OPEN
This is open, and cannot be resolved with a finite computation.
Let $A\subseteq \mathbb{N}$ be an infinite arithmetic progression and $f:A\to \{-1,1\}$ be a non-constant function. Must there exist a finite non-empty $S\subset A$ such that\[\sum_{n\in S}\frac{f(n)}{n}=0?\]What about if $A$ is an arbitrary set of positive density? What if $A$ is the set of squares excluding $1$?
Erdős and Straus
[ErSt75] proved this when $A=\mathbb{N}$. Sattler
[Sa75] proved this when $A$ is the set of odd numbers. For the squares $1$ must be excluded or the result is trivially false, since\[\sum_{k\geq 2}\frac{1}{k^2}<1.\]This is false for some sets $A$ of positive density - indeed, it fails for any set $A$ containing exactly one even number. (Sattler
[Sa82] credits this observation to Erdős, who presumably found this after
[ErGr80].)
Sattler
[Sa82b] proved the answer to the original question is yes, in that any arithmetic progression has this property.
The final question of the set of squares excluding $1$ appears to be open - Sattler announced a proof in
[Sa82] and
[Sa82b], but this never appeared.
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This page was last edited 20 December 2025.
Additional thanks to: Sarosh Adenwalla, Hayato Egami, Vjekoslav Kovac, and Desmond Weisenberg
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #318, https://www.erdosproblems.com/318, accessed 2026-01-16
