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Math-947-A2 · revision r1 · on Math-947

gpt pro 5.2 attempt on Math-947

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdos Problem #819
  2. FORMAL RESTATEMENT.
  3. QUICK LITERATURE/CONTEXT CHECK.
  4. ATTACK PLAN.
  5. WORK.
  6. VERIFICATION.
  7. FINAL.

Erdos Problem #819

FORMAL RESTATEMENT.
For each positive integer $N$, let \[ m := \lfloor N^{1/2}\rfloor. \] Define $f(N)$ to be the maximum of \[ |(A+A)\cap [1,N]| \] over all subsets $A\subseteq\{1,2,\dots,N\}$ with $|A|=m$, where $A+A:=\{a+b:\ a,b\in A\}$ and $[1,N]=\{1,2,\dots,N\}$. The problem asks to estimate $f(N)$.
QUICK LITERATURE/CONTEXT CHECK.
The statement records that Erdős and Freud proved \[ \left(\frac{3}{8}-o(1)\right)N \le f(N) \le \left(\frac{1}{2}+o(1)\right)N. \] I do not use any results not already stated in the problem text.
ATTACK PLAN.
Upper bounds: use counting of unordered pairs to control how many distinct sums can appear. Lower bounds: construct explicit $A$ producing many sums in $[1,N]$; as a baseline, compute small $N$ exactly to see what is achievable.
WORK.
Lemma 819.1 (unconditional counting upper bound). Let $m=\lfloor \sqrt N\rfloor$. For every $A\subseteq[1,N]$ with $|A|=m$, \[ |(A+A)\cap[1,N]| \le \binom{m+1}{2} = \frac{m(m+1)}{2} \le \frac{N}{2} + O(\sqrt N). \]

Proof. Every sum $s\in A+A$ can be represented as $s=a+b$ with $a,b\in A$. If we ignore order, each unordered pair $\{a,b\}$ with $a,b\in A$ produces at most one sum. There are exactly $\binom{m}{2}$ unordered pairs with $a\ne b$ and $m$ diagonal pairs $\{a,a\}$, so in total there are $\binom{m}{2}+m=\binom{m+1}{2}$ unordered pairs. Distinct sums are at most the number of unordered pairs, hence $|A+A|\le \binom{m+1}{2}$, and a fortiori $|(A+A)\cap[1,N]|\le \binom{m+1}{2}$. Since $m^2\le N < (m+1)^2$, we have $m^2\le N$ and $m\le \sqrt N$, giving $\binom{m+1}{2}\le \frac{1}{2}N + O(\sqrt N)$. $\square$

Lemma 819.2 (baseline lower bound). For $m=\lfloor\sqrt N\rfloor$, taking $A=\{1,2,\dots,m\}$ gives \[ |(A+A)\cap[1,N]| \ge 2m-1 = 2\lfloor\sqrt N\rfloor-1. \]

Proof. If $A=\{1,2,\dots,m\}$ then \[ A+A = \{2,3,\dots,2m\}. \] All these sums are $\le 2m\le 2\sqrt N\le N$ for $N\ge4$; in any case, $(A+A)\cap[1,N]$ contains $\{2,3,\dots,2m\}$. Thus $|(A+A)\cap[1,N]|\ge 2m-1$. $\square$

FAST REALITY CHECK (exact values for $N\le 30$). I exhaustively computed $f(N)$ for $1\le N\le 30$ by enumerating all subsets $A\subseteq[1,N]$ of size $m=\lfloor\sqrt N\rfloor$. The maximal values found were: \[ \begin{array}{c|c|c} N & m & f(N)\\\hline 1 & 1 & 0\\ 2\le N\le 3 & 1 & 1\\ 4\le N\le 8 & 2 & 3\\ 9\le N\le 15 & 3 & 6\\ 16\le N\le 24 & 4 & 10\\ 25\le N\le 30 & 5 & 15 \end{array} \] For example, at $N=16$ the set $A=\{1,2,4,8\}$ achieves $f(16)=10$.

VERIFICATION.
Lemma 819.1: the only step is the injection "distinct sums $\le$ unordered pairs"; this is valid because each unordered pair produces a single integer sum, though multiple pairs may collide. Lemma 819.2: sums of the initial segment form the full interval $[2,2m]$; checking the intersection with $[1,N]$ is immediate. The computed $f(N)$ values are finite enumerations and can be rechecked.
FINAL.
UNRESOLVED.
  1. (i) Strongest proved partial results here: the unconditional upper bound $f(N)\le \binom{\lfloor\sqrt N\rfloor+1}{2} = \frac{1}{2}N+O(\sqrt N)$ (Lemma 819.1), matching the scale of the upper bound stated in the problem, and the explicit lower bound $f(N)\ge 2\lfloor\sqrt N\rfloor-1$ (Lemma 819.2). Exact values for $N\le 30$ were computed.
  2. (ii) First gap: produce (or rigorously derive) a linear-in-$N$ lower bound $f(N)\ge cN$ with an explicit constant $c>0$ using only arguments developed here (i.e. without importing the stated Erdős--Freud result).
  3. (iii) Top 3 next moves: $1$ attempt a probabilistic construction of a size-$\sqrt N$ set $A$ for which many $s\le N$ have at least one representation $s=a+b$; $2$ search computationally for maximizers for larger $N$ (e.g. $N\le 60$) to guess structure; $3$ prove structural lemmas relating large $|(A+A)\cap[1,N]|$ to quasi-Sidon behaviour (few collisions) to approach the $\sim \tfrac12 N$ barrier.
  4. (iv) A minimal counterexample to any claimed linear lower bound would be a sequence $N_j\to\infty$ such that every $A\subseteq[1,N_j]$ with $|A|=\lfloor\sqrt{N_j}\rfloor$ satisfies $|(A+A)\cap[1,N_j]|=o(N_j)$; the small-$N$ data already contradict this behaviour at low scales, suggesting any counterexample would require subtle additive structure.