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Math-649-A4 · revision r1 · on Math-649

gpt pro 5.2 attempt on Math-649, version 2

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Manuscript

Contents
  1. Round 2 (Problem 30): Sidon sets in $\{1,\dots,N\}$
  2. 1) Round-2 Objective
  3. 2) Round-1 Foundation Used
  4. 3) New Insight / Tool (Round-2)
  5. 4) Attack Plan (Round-2)
  6. 5) Work (Round-2): Lindström bound
  7. 6) Adversarial Verification
  8. 7) Final
  9. 8) Completion Estimate
  10. 9) References

Round 2 (Problem 30): Sidon sets in $\{1,\dots,N\}$

1) Round-2 Objective

Path (A): proof / gap-closure. Building on Round 1, we close the main gap left by the elementary counting bound by proving the classical Erdős--Turán/Lindström upper bound \[ h(N)\le \sqrt N+N^{1/4}+1, \] thereby improving the Round-1 bound $h(N)\le (\sqrt2+o(1))\sqrt N$. The conjectural refinement $h(N)=\sqrt N+O_\varepsilon(N^\varepsilon)$ remains open.

2) Round-1 Foundation Used

We use only the following Round-1 result.
  • (R1 Lemma 30.1) If $S=\{a_1<\cdots<a_r\}\subseteq\{1,\dots,N\}$ is Sidon, then all positive differences $a_j-a_i$ $(1\le i<j\le r)$ are distinct.

3) New Insight / Tool (Round-2)

Order-of-differences summation + telescoping. For each order $\nu=j-i$ we sum the order-$\nu$ differences and bound this sum by a telescoping argument. Summing over $\nu\le m$ gives an upper bound on the sum of many distinct positive integers; comparing to the minimal possible sum $1+2+\cdots+D$ yields a quadratic inequality. Optimizing $m\asymp N^{1/4}$ produces the $N^{1/4}$ error term.

4) Attack Plan (Round-2)

Gap after Round 1: only a counting upper bound $\sim \sqrt{2N}$ was proved.

What to prove now: for every Sidon $S\subseteq\{1,\dots,N\}$, \[ |S|\le \sqrt N+N^{1/4}+1. \]

Why this closes the gap: it uses structure (telescoping constraints on low-order differences) rather than mere counting, forcing the correct leading constant $1$.

5) Work (Round-2): Lindström bound

Theorem (Lindström upper bound): If $S\subseteq\{1,\dots,N\}$ is Sidon and $|S|=r$, then \[ r\le \sqrt N+N^{1/4}+1. \] Hence $h(N)\le \sqrt N+N^{1/4}+1$.
Proof: Write $S=\{a_1<a_2<\cdots<a_r\}$. By (R1 Lemma 30.1), all positive differences $a_j-a_i$ $(i<j)$ are distinct.

Step 1: order-$\nu$ sums. For $1\le \nu\le r-1$, define \[ \Sigma_\nu:=\sum_{i=1}^{r-\nu}(a_{i+\nu}-a_i). \] Claim: $\Sigma_\nu\le \nu N$ for every $\nu$. Indeed, partition indices $i$ into residue classes mod $\nu$. For a fixed residue class $k\in\{1,\dots,\nu\}$, the partial sum over $i=k,k+\nu,k+2\nu,\dots$ telescopes to $a_{\text{last}+\nu}-a_k\le a_r-a_1\le N-1<N$. Summing over the $\nu$ residue classes gives $\Sigma_\nu<\nu N$, hence $\Sigma_\nu\le \nu N$.

Step 2: sum over $\nu\le m$ and compare two bounds. Fix an integer $m\ge 1$. If $m\ge r$, then $r\le m\le \lceil N^{1/4}\rceil\le N^{1/4}+1\le \sqrt N+N^{1/4}+1$ and we are done. So assume $m\le r-1$.

Let \[ \Sigma(1,m):=\sum_{\nu=1}^m \Sigma_\nu. \] Using $\Sigma_\nu\le \nu N$ gives \begin{equation}\label{wsref-0-65-71-3a-75-70-70-65-72} \Sigma(1,m)\le \left(\sum_{\nu=1}^m \nu\right)N=\frac{m(m+1)}{2}\,N. \end{equation}

On the other hand, $\Sigma(1,m)$ sums all differences of order at most $m$: $a_{i+\nu}-a_i$ with $1\le \nu\le m$ and $1\le i\le r-\nu$. The number of such differences is \[ D=\sum_{\nu=1}^m (r-\nu)=mr-\frac{m(m+1)}{2}. \] These are distinct positive integers (by (R1 Lemma 30.1)), so their sum is at least the minimum possible sum of $D$ distinct positive integers: \begin{equation}\label{wsref-0-65-71-3a-6c-6f-77-65-72} \Sigma(1,m)\ge 1+2+\cdots+D=\frac{D(D+1)}{2}. \end{equation} Combining (eq:upper) and (eq:lower) yields \[ D(D+1)\le m(m+1)N, \] hence in particular $D^2\le m(m+1)N$.

Step 3: isolate $r$. Write \[ D=mr-\frac{m(m+1)}{2}=m\left(r-\frac{m+1}{2}\right). \] Let $s:=r-\frac{m+1}{2}$, so $D=ms$. Then $D^2\le m(m+1)N$ implies \[ m^2s^2\le m(m+1)N\quad\Rightarrow\quad s^2\le N\left(1+\frac{1}{m}\right), \] so \[ r=s+\frac{m+1}{2}\le \sqrt N\sqrt{1+\frac{1}{m}}+\frac{m+1}{2}. \] Using $\sqrt{1+x}\le 1+\frac{x}{2}$ for $x\ge 0$ (with $x=1/m$) gives \[ r\le \sqrt N+\frac{\sqrt N}{2m}+\frac{m+1}{2}. \]

Step 4: optimize $m$. Choose $m=\lceil N^{1/4}\rceil$. Then $\frac{\sqrt N}{2m}\le \frac{N^{1/4}}{2}$ and $\frac{m+1}{2}\le \frac{N^{1/4}}{2}+1$, hence \[ r\le \sqrt N+\frac{N^{1/4}}{2}+\left(\frac{N^{1/4}}{2}+1\right)=\sqrt N+N^{1/4}+1. \] This proves the theorem. ∎

6) Adversarial Verification

Potential failure modes are checked as follows.
  • Quantifiers / edge cases: if the chosen $m$ exceeds $r-1$ the bound is trivial (handled explicitly).
  • Distinctness of differences: used only through (R1 Lemma 30.1).
  • Telescoping bound: each residue-class partial sum telescopes to at most $a_r-a_1\le N-1$; using $\nu N$ is safe.
  • Lower bound on sum: for $D$ distinct positive integers, the minimal sum is $1+\cdots+D$.

7) Final

UNRESOLVED (but strictly advanced). Round 2 proves the classical bound $h(N)\le \sqrt N+N^{1/4}+1$, a strict improvement over Round 1. The conjecture $h(N)=\sqrt N+O_\varepsilon(N^\varepsilon)$ remains open.

8) Completion Estimate

COMPLETION: 65%

9) References

No external references are required for the proof presented here; it is self-contained modulo (R1 Lemma 30.1).