gpt pro 5.4 attempt on Math-5897

Archive note. This extract contains the work on Problem #789 from a joint report on Problems #789, #873, and #875. The shared introduction, verification, conclusions, and references are retained. The source's 68% completion estimate applies to the joint report.

Round 2 continuation for Erdős problems #789, #873, #875

1) ROUND-2 OBJECTIVE

Because Round 1 bundled three Erdős problems into a single file, this continuation also treats the three problems together. I am pursuing path (C): I identify and correct a concrete Round 1 error for Problem #875 (a missed theorem from the literature), then propagate the corrected consequences. In the same pass I also strengthen the Round 1 upper bound for Problem #789 and prove a new fixed-$k$ obstruction for Problem #873. None of the three problems is fully solved, so the global status remains unresolved.

2) Round-1 FOUNDATION USED

I use the following Round 1 facts without reproving them.

1. For Problem #789, Round 1 already corrected the statement to the standard distinct-summand notion of admissibility, and proved the lower bound $h(n) \gg n^{1/3}$.
2. For Problem #873, Round 1 proved the monotonicity of $F(A,X,k)$ in $k$, namely that $F(A,X,k') \le F(A,X,k)$ whenever $k'\ge k$.
3. For Problem #875, Round 1 observed that every finite initial segment $A\cap [1,N]$ of an infinite admissible set is itself admissible, so any sharp finite bound inside $[1,N]$ immediately yields a universal density bound for infinite admissible sets.

3) NEW INSIGHT / TOOL (ROUND-2)

There are three genuinely new ingredients in this round.

1. Problem #789. I import the sharp Deshouillers--Freiman theorem for finite admissible subsets of $[1,N]$, which upgrades the Round 1 upper bound from $\sqrt{7n}$ to $2\sqrt{n}+O(1)$.
2. Problem #873. I introduce a new divisor-block construction: concatenate scaled divisor sets of primorials. This produces, for every fixed $k$, an infinite set $A$ for which $F(A,X,k)$ is larger than every power of $\log X$ infinitely often. This does not disprove the Erdős--Szemerédi conjecture, but it shows that any proof must use the freedom to let $k$ depend on $A$ and $\varepsilon$.
3. Problem #875. Round 1 missed Erdős--Nicolas--Sárkőzy's 1991 theorem constructing an infinite admissible set with polynomial density $A(x)\gg x^{5-2\sqrt6}$. After correcting this, I derive new consequences: existence of polynomial-growth admissible sets, existence of an admissible set with $\liminf a_{n+1}/a_n = 1$, a sharpened universal lower bound $a_n\ge n(n+2)/4$ for all large $n$, and a sharpened gap obstruction $\limsup (a_{n+1}-a_n)/n\ge 1/2$.

4) ATTACK PLAN (ROUND-2)

After Round 1, the exact remaining gaps were as follows.

1. For Problem #789, the upper bound was only $O(\sqrt n)$ with a weak constant. The natural gap was to import the sharp finite-interval extremal theorem and transfer it immediately to $h(n)$ by taking the special set $[1,n]$.
2. For Problem #873, Round 1 had no construction showing how badly a fixed value of $k$ can fail. The precise target now is to prove that for every fixed $k$, there is an infinite $A$ for which $F(A,X,k)$ is still very large infinitely often.
3. For Problem #875, the most serious issue is factual: Round 1 treated polynomial growth as if it might still be unknown. I first repair that mistake using the 1991 theorem, and then extract the strongest easy corollaries for growth and gap questions.

5) WORK (ROUND-2)

Problem #789: sharpening the universal upper bound

Let $h(n)$ be the corrected distinct-summand quantity from Round 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. Deshouillers and Freiman proved that there exists $N_0$ such that for every integer $N\ge N_0$, every admissible subset $C\subset [1,N]$ satisfies \[ |C| \le 2\sqrt{N+\tfrac14}-1. \] Now take the special $n$-element set \[ A_0 := \{1,2,\dots,n\}. \] Any admissible subset $B\subseteq A_0$ is an admissible subset of $[1,n]$, so for $n\ge N_0$, \[ |B| \le 2\sqrt{n+\tfrac14}-1. \] But $h(n)$ is the largest integer guaranteed inside every $n$-element set, so in particular it cannot exceed the largest admissible subset size inside this specific set $A_0$. Hence \[ h(n) \le 2\sqrt{n+\tfrac14}-1 \] for all sufficiently large $n$, which is exactly the claim. $\Box$

Consequence. Combining Claim 789.1 with the Round 1 lower bound $h(n)\gg n^{1/3}$, and with Choi's recorded improvement $h(n)\gg (n\log n)^{1/3}$, the corrected state of play is now \[ (n\log n)^{1/3} \ll h(n) \le 2\sqrt n + O(1). \] This is strictly sharper than the Round 1 upper bound $h(n)\le \sqrt{7n}$.

6) ADVERSARIAL VERIFICATION

Problem #789.

The only nontrivial hypothesis in Claim 789.1 is the Deshouillers--Freiman theorem, whose hypotheses are exactly satisfied by admissible subsets of $[1,n]$. The transfer from the interval problem to $h(n)$ is logically sound because $h(n)$ is a universal guarantee over all $n$-element sets, so one counterexample family of size $n$ immediately gives an upper bound.

Problem #873.

The divisor-block construction must satisfy three separate requirements: blocks must be ordered, the lcm inside each block must stay below the chosen threshold, and the quantitative lower bound must really exceed every power of $\log X$. The recursive condition $y_t>q_{t-1}P_{t-1}$ guarantees block ordering. The coprimality $\gcd(q_t,P_t)=1$ makes the lcm computation exact. Finally, the prime number theorem turns the divisor count $2^{\pi(y_t)}$ into the stated lower bound. No cross-block interaction is needed, because I only need a lower bound on $F(A,X_t,k)$.

Problem #875.

The main issue was factual: Round 1 overlooked a published theorem. After correcting that, all subsequent deductions are elementary. Claim 875.1 uses only the inversion of the counting-function lower bound and a standard exponential-growth contradiction. Claim 875.2 applies the finite Deshouillers--Freiman theorem to the initial segment $A\cap[1,N]$. Claim 875.3 sums an eventual linear gap bound and compares the quadratic leading coefficient against Claim 875.2; the contradiction is at the coefficient level $(1/4-\varepsilon/2)$ versus $1/4$, so there is no hidden constant issue. Claim 875.4 only claims existence for $c>5+2\sqrt6$, which is exactly the exponent one gets for free from the polynomial upper bound on $a_n$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advances over Round 1 are these.

1. Problem #789: the upper bound is sharpened from $\sqrt{7n}$ to $2\sqrt n+O(1)$.
2. Problem #873: a new fixed-$k$ obstruction is proved. For every fixed $k$, some infinite $A$ has $F(A,X,k)$ larger than every power of $\log X$ infinitely often.
3. Problem #875: a concrete Round 1 error is corrected. Polynomial-growth admissible sets already exist. As a result, the old final status for #875 must be replaced by the sharper quantitative bracket \[ c<\frac12 \text{ impossible}, \qquad c>5+2\sqrt6 \text{ possible}. \] Moreover there exists an admissible set with $\liminf a_{n+1}/a_n=1$.

8) COMPLETION ESTIMATE

COMPLETION: 68%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #789, problem page and discussion thread.
2. T. F. Bloom, Erdős Problem #873, problem page and discussion thread.
3. T. F. Bloom, Erdős Problem #875, problem page and discussion thread.
4. J.-M. Deshouillers and G. A. Freiman, On an additive problem of Erdős and Straus, 2, in Structure theory of set addition, Astérisque 258 $1999$, 141--148.
5. P. Erdős, J.-L. Nicolas, and A. Sárkőzy, Sommes de sous-ensembles, Séminaire de théorie des nombres de Bordeaux 3 $1991$, 55--72.
6. Prime Number Theorem and the asymptotic $\vartheta(x)\sim x$ for the Chebyshev function (used only in the quantitative corollary of Claim 873.1).

10) OUTPUT FORMAT NOTE

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