gpt pro 5.2 attempt on Math-758

Problem #319

1) FORMAL RESTATEMENT

For each integer $N\ge 1$, let $F(N)$ be the maximum cardinality of a set $A\subseteq \{1,2,\dots,N\}$ for which there exists a sign function \[ \delta:A\to\{-1,+1\},\qquad n\mapsto \delta_n, \] such that \[ \sum_{n\in A}\frac{\delta_n}{n}=0, \] and moreover \[ \sum_{n\in A'}\frac{\delta_n}{n}\neq 0 \quad\text{for every nonempty proper subset } \varnothing\neq A'\subsetneq A. \] The problem asks for good bounds/asymptotics for $F(N)$ as $N\to\infty$.

2) CONTEXT AND DEFINITIONS

Writing $A=P\sqcup Q$ where $P=\{n\in A:\delta_n=+1\}$ and $Q=\{n\in A:\delta_n=-1\}$, the condition is equivalent to \[ \sum_{p\in P}\frac1p=\sum_{q\in Q}\frac1q, \] together with a minimality condition: there do not exist nonempty proper subcollections $P'\subseteq P$, $Q'\subseteq Q$ with $\sum_{p\in P'}\frac1p=\sum_{q\in Q'}\frac1q$.

A particularly simple way to guarantee minimality is to take $P=\{1\}$ and $Q=B$ for some $B\subseteq\{2,\dots,N\}$ with $\sum_{b\in B}\frac1b=1$. Then any proper sub-sum over $B$ is $<1$.

3) QUICK LITERATURE/CONTEXT CHECK

The Erdős Problems page for #319 records an observation of Adenwalla that a linear-size lower bound follows from a theorem of Croot on Egyptian fractions in short intervals: there exists $B\subset [(\frac1e-o(1))N,N]$ with $\sum_{b\in B}1/b=1$, which yields $|A|\ge (1-\frac1e+o(1))N$ by taking $A=B\cup\{1\}$ with signs $+1$ on $1$ and $-1$ on $B$.

4) ATTACK PLAN

I will (i) state the needed consequence of Croot's theorem precisely; (ii) give a complete proof that it implies \[ F(N)\ge \left(1-\frac1e+o(1)\right)N; \] and (iii) check the "no proper subset sums to $0$" condition carefully.

This does not resolve the full asymptotic for $F(N)$, but it does give a fully rigorous linear lower bound of the claimed constant.

5) DETAILED WORK

Step 1: A usable consequence of Croot's theorem.

Croot proved (among other things) that for any fixed rational $r>0$ and sufficiently large $X$ there exist integers \[ X < n_1 < n_2 < \cdots < n_k < (e^r+o_r(1))X \] such that $\sum_{i=1}^k \frac1{n_i}=r$. Specializing to $r=1$ and choosing $X=X(N)$ with $(e+o(1))X\le N$ (e.g. $X=N/(e+\varepsilon_N)$ for any $\varepsilon_N\to 0^+$) gives:

Lemma 319.1 (existence of an interval representation of $1$). There exists a function $\alpha_N\to 1/e$ and, for all sufficiently large $N$, a finite set of integers \[ B\subseteq \{\lceil \alpha_N N\rceil,\lceil \alpha_N N\rceil+1,\dots,N\} \] such that \[ \sum_{b\in B}\frac1b = 1. \]

Step 2: This forces $B$ to have (almost) full size in the interval.

Let $I_N=\{\lceil \alpha_N N\rceil,\dots,N\}$. Using the integral test (or the asymptotic $H_n=\log n+\gamma+o(1)$), \[ \sum_{m\in I_N}\frac1m = \log\!\Big(\frac{N}{\alpha_N N}\Big) + o(1) = \log\!\Big(\frac1{\alpha_N}\Big)+o(1) = 1+o(1) \quad (N\to\infty), \] since $\alpha_N\to 1/e$.

Because $B\subseteq I_N$ and $\sum_{b\in B}\frac1b=1$, we have \[ \sum_{m\in I_N\setminus B}\frac1m = \sum_{m\in I_N}\frac1m - 1 = o(1). \] But every $m\in I_N\setminus B$ satisfies $m\le N$, hence $1/m\ge 1/N$. Therefore \[ \frac{|I_N\setminus B|}{N} \le \sum_{m\in I_N\setminus B}\frac1m = o(1), \] so $|I_N\setminus B|=o(N)$, i.e. \[ |B|=|I_N|-o(N)=\left(N-\lceil \alpha_N N\rceil +1\right)-o(N) = \left(1-\frac1e+o(1)\right)N. \]

Step 3: Build $A$ and verify minimality.

Let \[ A := B\cup\{1\}\subseteq\{1,\dots,N\}, \qquad \delta_1:=+1,\ \delta_b:=-1\ (b\in B). \] Then \[ \sum_{n\in A}\frac{\delta_n}{n} = 1-\sum_{b\in B}\frac1b = 0. \] Now let $\varnothing\neq A'\subsetneq A$.

• If $1\notin A'$, then $A'\subseteq B$ and $\sum_{n\in A'}\frac{\delta_n}{n}=-\sum_{b\in A'}\frac1b<0$, so it is nonzero.
• If $1\in A'$, write $A'=\{1\}\cup B'$ with $B'\subsetneq B$. Then \[ \sum_{n\in A'}\frac{\delta_n}{n} = 1-\sum_{b\in B'}\frac1b. \] Since $B'$ is a proper subset of $B$ and all terms are positive, $\sum_{b\in B'}\frac1b < \sum_{b\in B}\frac1b = 1$, hence the above sum is $>0$, in particular nonzero.

Thus $A$ satisfies all requirements and \[ |A|=|B|+1 \ge \left(1-\frac1e+o(1)\right)N. \]

6) VERIFICATION AND EDGE CASES

• The construction uses only positivity of the reciprocals to ensure strict inequality for proper subsets.
• For small $N$ this asymptotic statement is not informative; e.g. for $N=6$ one can take $A=\{1,2,3,6\}$ with signs $(+,-,-,-)$ since $1=\frac12+\frac13+\frac16$.

7) FINAL

UNRESOLVED.

• (i) Strongest proved partial result: $F(N)\ge (1-\frac1e+o(1))N$ via Croot's theorem and the explicit $A=B\cup\{1\}$ construction above.
• (ii) First remaining gap: obtain any nontrivial upper bound $F(N)\le (1-\varepsilon)N$ for some fixed $\varepsilon>0$, or determine the true asymptotic constant for $F(N)/N$.
• (iii) Next moves: study structure of minimal solutions of $\sum_{p\in P}1/p=\sum_{q\in Q}1/q$ with $P,Q\subseteq[1,N]$; try to rule out "too many" denominators by showing unavoidable sub-equalities among large sets of unit fractions.
• (iv) Minimal counterexample search: compute $F(N)$ for moderate $N$ by integer/rational exact search to guess the constant (if it exists), and to see whether extremals always have a "single positive term" form.

COMPLETION: 45%

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