GPT 6 Astra Ultra attempt on Math-6260
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Contents
Erdős Problem 1206 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
Is there a constant $c>0$ such that for every sufficiently large $N$, the first $N$ positive cubes contain a Sidon subset of at least $cN$ elements? Is there an infinite set $A\subseteq\mathbb N$ of positive density for which $\{a^3:a\in A\}$ is Sidon? Here Sidon means uniqueness of every two-term sum up to exchanging its summands.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 page leaves the convention for positive density implicit. Its remarks also raise higher-power analogues; these are not treated as already solved.
ATTACK PLAN AND WORK
A greedy argument works for every $N$-element set $X$ of distinct real numbers, and hence for the first $N$ cubes. Suppose a Sidon subset $S$ has size $m$. A new element $x$ can violate the property only through a collision $x+a=b+c$ with $a,b,c\in S$, or $2x=a+b$ with $a,b\in S$, or by already belonging to $S$. Equalities $x+a=x+b$ force $a=b$ and cause no new violation. Thus at most $m^3+m^2+m$ real values are forbidden.If $S$ is maximal within $X$, all of $X$ must be forbidden, giving \[N\le m^3+m^2+m\le3m^3\quad(m\ge1). \] Therefore every such $X$ contains a Sidon subset with at least $(N/3)^{1/3}$ elements. This proof handles repeated summands through the $2x$ condition.
An infinite example is $A=\{2^j:j\ge0\}$: its cubes are $8^j$, and uniqueness of their two-term sums follows from base-eight digits, including a single digit two when a summand repeats. But $A$ has density zero. Neither this infinite construction nor the finite cube-root bound supplies the positive proportion or positive-density conclusion asked for.