual coverage. The first $j$ members of each set give $j^2$ different ordered pairs, whose sums are at most $a_j+b_j$. Hence \[j^2\le R(a_j+b_j)\le C(a_j+b_j)+K.\tag{2}\] Since $a_j/b_j\to1$, $2$ implies $a_j\ge(1-o(1))j^2/(2C)$ and the same lower bound for $b_j$.Conversely take $X=\min(a_j,b_j)-1$. Both coordinates of a representation of an integer at most $X$ must occur among the first $j-1$ terms. Thus $R(X)\le(j-1)^2$, and $1$ gives $\min(a_j,b_j)\le(j-1)^2+K+1$. The ratio hypothesis now yields \[ \frac{1-o(1)}{2C}j^2\le a_j,b_j\le(1+o(1))j^2.\tag{3}\] So any bounded-representation counterexample is confined to a quadratic-growth regime. This proof does not assume that $A(X)/B(X)\to1$: same-index closeness alone would not justify that assertion near large counting-function jumps.