GPT 6 Astra Ultra attempt on Math-726
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Erdős Problem #241: independent continuation
Date: 2026-09-23. Research attempt; no claim of a new published result.1) TARGET AND SOURCE AUDIT
If $f(N)$ is the largest cardinality of a $B_3$-set in $[1,N]$, is $f(N)\sim N^{1/3}$?All supplied versions considered: 241.tex. The source supplies the known asymptotically sharp lower bound and quotes a stronger literature upper constant. I give an independent elementary upper-bound audit using mixed sums; this does not improve that quoted literature constant.
2) WORK
Let $A\subset[1,N]$ be $B_3$ and $m=|A|$. Consider all expressions $a+b-c$ with $a\le b$ in $A$ and $c\in A\setminus\{a,b\}$. They are pairwise distinct. Indeed an equality gives \[ a+b+f=d+e+c. \] Triple-sum uniqueness gives equality of the multisets $\{a,b,f\}$ and $\{d,e,c\}$. Since $c$ is neither $a$ nor $b$, it must equal $f$; cancellation then gives $\{a,b\}=\{d,e\}$. There are $m\binom m2=m^2(m-1)/2$ such expressions: fix $c$, then choose a two-element multiset from the remaining $m-1$ elements. They all lie in $[2-N,2N-1]$.Moreover none belongs to $A$. If $a+b-c=t\in A$, then $a+b=c+t$, and $B_3$ implies $B_2$ by adding any fixed member of $A$ to both sides. Thus $\{a,b\}=\{c,t\}$, impossible. Consequently \[ \frac{m^2(m-1)}2+m\le3N-2, \] which implies $m\le(6N)^{1/3}+O(1)$. This improves the most naive triple-counting constant $18^{1/3}$, but is weaker than the literature constant $(7/2)^{1/3}$ already quoted in the source.