GPT 6 Astra Ultra attempt on Math-6246
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Erdős Problem 1192 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
For each integer $r\geq2$, can one choose $A\subseteq\mathbb N$ and a constant $C$ with the following properties? If \[f_r(n)=\#\{(a_1,\ldots,a_r)\in A^r:a_1+\cdots+a_r=n\},\] then $f_r(n)>0$ for every sufficiently large $n$, and \[\sum_{n\leq x}f_r(n)^2\leq Cx\] for every $x\geq1$.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.Representations are written as ordered tuples. The existence question with fixed r is unchanged by using unordered representations, up to a factor at most r! in the count. The official page records a positive answer for r=2.
ATTACK PLAN AND WORK
Suppose such a set exists and write $S_r(x)=\sum_{n\le x}f_r(n)$. Eventual coverage and the proposed square-mean bound imply \[x-O(1)\le S_r(x)\le\sqrt{\lfloor x\rfloor\sum_{n\le x}f_r(n)^2} \le\sqrt C\,x.\tag{1}\] There are at most $A(x)^r$ contributing ordered tuples, so $1$ gives $A(x)\ge(x-O(1))^{1/r}$. Conversely every tuple from $A\cap[1,y]$ has sum at most $ry$, hence \[A(y)^r\le S_r(ry)\le r\sqrt C\,y.\tag{2}\] Any solution must therefore live in the critical density regime \[A(y)\asymp y^{1/r}.\tag{3}\] The proof treats ordered representations consistently; no factor of $r!$ is lost in either tuple count.The same energy bound controls large representation counts: \[|\{n\le x:f_r(n)\ge T\}|\le Cx/T^2.\tag{4}\] It does not prohibit rare large counts, and replacing it by a uniform bound on every $f_r(n)$ would change the problem. At the other extreme, proving $3$ alone does not prevent infinitely many uncovered integers.