gpt pro 5.4 attempt on Math-649

Erdős Problem #30 --- Round 2

1) ROUND-2 OBJECTIVE

I pursue path (A) proof on the upper-bound side. More precisely, I continue the Sidon-set argument from Round 1 and prove the strictly stronger estimate \[ h(N)\le \sqrt{N}+N^{1/4}+1. \] This is the most promising direction because Round 1 had already isolated the main gap as improving the crude bound \[ h(N)\le \frac{1+\sqrt{8N-7}}{2}<\sqrt{2N}+1 \] toward the conjectured form $\sqrt{N}+O_{\varepsilon}(N^{\varepsilon})$.

2) Round-1 FOUNDATION USED

I use the following Round 1 results as fixed input:

• Round-1 Lemma 30.1: if $S=\{a_1<\dots<a_m\}\subseteq\{1,\dots,N\}$ is Sidon, then all positive differences $a_j-a_i$ ($i<j$) are distinct.
• Round-1 exact computation for $1\le N\le 40$: the values of $h(N)$ were determined exactly by backtracking; I reuse this as a consistency check and rerun a smaller verification below.

I do not re-prove these items.

3) NEW INSIGHT / TOOL (ROUND-2)

Round 1 counted only how many positive differences a Sidon set has. The new ingredient is to count the sum of a carefully chosen subfamily of differences, \[ a_{i+k}-a_i\qquad (1\le k\le u,\ 1\le i\le m-k). \] There are \[ T=\sum_{k=1}^u (m-k)=um-\frac{u(u+1)}{2} \] such differences. Because they are distinct positive integers, their total sum is at least $1+2+\cdots+T$. On the other hand, for each fixed $k$, the sum telescopes and is at most $kN$. Balancing these two bounds and optimizing $u$ at scale $N^{1/4}$ gives the sharp classical upper bound.

4) ATTACK PLAN (ROUND-2)

Round 1 left the following exact gap on Problem #30:

Improve the upper bound from $(\sqrt{2}+o(1))\sqrt{N}$ to something of the form $\sqrt{N}+o(\sqrt{N})$.

To close a substantial part of that gap, it is enough to prove the following two claims.

1. For every integer $u$ with $1\le u\le m-1$, if \[ T:=um-\frac{u(u+1)}{2}, \] then \[ T(T+1)\le Nu(u+1). \]
2. Choosing $u=\lceil N^{1/4}\rceil$ converts this quadratic inequality into \[ m<\sqrt{N}+N^{1/4}+1. \]

This overcomes the Round 1 obstruction by using not just existence of many distinct differences, but also the fact that distinct positive integers must have large total sum.

5) WORK (ROUND-2)

Let \[ S=\{a_1<a_2<\cdots<a_m\}\subseteq \{1,2,\dots,N\} \] be a Sidon set, so $m=|S|$.

Fast reality check. I reran an exact backtracking search for $h(N)$ on the range $1\le N\le 40$ and verified that the new bound below holds in every case. Representative exact values are \[ \begin{array}{c|cccc} N&10&20&30&40\\\hline h(N)&4&6&7&8 \end{array} \] while \[ \begin{array}{c|cccc} N&10&20&30&40\\\hline \sqrt{N}+N^{1/4}+1&5.9405\ldots&7.5868\ldots&8.8175\ldots&9.8394\ldots \end{array} \] so there are no small-$N$ violations in this exact check.

Case split. If \[ m\le \lceil N^{1/4}\rceil+1, \] then certainly \[ m\le \lceil N^{1/4}\rceil+1\le N^{1/4}+2\le \sqrt{N}+N^{1/4}+1 \] for every $N\ge 1$, and we are done. Thus the only interesting case is \[ m>\lceil N^{1/4}\rceil+1. \] Set \[ u:=\lceil N^{1/4}\rceil. \] Then $1\le u\le m-1$.

Lemma 30.R2.1 (lower bound for the short-difference sum). Define \[ D_u:=\{a_{i+k}-a_i: 1\le k\le u,\ 1\le i\le m-k\}. \] Then \[ |D_u|=T:=um-\frac{u(u+1)}{2}, \] and \[ \sum_{k=1}^{u}\sum_{i=1}^{m-k}(a_{i+k}-a_i)\ge \frac{T(T+1)}{2}. \]

Proof. By Round-1 Lemma 30.1, all positive differences in a Sidon set are distinct. Therefore all members of $D_u$ are distinct. For each fixed $k$ there are exactly $m-k$ terms of the form $a_{i+k}-a_i$, so \[ |D_u|=\sum_{k=1}^{u}(m-k)=um-\frac{u(u+1)}{2}=T. \] Because $D_u$ consists of $T$ distinct positive integers, its sum is at least the sum of the first $T$ positive integers: \[ \sum_{d\in D_u} d\ge 1+2+\cdots+T=\frac{T(T+1)}{2}. \] But \[ \sum_{d\in D_u} d=\sum_{k=1}^{u}\sum_{i=1}^{m-k}(a_{i+k}-a_i), \] which gives the stated lower bound. $\square$

Lemma 30.R2.2 (telescoping upper bound for the short-difference sum). For each integer $k$ with $1\le k\le u$, \[ \sum_{i=1}^{m-k}(a_{i+k}-a_i)\le kN. \] Consequently, \[ \sum_{k=1}^{u}\sum_{i=1}^{m-k}(a_{i+k}-a_i)\le \frac{Nu(u+1)}{2}. \]

Proof. Fix $k$. Then \[ \sum_{i=1}^{m-k}(a_{i+k}-a_i) =\sum_{i=1}^{m-k} a_{i+k}-\sum_{i=1}^{m-k} a_i =\sum_{j=m-k+1}^{m} a_j-\sum_{j=1}^{k} a_j. \] The first sum has $k$ terms, each at most $N$, so it is at most $kN$. The second sum is nonnegative. Hence \[ \sum_{i=1}^{m-k}(a_{i+k}-a_i)\le kN. \] Summing this inequality over $k=1,2,\dots,u$ gives \[ \sum_{k=1}^{u}\sum_{i=1}^{m-k}(a_{i+k}-a_i) \le \sum_{k=1}^{u} kN =\frac{Nu(u+1)}{2}. \] $\square$

Corollary 30.R2.3 (quadratic inequality). With $T=um-u(u+1)/2$, \[ T(T+1)\le Nu(u+1). \]

Proof. Combine Lemma 30.R2.1 and Lemma 30.R2.2. $\square$

Theorem 30.R2.4 (strict advance: Erdős--Turán upper bound). Every Sidon subset of $\{1,2,\dots,N\}$ has size \[ m<\sqrt{N}+N^{1/4}+1. \] Hence \[ h(N)\le \sqrt{N}+N^{1/4}+1. \]

Proof. From Corollary 30.R2.3 we obtain \[ T^2<T(T+1)\le Nu(u+1). \] Therefore \[ T<\sqrt{Nu(u+1)}. \] Since \[ \left(u+\frac12\right)^2=u(u+1)+\frac14>u(u+1), \] we have \[ \sqrt{u(u+1)}<u+\frac12. \] Thus \[ T<\sqrt{N}\left(u+\frac12\right). \] Substituting $T=um-u(u+1)/2$ yields \[ um-\frac{u(u+1)}{2}<\sqrt{N}\left(u+\frac12\right). \] Divide by $u$: \[ m<\sqrt{N}+\frac{\sqrt{N}}{2u}+\frac{u+1}{2}. \] Now use $u=\lceil N^{1/4}\rceil$. Since $u\ge N^{1/4}$, \[ \frac{\sqrt{N}}{2u}\le \frac{N^{1/4}}{2}. \] Since $u\le N^{1/4}+1$, \[ \frac{u+1}{2}\le \frac{N^{1/4}+2}{2}=\frac{N^{1/4}}{2}+1. \] Hence \[ m<\sqrt{N}+\frac{N^{1/4}}{2}+\frac{N^{1/4}}{2}+1=\sqrt{N}+N^{1/4}+1. \] This proves the theorem. $\square$

6) ADVERSARIAL VERIFICATION

I checked the new argument against the main failure modes.

• Hidden use of an old proof. The only Round 1 fact reused is the distinctness of positive differences in a Sidon set. I cite that result explicitly and do not re-prove it.
• Range of $u$. The family $a_{i+k}-a_i$ is only defined for $k\le m-1$. I handled this by a case split: if $m\le \lceil N^{1/4}\rceil+1$, the target inequality is immediate; otherwise $u=\lceil N^{1/4}\rceil$ satisfies $u\le m-1$.
• Telescoping identity. For fixed $k$, \[ \sum_{i=1}^{m-k}(a_{i+k}-a_i)=\sum_{j=m-k+1}^{m} a_j-\sum_{j=1}^{k} a_j, \] so no term has been dropped incorrectly. The upper bound $\le kN$ is valid because the subtracted sum is nonnegative.
• Small $N$. The preliminary case split covers all small values automatically. The exact computation for $N\le 40$ gives an independent sanity check.
• Quantifiers. The theorem is proved for an arbitrary Sidon set $S\subseteq\{1,\dots,N\}$, so the conclusion for $h(N)$ follows immediately.

No contradiction with the Round 1 lower-bound constructions arises: Round 1 only gave lower bounds near $\sqrt{N}$ along special values of $N$, whereas the new theorem is a stronger universal upper bound.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

This round proves the classical bound \[ h(N)\le \sqrt{N}+N^{1/4}+1, \] which is a strict improvement over the Round 1 estimate \[ h(N)\le \frac{1+\sqrt{8N-7}}{2}<\sqrt{2N}+1. \] So the investigation has advanced substantially on the upper-bound side. However, the original open question---whether the error term can be sharpened all the way to $O_{\varepsilon}(N^{\varepsilon})$ for every $\varepsilon>0$---remains open.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 62%

9) REFERENCES

No external references were used. I relied only on the carried-over Round 1 file already present in the workspace (specifically its Problem #30 section) and on a local exact Python backtracking computation for $1\le N\le 40$.

10) OUTPUT FORMAT

The problem number identified for this continuation is $x=30$, so this export is the requested 30_solutions.tex.