gpt pro 5.4 attempt on Math-655

Erdős Problem 41 (Round 4)

Problem

An infinite set $A\subset \mathbb{N}$ is a $B_3$-set if all sums \[ a_1+a_2+a_3\qquad (a_1\le a_2\le a_3,\ a_i\in A) \] are distinct. The target statement is \[ \liminf_{N\to\infty}\frac{|A\cap[1,N]|}{N^{1/3}}=0. \] Write \[ A(N):=|A\cap[1,N]|. \]

1) ROUND-4 OBJECTIVE

Path (C): obstruction/correction.

Round 3 showed that the Round 2 "full proof" was not gap-free, and reduced the original Erdős problem to the oscillatory regime \[ 0<\liminf_{N\to\infty}\frac{A(N)}{N^{1/3}} < \limsup_{N\to\infty}\frac{A(N)}{N^{1/3}} <\infty. \] The most promising Round 4 advance is therefore not to restart the whole proof, but to sharpen that obstruction: we prove that no such oscillation can stabilize on an asymptotically dense multiplicative scale. This rules out an entire class of possible counterexamples.

2) Round-3 FOUNDATION USED

We use the following vetted results from Round 3.

• Round 1 Lemma 41.1 (retained in Round 3). For every infinite $B_3$-set, \[ A(N)\le (18N)^{1/3}+2, \] hence \[ \limsup_{N\to\infty}\frac{A(N)}{N^{1/3}}<\infty. \]
• Round 3 Proposition 5.1. If the limit \[ \lim_{N\to\infty}\frac{A(N)}{N^{1/3}} \] exists for an infinite $B_3$-set, then it must equal $0$. Equivalently, no infinite $B_3$-set satisfies \[ A(N)\sim \alpha N^{1/3} \qquad (\alpha>0). \]
• Round 3 Proposition 5.2. Given only the verified inputs, the original Erdős problem is equivalent to ruling out the oscillatory regime \[ 0<\liminf_{N\to\infty}\frac{A(N)}{N^{1/3}} < \limsup_{N\to\infty}\frac{A(N)}{N^{1/3}} <\infty. \]

3) NEW INSIGHT / TOOL (ROUND-4)

Dense-subsequence rigidity by monotonicity.

Let $B(N)$ be any nondecreasing counting function. If $B(N)N^{-\gamma}$ converges along a sequence $N_k$ with $N_{k+1}/N_k\to 1$, then in fact $B(N)N^{-\gamma}$ converges along all integers $N$. Applied with $\gamma=1/3$, this means that a $B_3$-set cannot even look pseudo-cubic along an asymptotically dense multiplicative subsequence.

This is genuinely new in Round 4: Round 3 ruled out only global pseudo-cubic asymptotics; here we rule out subsequential pseudo-cubic asymptotics on scales $N_{k+1}=(1+o(1))N_k$.

4) ATTACK PLAN (ROUND-4)

The exact gap left after Round 3 is the possibility of a $B_3$-set with \[ 0<\liminf f(N)<\limsup f(N)<\infty, \qquad f(N):=\frac{A(N)}{N^{1/3}}, \] but with no full limit.

Round 4 proves the following claims.

1. A general rigidity lemma: if $f(N_k)$ converges along a sequence with $N_{k+1}/N_k\to 1$, then $f(N)$ converges globally.
2. Using Round 3 Proposition 5.1, this implies that no infinite $B_3$-set admits a subsequence $N_k$ with $N_{k+1}/N_k\to 1$ and $f(N_k)\to \alpha>0$.
3. Therefore any hypothetical counterexample must oscillate on every asymptotically dense multiplicative scale; equivalently, every positive cluster value of $f$ can only be attained along multiplicatively sparse subsequences.

This does not settle the full problem, but it excludes a broad family of oscillatory profiles that remained possible after Round 3.

5) WORK (ROUND-4)

Lemma 5.1 (dense-subsequence rigidity).

Let $B:\mathbb{N}\to[0,\infty)$ be nondecreasing, let $\gamma>0$, and define \[ g(N):=\frac{B(N)}{N^{\gamma}}. \] Suppose there is an increasing sequence $N_1<N_2<\cdots$ such that \[ \frac{N_{k+1}}{N_k}\to 1 \qquad\text{and}\qquad g(N_k)\to \alpha \] for some $\alpha\in[0,\infty)$. Then \[ g(N)\to \alpha\qquad (N\to\infty). \]

Proof.

Fix large $N$, and choose the unique index $k=k(N)$ such that \[ N_k\le N < N_{k+1}. \] Since $B$ is nondecreasing, \[ B(N_k)\le B(N)\le B(N_{k+1}). \] Dividing by $N^{\gamma}$ gives \[ g(N_k)\left(\frac{N_k}{N}\right)^{\gamma} \le g(N) \le g(N_{k+1})\left(\frac{N_{k+1}}{N}\right)^{\gamma}. \] Because $N_k\le N < N_{k+1}$, we also have \[ \frac{N_k}{N}\ge \frac{N_k}{N_{k+1}} \qquad\text{and}\qquad \frac{N_{k+1}}{N}\le \frac{N_{k+1}}{N_k}. \] Hence \[ g(N_k)\left(\frac{N_k}{N_{k+1}}\right)^{\gamma} \le g(N) \le g(N_{k+1})\left(\frac{N_{k+1}}{N_k}\right)^{\gamma}. \] Now $N\to\infty$ implies $k(N)\to\infty$, and by hypothesis \[ g(N_k)\to \alpha, \qquad g(N_{k+1})\to \alpha, \qquad \frac{N_{k+1}}{N_k}\to 1. \] Therefore both the lower and upper bounds tend to $\alpha$. By the squeeze theorem, \[ g(N)\to \alpha. \] This proves the lemma.

Corollary 5.2 (no positive asymptotic on asymptotically dense scales).

Let $A$ be an infinite $B_3$-set and define \[ f(N):=\frac{A(N)}{N^{1/3}}. \] Then there is no increasing sequence $N_1<N_2<\cdots$ such that \[ \frac{N_{k+1}}{N_k}\to 1 \qquad\text{and}\qquad f(N_k)\to \alpha>0. \]

Proof.

Assume such a sequence exists. Since $A(N)$ is nondecreasing, Lemma 5.1 applies with $B=A$ and $\gamma=1/3$, and yields \[ \frac{A(N)}{N^{1/3}}\to \alpha. \] But Round 3 Proposition 5.1 says that if the full limit $\lim_{N\to\infty}A(N)N^{-1/3}$ exists for an infinite $B_3$-set, then it must equal $0$. This contradicts $\alpha>0$. Therefore no such sequence can exist.

Corollary 5.3 (positive cluster values are multiplicatively sparse).

Let $A$ be an infinite $B_3$-set and let $\alpha>0$. If $N_k$ is any increasing sequence satisfying \[ \frac{A(N_k)}{N_k^{1/3}}\to \alpha, \] then \[ \frac{N_{k+1}}{N_k}\not\to 1. \] Equivalently, \[ \limsup_{k\to\infty}\frac{N_{k+1}}{N_k}>1. \]

Proof.

If $N_{k+1}/N_k\to 1$, then Corollary 5.2 gives a contradiction. So $N_{k+1}/N_k\not\to 1$. Since every ratio is at least $1$, this is equivalent to \[ \limsup_{k\to\infty}\frac{N_{k+1}}{N_k}>1. \]

Proposition 5.4 (sharpened obstruction to a counterexample).

Assume, for contradiction, that the original Erdős statement fails for some infinite $B_3$-set $A$. Then, by Round 3 Proposition 5.2, \[ 0<\ell:=\liminf_{N\to\infty}f(N) < L:=\limsup_{N\to\infty}f(N) <\infty. \] Let \[ K:=\left\{\beta\in\mathbb{R}:\ \exists N_k\to\infty\text{ with }f(N_k)\to\beta\right\} \] be the cluster set of $f$. Then $K\subset[\ell,L]\subset(0,\infty)$, and for every $\alpha\in K$, every increasing sequence $N_k$ with $f(N_k)\to\alpha$ must be multiplicatively sparse in the sense of Corollary 5.3.

In particular, no hypothetical counterexample can stabilize near any positive level $\alpha$ on a scale $N_{k+1}=(1+o(1))N_k$.

Proof.

The inequality \[ 0<\ell<L<\infty \] is exactly the unresolved regime identified in Round 3 Proposition 5.2. By definition of cluster set, every $\alpha\in K$ is obtained as the limit of $f(N_k)$ along some sequence $N_k\to\infty$. Since $K\subset[\ell,L]\subset(0,\infty)$, Corollary 5.3 applies to every such sequence and shows that $N_{k+1}/N_k\not\to1$. This is precisely the asserted multiplicative sparseness, and the final sentence is just a restatement.

6) ADVERSARIAL VERIFICATION

We now test the new argument for weaknesses.

• Use of Round 3 results. The only inherited nontrivial input beyond monotonicity is Round 3 Proposition 5.1, i.e. the verified "no pseudo-cubes" consequence of Helm. We do not reuse the unsupported stronger oscillation theorem from Round 2.
• Hidden regularity assumptions. Lemma 5.1 uses only that $B(N)$ is nondecreasing. No convexity, differentiability, or regular variation is assumed.
• Boundary case $\alpha=0$. Lemma 5.1 remains true for $\alpha=0$. Corollary 5.2 only excludes $\alpha>0$, exactly because Round 3 Proposition 5.1 leaves $\alpha=0$ possible.
• Could the ratios oscillate? Yes, and our statement allows that. Corollary 5.3 says only that the ratios cannot converge to $1$ if a positive cluster value is present. This is the precise conclusion proved.
• Does this settle the original Erdős problem? No. A counterexample could still oscillate between positive levels along sufficiently sparse subsequences. The remaining open regime is now narrower: any such oscillation must occur on genuinely macroscopic multiplicative jumps.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 3 reduced the problem to the regime \[ 0<\liminf_{N\to\infty}\frac{A(N)}{N^{1/3}} < \limsup_{N\to\infty}\frac{A(N)}{N^{1/3}} <\infty. \] Round 4 proves a new rigidity theorem showing that such a regime cannot stabilize along any asymptotically dense multiplicative subsequence $N_{k+1}=(1+o(1))N_k$. Equivalently, every positive cluster value of $A(N)/N^{1/3}$ must be realized only along multiplicatively sparse subsequences. This rules out an entire class of possible counterexamples and is strictly stronger than any result established in earlier rounds.

8) COMPLETION ESTIMATE

COMPLETION: 78%

9) REFERENCES

• P. Erdős, Problems and results on combinatorial number theory III, in Number Theory Day (Rockefeller Univ., 1976), Lecture Notes in Math. 626, Springer, 1977, pp. 43--72.
• M. Helm, On the Distribution of $B_3$-Sequences, J. Number Theory 58 $1996$, 124--129.
• K. O'Bryant, A Complete Annotated Bibliography of Work Related to Sidon Sequences, Electron. J. Combin. DS11 $2004$. Used for the recorded abstract of Helm's paper.
• J. Cilleruelo, Conjuntos de Sidon, in Aritmética, Grupos y Análisis, AGRA II lecture notes (2015; later online circulation checked in 2022). Used for the statement that the odd-$h$ liminf problem remains open there, and that Helm proves only the pseudo-cube obstruction.