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Math-5576 · mathematics · Mathematics

Erdős Problem #318

Status Solved · erdosproblems.comLLM status Solved from source status

Statement

Let $A\subseteq\mathbb{N}$ be infinite and let $f:A\to\{-1,1\}$ be non-constant. Say that $(A,f)$ has the “zero-sum” property if there exists a finite nonempty $S\subset A$ such that \[\sum_{n\in S}\frac{f(n)}{n}=0.\tag{$\ast$}\]

Question 1: If $A$ is an infinite arithmetic progression, must $(A,f)$ have the zero-sum property for every non-constant $f$?

Question 2: If $A$ has positive density, must $(A,f)$ have the zero-sum property for every non-constant $f$? (Statement says: no.)

Question 3: If $A=\{k^2:k\ge 2\}$ (squares excluding $1$), must $(A,f)$ have the zero-sum property?

Restatement from Math-5576-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 6

Math-5576-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-5576

Erdős Problem #318: signed reciprocal sums

Partial reconstruction and reductions, 24 September 2026.

1) Statement and scope

Let $A\subseteq\mathbb N$ be an infinite arithmetic progression, and let $f:A\to\{-1,1\}$ be nonconstant. Must a finite nonempty $S\subseteq A$ satisfy \[\sum_{n\in S}\frac{f(n)}n=0?\] The source also asks about arbitrary positive-density sets and about $A=\{k^2:k\ge2\}$. Below the progression question is proved using an explicitly stated density theorem, and the arbitrary-density question is disproved. The general square-denominator question is not reconstructed.

2) External input

We use Bloom's positive-density unit-fraction theorem: if $D\subseteq\mathbb N$ has positive upper asymptotic density, some finite nonempty $T\subseteq D$ satisfies $\sum_{t\in T}1/t=1$. This is Theorem 2 of On a density conjecture about unit fractions, https://arxiv.org/abs/2112.03726. Its proof is an external dependency; the progression reduction below supplies all additional steps.

3) Every progression: a common-multiple reduction

Write $A=\{a+dk:k\ge0\}$, with $a,d\ge1$. Choose $x,y\in A$ with $f(x)=1$ and $f(y)=-1$, and put $L=\operatorname{lcm}(x,y)$. Since \[\gcd(x,d)=\gcd(y,d)=\gcd(a,d),\] comparison of prime valuations gives $\gcd(L,d)=\gcd(a,d)$ as well. Thus the congruences $z\equiv a\pmod d$, $z\equiv0\pmod L$ are compatible. After omitting a finite initial segment, their intersection $C=A\cap L\mathbb N$ is an arithmetic progression of positive density.

At least one sign class $C_\sigma=\{z\in C:f(z)=\sigma\}$ has positive upper density: otherwise their union would have upper density zero. Choose $n\in\{x,y\}$ with $f(n)=-\sigma$. Every member of $C_\sigma$ is divisible by $n$, so \[D=\{z/n:z\in C_\sigma\}\subseteq\mathbb N\] has positive upper density. Explicitly $D(X)=C_\sigma(nX)$, which gives this assertion along a subsequence witnessing the positive upper density of $C_\sigma$.

Apply the external theorem to $D$, obtaining $T$ with reciprocal sum one. The set $S=\{n\}\cup\{nt:t\in T\}$ is a disjoint union because all $nt$ have sign $\sigma$, whereas $n$ has sign $-\sigma$. It is a subset of $A$, and \[\sum_{z\in S}\frac{f(z)}z=-\frac\sigma n+\frac\sigma n\sum_{t\in T}\frac1t=0.\] This proves the full progression assertion, including nonprimitive progressions and arbitrary finite starting points.

4) Positive density alone is insufficient

Take $A=\{1,3,5,\ldots\}\cup\{2\}$, assigning sign $-1$ to 2 and $+1$ to every odd integer. A zero sum would have to contain 2 and satisfy $\sum_{n\in S\setminus\{2\}}1/n=1/2$. The left side has odd denominator after reduction, since the least common multiple of its denominators is odd. It cannot equal $1/2$. Thus $A$ has density $1/2$ and gives a counterexample.

More generally, if a set has one element whose $p$-adic valuation strictly exceeds that of every other element, the same denominator argument prevents a zero sum when that element is the only negative term. This observation is used only for finite candidate sums, so no convergence assumption is involved.

5) What can be checked for squares

Excluding 1 is necessary: $\sum_{k\ge2}1/k^2\le1/4+\int_2^\infty x^{-2}\,dx=3/4<1$. Therefore sign $-1$ at 1 and sign $+1$ at all other squares cannot give a zero sum. With 1 excluded, at least the analogous coloring with only 4 negative does admit one, by the exact identity \[\frac14=\frac1{3^2}+\frac1{4^2}+\frac1{5^2}+\frac1{7^2}+\frac1{12^2}+\frac1{15^2}+\frac1{20^2}+\frac1{28^2}+\frac1{35^2}.\] Here the $1/4^2$ term has denominator 16, distinct from the negative term with denominator 4. Multiplication by $420^2$ verifies the identity in integers. This single coloring is not a proof for arbitrary assignments of signs to the squares. The density theorem cannot be applied directly to the squares, which have density zero.

6) Outcome and attribution

UNRESOLVED in this attempt for the full three-part problem. The progression part is complete conditional on the stated published density theorem; the positive-density variant has a complete elementary counterexample. The remaining gap is the affirmative assertion for every nonconstant coloring of squares excluding 1. The source reports a proof by Larsen, which is not reproduced here. No new theorem or independent formal verification is claimed.

Sources: https://www.erdosproblems.com/318 and its discussion, checked 23 September 2026; Bloom's paper above. The progression result is historically due to Sattler, and the density counterexample is credited to Erdős. The common-multiple reduction is also suggested in the source discussion.

The estimate measures coverage of the full multipart question, not correctness probability.

7) COMPLETION ESTIMATE

COMPLETION: 70%

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Math-5576-A6 · No claim recorded · LLM-generated · gpt pro 5.4 ·

gpt pro 5.4 attempt on Math-5576, version 3

Erdős Problem #318

Clean write-up from two explicit external inputs

This note is not self-contained in the sense of reproving Bloom's theorem or Larsen's theorem. Those two results are stated explicitly as inputs and then used to deduce all three answers to Erdős Problem #318. Apart from those two inputs, the argument is elementary.

Problem statement

Let $A\subseteq \mathbb{N}$ be an infinite arithmetic progression and let $f\colon A\to\{-1,1\}$ be non-constant. Must there exist a finite nonempty set $S\subseteq A$ such that \[ \sum_{n\in S}\frac{f(n)}{n}=0? \] What happens if $A$ is merely a set of positive density? What happens if \[ A=\{4,9,16,25,\dots\} \] is the set of perfect squares greater than $1$?

Density notation

For any $B\subseteq \mathbb{N}$, write \[ \overline d(B):=\limsup_{x\to\infty}\frac{|B\cap[1,x]|}{x}, \qquad \underline d(B):=\liminf_{x\to\infty}\frac{|B\cap[1,x]|}{x}. \]

External inputs

We use the following two results as black boxes.
Input (Bloom): If $B\subseteq \mathbb{N}$ has positive upper density, then there exists a finite set $T\subseteq B$ such that \[ \sum_{t\in T}\frac1t=1. \]
Input (Larsen): If $X$ and $Y$ form a partition of the set of perfect squares greater than $1$ into two nonempty parts, then there exist finite nonempty sets $X'\subseteq X$ and $Y'\subseteq Y$ such that \[ \sum_{x\in X'}\frac1x=\sum_{y\in Y'}\frac1y. \]

Main theorem

Theorem: For Erdős Problem #318, the answers are:
  1. for an infinite arithmetic progression $A$, yes;
  2. for an arbitrary set $A$ of positive density, no;
  3. for the set $A=\{4,9,16,25,\dots\}$ of squares excluding $1$, yes.
Proof: We prove the three assertions in order.

1. Arithmetic progressions. Let \[ A=\{a+kd:k\ge 0\} \] with $a,d\in\mathbb{N}$, and let $f\colon A\to\{-1,1\}$ be non-constant. Define \[ P:=\{n\in A:f(n)=1\}, \qquad N:=\{n\in A:f(n)=-1\}. \] Since $A$ has natural density $1/d$, at least one of $P$ and $N$ has positive upper density.

We first need a simple lemma.

Lemma: For every $n_0\in A$, the set \[ A_{n_0}:=\{m\in A:n_0\mid m\} \] is an infinite arithmetic progression, and hence has positive natural density.
Proof: Write \[ n_0=a+k_0d \] for some $k_0\ge 0$. An element $m=a+kd\in A$ belongs to $A_{n_0}$ if and only if \[ n_0\mid a+kd. \] Since $n_0=a+k_0d$, this is equivalent to \[ n_0\mid d(k-k_0). \] Let $g=\gcd(d,n_0)$. Then \[ n_0\mid d(k-k_0) \iff \frac{n_0}{g}\mid (k-k_0). \] Hence the admissible values of $k$ form an infinite arithmetic progression with common difference $n_0/g$. Therefore the corresponding values of $m=a+kd$ form an arithmetic progression with common difference \[ d\cdot\frac{n_0}{g}=\frac{dn_0}{g}=\operatorname{lcm}(d,n_0). \] So $A_{n_0}$ is an infinite arithmetic progression and in particular has positive natural density. ∎

We now consider two cases.

Case 1: both $P$ and $N$ have positive upper density. By Bloom's theorem, there exist finite sets \[ S_+\subseteq P, \qquad S_-\subseteq N \] such that \[ \sum_{n\in S_+}\frac1n=1, \qquad \sum_{n\in S_-}\frac1n=1. \] Since $P\cap N=\varnothing$, the set $S:=S_+\cup S_-$ is finite and nonempty, and \[ \sum_{n\in S}\frac{f(n)}{n} =\sum_{n\in S_+}\frac1n-\sum_{n\in S_-}\frac1n =1-1=0. \] This proves the claim in Case 1.

Case 2: one sign class has upper density $0$. Without loss of generality, assume \[ \overline d(N)=0. \] Because $f$ is non-constant and takes values only in $\{-1,1\}$, the set $N$ is nonempty; choose $n_0\in N$. By the lemma, $A_{n_0}$ has positive natural density; denote this density by $\delta>0$. Set \[ M:=P\cap A_{n_0}. \] Then \[ M=A_{n_0}\setminus (N\cap A_{n_0}). \] Since $N\cap A_{n_0}\subseteq N$, we have \[ \overline d(N\cap A_{n_0})\le \overline d(N)=0. \] Also $A_{n_0}$ has natural density $\delta$, so \[ \underline d(M) \ge \underline d(A_{n_0})-\overline d(N\cap A_{n_0}) =\delta-0 =\delta>0. \] Thus $M$ has positive lower density.

Now define \[ B:=\left\{\frac{m}{n_0}:m\in M\right\}\subseteq \mathbb{N}. \] Because $M=n_0B$, for every $x\ge 1$ we have \[ |B\cap[1,x]|=|M\cap[1,n_0x]|. \] Hence \[ \underline d(B) =\liminf_{x\to\infty}\frac{|B\cap[1,x]|}{x} = n_0\liminf_{x\to\infty}\frac{|M\cap[1,n_0x]|}{n_0x} \ge n_0\,\underline d(M)>0. \] In particular, $B$ has positive upper density. By Bloom's theorem, there exists a finite set $T\subseteq B$ such that \[ \sum_{t\in T}\frac1t=1. \] Define \[ S_+:=\{n_0t:t\in T\}. \] Then $S_+\subseteq M\subseteq P$, and \[ \sum_{m\in S_+}\frac1m =\sum_{t\in T}\frac{1}{n_0t} =\frac1{n_0}\sum_{t\in T}\frac1t =\frac1{n_0}. \] Now set \[ S:=S_+\cup\{n_0\}. \] Since $f(n_0)=-1$ and $f(m)=1$ for every $m\in S_+$, we obtain \[ \sum_{n\in S}\frac{f(n)}{n} =\sum_{m\in S_+}\frac1m-\frac1{n_0} =\frac1{n_0}-\frac1{n_0} =0. \] This proves the arithmetic-progression case.

2. Arbitrary positive-density sets. Here the statement is false. Take \[ A:=\{2\}\cup\{\text{odd positive integers}\}. \] This set has natural density $1/2$. Define $f\colon A\to\{-1,1\}$ by \[ f(2)=-1, \qquad f(n)=1 \quad\text{for every odd } n. \] Let $S\subseteq A$ be finite and nonempty.

If $2\notin S$, then \[ \sum_{n\in S}\frac{f(n)}{n}=\sum_{n\in S}\frac1n>0. \]

If $2\in S$, write \[ T:=S\setminus\{2\}. \] Then every element of $T$ is odd, and \[ \sum_{n\in S}\frac{f(n)}{n} =\sum_{n\in T}\frac1n-\frac12. \] Now $\sum_{n\in T}1/n$ is a rational number whose reduced denominator is odd: indeed, with common denominator \[ D:=\prod_{n\in T} n, \] which is odd, we can write \[ \sum_{n\in T}\frac1n=\frac{M}{D} \] for some integer $M$; cancelling common factors cannot introduce a factor $2$ into the denominator. Therefore \[ \sum_{n\in T}\frac1n\ne \frac12. \] So in this case as well, \[ \sum_{n\in S}\frac{f(n)}{n}\ne 0. \] Hence there exists a set of positive density for which no such finite set $S$ exists.

3. The set of squares greater than $1$. Let \[ A=\{4,9,16,25,\dots\}, \] and let $f\colon A\to\{-1,1\}$ be non-constant. Define \[ X:=\{n\in A:f(n)=1\}, \qquad Y:=\{n\in A:f(n)=-1\}. \] Because $f$ is non-constant and its codomain is exactly $\{-1,1\}$, both $X$ and $Y$ are nonempty, and they form a partition of $A$. By Larsen's theorem, there exist finite nonempty sets \[ X'\subseteq X, \qquad Y'\subseteq Y \] such that \[ \sum_{x\in X'}\frac1x=\sum_{y\in Y'}\frac1y. \] Let \[ S:=X'\cup Y'. \] Then $S\subseteq A$ is finite and nonempty, and \[ \sum_{n\in S}\frac{f(n)}{n} =\sum_{x\in X'}\frac1x-\sum_{y\in Y'}\frac1y =0. \] This proves the squares case.

Combining the three parts, the answers to Erdős Problem #318 are:

  1. arithmetic progressions: yes;
  2. arbitrary positive-density sets: no;
  3. squares greater than $1$: yes.
∎

References

  • [BloomJEMS] Thomas F. Bloom, On a density conjecture about unit fractions (with an appendix by Thomas F. Bloom and Bhavik Mehta), Journal of the European Mathematical Society 27 $2025$, no. 11, 4563--4589. DOI: 10.4171/JEMS/1456.
  • [ErdosGraham] Paul Erdős and Ronald L. Graham, Old and New Problems and Results in Combinatorial Number Theory, Monographies de L'Enseignement Mathématique, vol. 28, L'Enseignement Mathématique, Geneva, 1980.
  • [Larsen] Daniel Larsen, Sufficiently abundant numbers are pseudoperfect, 2026 note. The theorem on partitions of the perfect squares greater than $1$ appears in the public source file GitHub source file.
  • [Sattler] R. Sattler, On Erdős property $P_1$ for the arithmetical sequence, Indagationes Mathematicae (Proceedings) 85 $1982$, 347--352. DOI: 10.1016/1385-7258$82$90026-9.

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Math-5576-A5 · No claim recorded · LLM-generated · gpt pro 5.4 ·

gpt pro 5.4 attempt on Math-5576, version 2

Erdős Problem #318

1) ROUND-2 OBJECTIVE

Path pursued: (A) full proof. Round 1 already settled the arithmetic-progression case and the positive-density counterexample case, leaving only the squares case open. In this round I close that exact gap by applying a new external theorem of Daniel Larsen whose statement matches the missing squares case precisely.

2) Round-1 FOUNDATION USED

I use the following Round 1 conclusions without re-proving them.
  1. For an infinite arithmetic progression $A$ and a non-constant function $f:A\to\{-1,1\}$, there does exist a finite nonempty $S\subset A$ with \[ \sum_{n\in S} \frac{f(n)}{n}=0. \]
  2. For arbitrary positive-density sets, the statement is false; Round 1 gave an explicit counterexample of parity type.
  3. The only unresolved part after Round 1 was the case where $A$ is the set of perfect squares greater than $1$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is the following theorem from Daniel Larsen's 2026 note Sufficiently abundant numbers are pseudoperfect.
Larsen's theorem on partitions of the squares.
Let $X$ and $Y$ be any partition of the set of perfect squares greater than $1$ into two nonempty parts. Then there exist nonempty finite subsets $X'\subset X$ and $Y'\subset Y$ such that \[ \sum_{x\in X'} \frac1x = \sum_{y\in Y'} \frac1y. \] This theorem was not available in Round 1; it closes the only missing case.

4) ATTACK PLAN (ROUND-2)

The gap left after Round 1 is sharply formulated: given a non-constant sign function on the squares greater than $1$, prove the existence of a finite zero-sum. Larsen's theorem is tailor-made for this. The plan is simply to translate the sign function into a two-colouring of the squares and then apply Larsen's theorem.

5) WORK (ROUND-2)

Let \[ A=\{4,9,16,25,\dots\} \] be the set of perfect squares greater than $1$, and let \[ f:A\to\{-1,1\} \] be non-constant. Define \[ X:=\{n\in A: f(n)=1\}, \qquad Y:=\{n\in A: f(n)=-1\}. \] Because $f$ is non-constant and takes values only in $\{-1,1\}$, both $X$ and $Y$ are nonempty, and they form a partition of $A$.

Apply Larsen's theorem to this partition. We obtain nonempty finite sets \[ X'\subset X, \qquad Y'\subset Y \] such that \[ \sum_{x\in X'} \frac1x = \sum_{y\in Y'} \frac1y. \] Set \[ S:=X'\cup Y'. \] Then $S$ is a finite nonempty subset of $A$, and \[ \sum_{n\in S} \frac{f(n)}{n} = \sum_{x\in X'} \frac{1}{x} - \sum_{y\in Y'} \frac{1}{y} =0. \] So the squares case has an affirmative answer.

Combining this with the Round 1 foundation, the complete resolution of Problem #318 is now:

  1. for infinite arithmetic progressions: yes;
  2. for arbitrary positive-density sets: no;
  3. for the set of perfect squares greater than $1$: yes.

6) ADVERSARIAL VERIFICATION

I checked the new deduction against the likely failure modes.
  1. Does non-constant imply both colour classes are nonempty? Yes, because the codomain is exactly $\{-1,1\}$.
  2. Does Larsen's theorem apply to exactly this domain? Yes. Its statement is about partitions of the set of perfect squares greater than $1$, which is exactly the set $A$.
  3. Are the subsets produced nonempty and finite? Yes; that is part of Larsen's theorem.
  4. Is there any overlap issue in $S=X'\cup Y'$? No, because $X$ and $Y$ are disjoint parts of a partition.
  5. Am I re-proving Larsen's theorem? No. The Round 2 advance here is a gap-free application of an external theorem whose hypotheses match the remaining open case exactly.

7) FINAL

FULL SOLUTION -- FULL PROOF.

Using Larsen's theorem on partitions of the perfect squares greater than $1$, the squares case is affirmative. Together with the Round 1 results for arithmetic progressions and for positive-density sets, this completely resolves Erdős Problem #318.

8) COMPLETION ESTIMATE

COMPLETION: 100%

9) REFERENCES

  1. T. F. Bloom, Erdős Problem #318, problem page, accessed 2026-03-07.
  2. D. Larsen, Sufficiently abundant numbers are pseudoperfect, 2026 note; theorem on partitions of the perfect squares greater than $1$ checked from the publicly posted source file.

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