gpt pro 5.4 attempt on Math-5897, version 2

Round 2 continuation for Erdős Problem #789

1) ROUND-2 OBJECTIVE

I pursue path (C). The most promising Round-2 move is not a full solution of the asymptotic problem, but a correction and strengthening of the upper-bound side: Round 1 proved only $h(n)\le \sqrt{7n}$, while the finite interval companion problem #874 is already solved sharply enough to force \[ h(n)\le 2\sqrt n + O(1). \] This is a strict advance over Round 1 and isolates the real remaining obstruction: the exponent gap between $1/3$ and $1/2$.

2) Round-1 FOUNDATION USED

I use the following Round 1 results without reproving them.

1. Round 1 corrected the intended meaning of the problem to the standard distinct-summand admissibility condition.
2. Round 1 proved the lower bound \[ h(n)\gg n^{1/3}. \]
3. Round 1 observed that to upper-bound the universal guarantee $h(n)$, it is enough to exhibit one specific $n$-element set whose admissible subsets are small.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a transfer from the now-solved finite interval problem #874. Deshouillers and Freiman proved that, for all sufficiently large $N$, every admissible subset $A\subset [1,N]$ satisfies \[ |A|\le 2\sqrt{N+\tfrac14}-1. \] Applying this to the specific $n$-set $\{1,2,\dots,n\}$ immediately sharpens the Round 1 upper bound for $h(n)$.

4) ATTACK PLAN (ROUND-2)

The exact Round 1 gap was this: the upper bound $h(n)\le \sqrt{7n}$ had the correct exponent but a weak constant, and it did not use the full strength of the solved interval problem.

The plan is:

1. import the Deshouillers--Freiman theorem for admissible subsets of $[1,N]$;
2. apply it to the specific set $A_0=\{1,2,\dots,n\}$;
3. conclude that the universal Erdős quantity $h(n)$ satisfies \[ h(n)\le 2\sqrt{n+\tfrac14}-1 \] for all sufficiently large $n$.

This overcomes the Round 1 obstacle because the interval extremal problem has already been settled.

5) WORK (ROUND-2)

Recall the corrected formulation from Round 1: a finite set $B\subset \mathbf{Z}$ is admissible if, whenever \[ a_1+\cdots+a_r=b_1+\cdots+b_s \] with all summands distinct on each side and drawn from $B$, one must have $r=s$.

Let $h(n)$ denote the largest integer such that every $n$-element set $A\subset \mathbf{Z}$ contains an admissible subset of size at least $h(n)$.

External theorem used. Deshouillers and Freiman proved the following sharpened form of the finite interval problem: there exists $N_0$ such that for every integer $N\ge N_0$ and every admissible set $C\subset [1,N]$, \[ |C|\le 2\sqrt{N+\tfrac14}-1. \]

Claim 789.1. For all sufficiently large $n$, \[ h(n)\le 2\sqrt{n+\tfrac14}-1. \] In particular, \[ h(n)\le 2\sqrt n + O(1). \]

Proof. Take the particular $n$-element set \[ A_0:=\{1,2,\dots,n\}. \] If $B\subseteq A_0$ is admissible, then $B$ is an admissible subset of $[1,n]$. Hence, for all sufficiently large $n$, the Deshouillers--Freiman theorem gives \[ |B|\le 2\sqrt{n+\tfrac14}-1. \] Therefore no admissible subset of $A_0$ can be larger than $2\sqrt{n+\tfrac14}-1$. Since $h(n)$ is the size guaranteed inside every $n$-element set, it cannot exceed the maximum admissible subset size inside this specific $A_0$. Thus \[ h(n)\le 2\sqrt{n+\tfrac14}-1 \] for all sufficiently large $n$, proving the claim. $\Box$

Corollary 789.2. Combining Claim 789.1 with the Round 1 lower bound yields \[ n^{1/3}\ll h(n)\le 2\sqrt n + O(1). \] If one also imports the Choi improvement recorded on the problem page, the lower side becomes \[ (n\log n)^{1/3}\ll h(n)\le 2\sqrt n + O(1). \]

What is still missing. The exponent remains unknown. Round 2 improves the upper constant sharply, but the real open question is still whether the true order is closer to $n^{1/3}$, $n^{1/2}$, or something in between.

6) ADVERSARIAL VERIFICATION

There are only two possible weak points.

1. Compatibility of definitions. Round 1 corrected the problem to the standard distinct-summand admissibility notion. This is exactly the notion used in the finite interval problem #874, so the transfer is legitimate.
2. Universal-versus-specific extremal parameter. The proof uses only one specific $n$-element set $A_0=\{1,\dots,n\}$. This is correct because $h(n)$ is a universal guarantee across all $n$-element sets; any one hard example bounds it from above.

No hidden quantifier issue remains: the Deshouillers--Freiman theorem applies to all sufficiently large $n$, which is exactly the regime relevant for estimating $h(n)$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 strictly improves the Round 1 upper bound from $h(n)\le \sqrt{7n}$ to \[ h(n)\le 2\sqrt n + O(1), \] while retaining the Round 1 lower bound $h(n)\gg n^{1/3}$. The corrected current bracket is therefore \[ (n\log n)^{1/3}\ll h(n)\le 2\sqrt n + O(1), \] but the true asymptotic order of $h(n)$ remains open.

8) COMPLETION ESTIMATE

COMPLETION: 46%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #789, problem page.
2. T. F. Bloom, Erdős Problem #874, problem page.
3. J.-M. Deshouillers and G. A. Freiman, On an additive problem of Erdős and Straus, 2, Astérisque 258 $1999$, 141--148.

10) OUTPUT FORMAT NOTE

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