gpt pro 5.2 attempt on Math-5576

Erdős Problem #318

1) RESTATE

Let $A\subset\mathbb{N}$ be an infinite arithmetic progression and let $f:A\to\{-1,1\}$ be non-constant.

1. Must there exist a finite nonempty set $S\subset A$ such that \[ \sum_{n\in S}\frac{f(n)}{n}=0\ ? \]
2. What if $A\subset\mathbb{N}$ is any set of positive (upper) density?
3. What if $A=\{4,9,16,25,\dots\}$ is the set of squares excluding $1$?

2) KNOWN FACTS

1. For $A=\mathbb{N}$ the answer to $1a$ is yes (Erdős--Straus).
2. For $A$ an arithmetic progression, the answer to $1a$ is known to be yes (Sattler).
3. For general positive-density sets, the answer to $1b$ is no: there are counterexamples.
4. For squares $\{n^2:n\ge 2\}$, the problem appears open.
5. (Modern tool) Bloom proved that any subset $B\subset\mathbb{N}$ of positive upper density contains a finite $T\subset B$ with $\sum_{t\in T} 1/t=1$.

3) PROOF STRATEGY

Provide a direct proof of $1a$ for arithmetic progressions using Bloom's density theorem on unit fractions. Then give a parity-based counterexample for $1b$. Finally, record the (apparently) open status for squares.

4) ATTEMPTED PROOF

Part $1a$: arithmetic progressions.

Let $A=\{a+kd:k\ge 0\}$ with $a,d\in\mathbb{N}$. Partition $A$ into \[ P:=\{n\in A:f(n)=+1\},\qquad N:=\{n\in A:f(n)=-1\}. \] Since $A$ has positive natural density $1/d$, at least one of $P,N$ has positive upper density in $\mathbb{N}$.

Lemma 4.1 (multiples of an element stay dense in an arithmetic progression). Fix $n_0\in A$. Then \[ A_{n_0}:=\{m\in A: n_0\mid m\} \] is an infinite arithmetic progression, hence has positive density. Proof. We need solutions to the congruence $a+kd\equiv 0\pmod{n_0}$. Because $n_0\in A$, we have $n_0\equiv a\pmod d$, so $\gcd(d,n_0)\mid a$ and the congruence has a solution $k_0$. All solutions are $k\equiv k_0\pmod{n_0/\gcd(d,n_0)}$, hence $A_{n_0}$ is an arithmetic progression with common difference $\operatorname{lcm}(d,n_0)=dn_0/\gcd(d,n_0)$. $\square$

Lemma 4.2 (Bloom's density theorem, used as a black box). If $B\subset\mathbb{N}$ has positive upper density, then there exists a finite $T\subset B$ with $\sum_{t\in T}1/t=1$. $\square$

Case 1: both $P$ and $N$ have positive upper density. By Lemma 4.2 choose finite $S_+\subset P$ and $S_-\subset N$ with $\sum_{n\in S_+}1/n=1$ and $\sum_{n\in S_-}1/n=1$. Then for $S:=S_+\cup S_-$ we have \[ \sum_{n\in S}\frac{f(n)}{n}=\sum_{n\in S_+}\frac{1}{n}-\sum_{n\in S_-}\frac{1}{n}=1-1=0. \]

Case 2: one sign class has zero upper density. Assume without loss of generality that $\overline d(N)=0$ (upper density $0$), so $P$ has the same positive upper density as $A$. Pick any $n_0\in N$. By Lemma 4.1, $A_{n_0}$ has positive density. Since $N\cap A_{n_0}\subset N$ has upper density $0$, the set \[ M:=P\cap A_{n_0}=A_{n_0}\setminus (N\cap A_{n_0}) \] has the same positive density as $A_{n_0}$. Define the scaled set \[ B:=\left\{\frac{m}{n_0}: m\in M\right\}\subset \mathbb{N}. \] Because $M=n_0 B$, $B$ also has positive upper density. By Lemma 4.2 there exists finite $T\subset B$ with $\sum_{t\in T}1/t=1$. Let $S_+:=\{n_0 t: t\in T\}\subset M\subset P$. Then \[ \sum_{m\in S_+}\frac{1}{m}=\sum_{t\in T}\frac{1}{n_0 t}=\frac{1}{n_0}\sum_{t\in T}\frac{1}{t}=\frac{1}{n_0}. \] Finally, set $S:=S_+\cup\{n_0\}$. Since $f(n_0)=-1$ and $f\equiv +1$ on $S_+$, \[ \sum_{n\in S}\frac{f(n)}{n}=\sum_{m\in S_+}\frac{1}{m}-\frac{1}{n_0}=\frac{1}{n_0}-\frac{1}{n_0}=0. \] This completes a proof of $1a$ for any arithmetic progression $A$.

Part $1b$: positive density sets. Counterexample.

Let $A:=\{2\}\cup\{\text{all odd positive integers}\}$. Then $A$ has natural density $1/2$. Define $f(2)=-1$ and $f(n)=+1$ for all odd $n$. Take any finite nonempty $S\subset A$. If $2\notin S$, then $\sum_{n\in S} f(n)/n>0$. If $2\in S$, then \[ \sum_{n\in S}\frac{f(n)}{n}=\sum_{\substack{n\in S\\ n\ \text{odd}}}\frac{1}{n}-\frac{1}{2}. \] But a sum of reciprocals of odd integers is a rational number whose reduced denominator is odd (a common denominator is the product of the odd integers, hence odd; cancelling a gcd cannot introduce a factor $2$). Therefore it cannot equal $1/2$. Hence the displayed quantity cannot be $0$. So $1b$ has a negative answer.

Part $1c$: squares.

For $A=\{n^2:n\ge 2\}$, the answer appears to be open.

5) OBSTACLES

1. Part $1a$ above relies on Bloom's deep theorem about unit fractions in positive density sets; obtaining an elementary proof specialized to arithmetic progressions would be interesting but is not provided here.
2. For squares, current methods (including density-based ones) do not apply because the squares have zero density and the reciprocal sum $\sum 1/n^2$ converges.

6) FINAL

UNRESOLVED.

1. What I tried: Gave a complete proof of the arithmetic progression case using Bloom's density theorem; gave an explicit counterexample for the positive-density variant.
2. Where it fails / stuck: The squares case remains open; I do not have a proof or counterexample there.
3. What might work next: Look for square-specific Egyptian-fraction identities (reciprocal-square identities) that could simulate the "density" argument, or find an obstruction (e.g. via denominator parity/valuation) robust enough to rule out zero sums.
4. Confidence: High confidence in the proof of $1a$ conditional on Bloom's theorem, and in the counterexample for $1b$. The open status of $1c$ is based on current literature reports.

7) WRITE-UP (clean, complete)

For an arithmetic progression $A$ and a non-constant $f:A\to\{-1,1\}$, let $P$ and $N$ be the $+1$ and $-1$ classes. If both have positive upper density, Bloom's theorem gives finite subsets of each whose reciprocal sums equal $1$, and their signed union has sum $0$. If one class (say $N$) has zero upper density, choose $n_0\in N$. The set of multiples of $n_0$ inside $A$ is itself an arithmetic progression, hence has positive density. Removing the density-zero subset of those multiples lying in $N$ leaves a positive-density set $M\subset P$ of positive multiples of $n_0$. Scaling by $n_0$ produces a positive-density set $B\subset\mathbb{N}$, so Bloom's theorem gives a finite $T\subset B$ with $\sum_{t\in T}1/t=1$. Scaling back gives a finite $S_+\subset P$ with $\sum_{m\in S_+}1/m=1/n_0$, so $S_+\cup\{n_0\}$ yields a signed reciprocal sum $0$.

For general positive-density sets, the claim is false: taking $A=\{2\}\cup\{\text{odds}\}$ and $f(2)=-1$, $f(\text{odd})=+1$ gives a parity obstruction, since a sum of reciprocals of odd integers can never equal $1/2$.

For the set of squares $\{n^2:n\ge 2\}$, the problem appears open.