gpt pro 5.4 attempt on Math-923

Round-2 continuation for Erdős Problems 741, 749, 750

This is a genuine continuation of the previous round. I only use Round-1 material when explicitly indicated below, and I only re-prove statements when they are strengthened.

Conventions for this continuation: $\overline d(\cdot)$ and $\underline d(\cdot)$ denote upper and lower asymptotic density in $\mathbb N$, and when discussing Problem 750 it is enough to work with induced subgraphs, since deleting edges can only increase the independence number.

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Problem 749

1) ROUND-2 OBJECTIVE

Path (C): prove a rigorous asymptotic obstruction that is substantially sharper than the Round-1 bound $A(x)\ll \sqrt{x}$.

The goal of this round is to show that any hypothetical solution must in fact be quadratically parameterized: if $A=\{a_1<a_2<\cdots\}$, then necessarily \[ a_n=\Theta(n^2), \] with explicit constants depending only on $\epsilon$ and the representation bound.

2) Round-1 FOUNDATION USED

I use exactly two Round-1 facts.

• Round-1 proved the coarse necessary bound $A(x)\ll \sqrt{x}$ from bounded representation.
• Round-1 observed that the real obstacle is compatibility across scales, not finite-model feasibility.

In this round I sharpen the first point from a coarse counting estimate in $x$ to a precise estimate in terms of the ordered sequence $a_n$.

3) NEW INSIGHT / TOOL (ROUND-2)

The key observation is to evaluate the density condition at the special cutoff \[ N=a_{n+1}-1. \] Since every sum $\le N$ uses only the first $n$ elements of $A$, one gets a direct upper bound on $a_{n+1}$ from the maximum possible number of distinct sums of an $n$-element set. Combined with the bounded-representation lower bound, this traps $a_n$ between two quadratic curves.

4) ATTACK PLAN (ROUND-2)

The exact gap left by Round-1 was:

Can we say anything more precise than $A(x)\asymp \sqrt{x}$ about the shape of a candidate set?

I will prove two complementary bounds.

Theorem 749.1. Let $A=\{a_1<a_2<\cdots\}\subseteq\mathbb N$.

1. If $\underline d(A+A)\ge 1-\epsilon$, then for every $\eta>0$ there exists $n_0(\eta)$ such that \[ a_{n+1}\le \frac{n(n+1)}{2(1-\epsilon-\eta)}+1 \qquad (n\ge n_0(\eta)). \] In particular, \[ \limsup_{n\to\infty}\frac{2a_n}{n^2}\le \frac{1}{1-\epsilon}. \]

2. If moreover $(\mathbf 1_A\ast \mathbf 1_A)(m)\le g$ for all $m$, then for every $n$, \[ a_n\ge \frac{n^2}{2g}. \] In particular, \[ \liminf_{n\to\infty}\frac{2a_n}{n^2}\ge \frac1g. \]

Hence any candidate for Problem 749 must satisfy \[ \frac1g\le \liminf_{n\to\infty}\frac{2a_n}{n^2} \le \limsup_{n\to\infty}\frac{2a_n}{n^2} \le \frac1{1-\epsilon}, \] and therefore \[ a_n=\Theta(n^2), \] while the counting function \[ A(x):=|A\cap[1,x]| \] satisfies \[ A(x)=\Theta(\sqrt x). \]

5) WORK (ROUND-2)

Proof of Theorem 749.1$1$. Fix $\eta>0$. Since $\underline d(A+A)\ge 1-\epsilon$, there exists $N_0(\eta)$ such that for all $N\ge N_0(\eta)$, \[ |(A+A)\cap[1,N]|\ge (1-\epsilon-\eta)N. \] Now take $N=a_{n+1}-1$. Any representation \[ m=a_i+a_j\le a_{n+1}-1 \] must use only indices $i,j\le n$, because $a_{n+1}+a_1>a_{n+1}-1$. Therefore \[ (A+A)\cap[1,a_{n+1}-1] \subseteq \{a_i+a_j:1\le i\le j\le n\}. \] The set on the right has cardinality at most the number of unordered pairs from an $n$-element set, namely $n(n+1)/2$. Hence, for all sufficiently large $n$, \[ (1-\epsilon-\eta)(a_{n+1}-1) \le \left|(A+A)\cap[1,a_{n+1}-1]\right| \le \frac{n(n+1)}{2}. \] So \[ a_{n+1}\le \frac{n(n+1)}{2(1-\epsilon-\eta)}+1. \] Dividing by $n^2/2$ and letting $n\to\infty$ yields \[ \limsup_{n\to\infty}\frac{2a_n}{n^2}\le \frac{1}{1-\epsilon-\eta}. \] Since $\eta>0$ is arbitrary, \[ \limsup_{n\to\infty}\frac{2a_n}{n^2}\le \frac{1}{1-\epsilon}. \] This proves $1$. $\square$

Proof of Theorem 749.1$2$. Assume \[ (\mathbf 1_A\ast \mathbf 1_A)(m)\le g \qquad\text{for all }m. \] Consider the $n^2$ ordered pairs $(a_i,a_j)$ with $1\le i,j\le n$. Each such pair produces a sum in $[2,2a_n]$. Since each integer is assumed to have at most $g$ ordered representations, the number of distinct sums produced by these $n^2$ pairs is at least $n^2/g$. But there are at most $2a_n-1$ positive integers in $[2,2a_n]$, so certainly \[ 2a_n\ge \frac{n^2}{g}. \] Equivalently, \[ a_n\ge \frac{n^2}{2g}. \] Dividing by $n^2/2$ gives \[ \frac{2a_n}{n^2}\ge \frac1g \] for every $n$, hence \[ \liminf_{n\to\infty}\frac{2a_n}{n^2}\ge \frac1g. \] This proves $2$. $\square$

Combining $1$ and $2$ gives the stated quadratic window.

6) ADVERSARIAL VERIFICATION

• Why $N=a_{n+1}-1$? This is essential. At that cutoff, no sum involving $a_{n+1}$ can appear, so only the first $n$ elements matter.
• Ordered vs. unordered representations. The upper bound in part $1$ counts distinct sums, so unordered pairs are correct. The lower bound in part $2$ uses the hypothesis on $(\mathbf 1_A\ast \mathbf 1_A)(m)$, which counts ordered representations, so $n^2$ ordered pairs are correct there.
• Lower density. Part $1$ uses only the definition of $\underline d(A+A)$ with an arbitrary $\eta>0$; no unjustified subsequence argument appears.
• No hidden assumption on $A(0)$. The argument uses only positivity of the elements and monotonicity of the sequence $(a_n)$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round-2 proves a sharp structural obstruction:

Any solution of Problem 749 must come from a sequence $A=\{a_n\}$ with genuinely quadratic growth $a_n=\Theta(n^2)$, and the quadratic constant is trapped between $1/(2g)$ and $1/(2(1-\epsilon))$.

So the open problem is not merely about finding a sparse set of size $\asymp \sqrt{x}$; it is about building an almost extremal quadratic generalized Sidon-type sequence whose initial segments remain globally compatible.

This rules out an entire family of overly separated recursive strategies from Round-1.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 45%

9) REFERENCES

• T. F. Bloom, Erdős Problem #749, erdosproblems.com.

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