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. Sattler also announced a proof of the final question, the set of squares excluding $1$, in [Sa82] and [Sa82b], but this never appeared.

Larsen has proved that the answer is yes in the case of squares excluding $1$.

View the LaTeX source

This page was last edited 01 April 2026. View history

• For the positive density case: if there exists some $n \in A$ such that (say) f(n) = -1, and the set B of multiples of n in A such that f(m) = 1 for all $m \in B$ has positive upper density, then the result follows from theorem 2 of Bloom in [1]: if we consider $C := \{ \frac{m}{n} \in \mathbb{N}/ m \in B\}$, there exists a finite $S' \subset C $ such that $\sum_{c \in S'} \frac{1}{c} = 1$, so that $\frac{f(n)}{n} + \sum_{c \in S'}\frac{f(cn)}{cn} = \frac{-1}{n} + \sum_{c \in S'}\frac{1}{cn} = \frac{-1}{n} + \frac{1}{n} = 0$, since we assume that f(cn) always equals 1. Therefore, the finite set $\{n\} \cup \{nc \in A / c \in S'\} \subset A$ satisfies the condition.
  I think this approach might give a proof of the arithmetic progressions case as well by taking any function f to this case.
  [1]: Bloom, T. F. (2021). On a density conjecture about unit fractions. arXiv preprint arXiv:2112.03726, 1(3), 4.

  Agustín Meza — 12:19 on 20 Dec 2025

  👍 0 📝 0 🤖 0

  • It is because of the simiral reasoning that I find the question for squares difficult. One has too few chances to make up for the opposite sign. But, on the other hand, there is still quite a lot of arithmetic identities involving sums of reciprocal squares, so nothing is finitely decideable.

    Vjeko_Kovac — 05:42 on 12 Jan 2026

    👍 0 📝 0 🤖 0

  • A kind of set which satisfy the conditions of the statement is the following: let $A^{+} \subset A$ be the elements of A for which f(a) and $A^{-} \subset A$ the ones for which f(a) = -1. If the sets $P^{+},P^{-}$ consisting of the primes in both $A^{+}$ and $A^{-}$ satisfy $\sum_{p \in P^{+}}\frac{1}{p}, \sum_{p \in P^{-}}\frac{1}{p} \to \infty$, then A satisfies the hypothesis of the problem.
    Assuming that both $A^{+},A^{-}$ have positive lower natural density (I think the method could apply as well in the general case), from the previous comment, if the proposition is false, then for any $p \in P^{-}$ the set of multiples of p in $A^{+}$ has zero natural density. Therefore, for any fixed $p \in P^{-}$ we would have $\mathbb{E}_{n \le x} 1_{pn \in A^{+}} \to_{x} 0$. Taking a finite subset $S \subset P^{-}$ for which $\sum_{p \in S}\frac{1}{p} \ge C\epsilon^{-2}$ for some arbitrarily small $\epsilon > 0$ and some constant $C > 0$, due to Elliott's inequality we can say that there exists at least some $p \in S$ for which $\mathbb{E}_{n \le \frac{x}{p}}1_{pn \in A^{+}} = \mathbb{E}_{n \le x}1_{n \in A^{+}} + O(\epsilon)$. But since $\mathbb{E}_{n \le \frac{x}{p}}1_{pn \in A^{+}} \to_{x} 0$, if we take some $\epsilon$ sufficiently smaller than the lower density of A, it would be a contradiction.

    Agustín Meza — 14:55 on 02 Feb 2026

    👍 0 📝 0 🤖 0

• I have uploaded a note which shows how the technical result from [825] can be applied to the case where $A$ is the set of squares, excluding $1$. The greedy method works quite well here as a prelude to applying the circle method. The same method should also work for $k$th powers, once the first few are excluded.

  (The site has been updated to address this comment.)
  DanielLarsen — 19:50 on 01 Feb 2026

  👍 3 📝 1 🤖 0

• The case of squares (the last remaining one) can probably be solved by brute-force using calculations on various square identities. Here's a note on it for the potential computer attack and a short explanation below. I couldn't do it on a laptop, will try to run it on a more powerful machine - if anyone is willing to try it too, they are most welcome. Also currently trying to run it through Aristostle to formalize if possible.
  
  To obtain a computer-verifiable proof one must produce a finite certificate of unsatisfiability for a finite universe. Concretely, fix a bound $N$ and consider the Boolean variables $X_n$ for $2\le n\le N$, interpreted as $X_n=\text{true}\iff g(n)=+1$ and $X_n=\text{false}\iff g(n)=-1$. For each exact reciprocal-square identity\[
  \sum_{u\in U}\frac1{u^2}=\sum_{v\in V}\frac1{v^2}
  \]in a chosen finite library $B$, and for each scale $m$ such that $mU\cup mV\subseteq\{2,\dots,N\}$, add two CNF clauses forbidding the two opposite-monochromatic sign patterns on the scaled sets:\[
  \neg\bigl(\forall u\in U\, X_{mu}\ \wedge\ \forall v\in V\, \neg X_{mv}\bigr),\qquad
  \neg\bigl(\forall u\in U\, \neg X_{mu}\ \wedge\ \forall v\in V\, X_{mv}\bigr),
  \]equivalently the clauses\[
  \Big(\bigvee_{u\in U}\neg X_{mu}\Big)\ \vee\ \Big(\bigvee_{v\in V}X_{mv}\Big),\qquad
  \Big(\bigvee_{u\in U}X_{mu}\Big)\ \vee\ \Big(\bigvee_{v\in V}\neg X_{mv}\Big).
  \]Finally, enforce that the coloring is non-constant by adding $\bigvee_{n=2}^N X_n$ and $\bigvee_{n=2}^N \neg X_n$. If a SAT solver returns UNSAT for the resulting CNF, then this constitutes a finite certificate that every non-constant $g:\{2,\dots,N\}\to\{\pm1\}$ realizes some scaled identity in $B$ with opposite constant signs on the two sides, hence yields a finite set $T\subset\{2,\dots,N\}$ with $\sum_{m\in T}g(m)/m^2=0$. A fully checkable proof is obtained by recording (i) the explicit identity library $B$ (each identity verified independently by exact rational arithmetic), (ii) the bound $N$, and (iii) the solver-produced unsatisfiability certificate (e.g. a DRAT/FRAT proof) together with a standard proof checker verifying the certificate against the CNF.

  Przemek Chojecki — 15:34 on 24 Jan 2026

  👍 0 📝 0 🤖 0

  • I think this approach shows that the problem is "verifiable" - checkable in finite time if true - but does not demonstrate that it is "decidable" - resolvable in finite time both if the statement is true and if it is false. The compactness theorem shows that if the claim for squares is true, then there is a finite bound $N$ on the coloring statement one needs to check, but if the claim is false, one has to check satisfiability for all $N$, and this cannot be done in finite time without some additional input such as a bound on $N$.
    
    One can speculate that Sattler had an approach like this but found out that his SAT problems were actually solvable for the $N$ that he tried, leaving him with no conclusion.

    TerenceTao — 16:23 on 24 Jan 2026

    👍 1 📝 0 🤖 0

    • I don't understand the reduction to finite $N$.
      
      It seems to me that the "non-constant" condition messes with things. No finite subset of the squares suffices to witness "non-constant" behavior, or on the flip side, it's easy to avoid solutions below any finite bound by setting $f(n)=1$ for all $n\le N$.
      
      I may be missing a trick with "rescaling", but I'm suspicious. One test case I'm considering is the function that is $1$ everywhere except $f(p^2)=-1$ for a single large prime $p$. I feel like finite searches can't account for that?
      
      [Update: maybe a better test case is $f(n^2)=-1$ iff $p\mid n$.]

      BorisAlexeev — 02:12 on 25 Jan 2026

      👍 0 📝 0 🤖 0

      • Oh, good point. I withdraw my previous remark, and will revert the classification of the problem to "open".

        TerenceTao — 06:08 on 25 Jan 2026

        👍 0 📝 0 🤖 0

• No.
  
  General counterexample (works for ANY infinite A⊆ℕ, hence for infinite arithmetic progressions, positive-density sets, and the squares ≥4):
  
  Let m := min(A). Define f:A→{−1,1} by
  f(m) = −1 and f(n)=+1 for all n∈A\{m}.
  Then f is non-constant.
  
  Claim: there is no finite nonempty S⊂A with Σ_{n∈S} f(n)n = 0.
  
  Proof:
  - If m∉S, then Σ_{n∈S} f(n)n = Σ_{n∈S} n > 0.
  - If m∈S, write S = {m} ∪ T with T⊂A\{m}.
  Then Σ_{n∈S} f(n)n = −m + Σ_{n∈T} n.
  If T=∅, the sum is −m < 0.
  If T≠∅, every n∈T satisfies n ≥ m+1 (in fact >m), so Σ_{n∈T} n ≥ m+1, hence the total is ≥ −m + (m+1) = 1 > 0.
  So the sum is never 0.
  
  Consequences:
  1) If A is an infinite arithmetic progression in ℕ (A={a+kd : k≥0}), take m=a and the same f. No such S exists.
  2) If A is any set of positive density (or any infinite subset of ℕ), same construction kills it.
  3) If A is the set of squares excluding 1, then m=4; set f(4)=−1 and f(n)=+1 for all other squares. Any S either sums to >0 (if 4∉S) or to −4 or at least −4+9>0 (if 4∈S). So again no such S exists.

  None — 02:22 on 12 Jan 2026

  👍 0 📝 0 🤖 5

  • The question involves $\sum_{n \in S} f(n)/n$, not $\sum_{n \in S} f(n)n$.
    
    Also,

    • It is recommended to use LaTeX formatting for math expressions on this site.
    • If you find that a question has an suspiciously simple proof or disproof (and particularly if it does not use the full strength of the hypotheses), one should consider the possibility that some misinterpretation of the problem has taken place (either by the reader, the website maintainer, or even by Erdos himself).
    • Any use of AI to generate content needs to be disclosed as per forum rules.

    TerenceTao — 03:04 on 12 Jan 2026

    👍 3 📝 0 🤖 0

• The most obvious counterexample that Adenwalla mentions (one even number and other numbers odd) is credited to Erdős in this paper of Sattler at the very beginning (p. 341). In that light I do not understand the omission about arbitrary sets of positive density. It seems that Erdős corrected himself between 1980 (publication year of the Erdős-Graham problem book) and 1982 (publication year of the above Sattler's paper).

  
  

  The first question on arithmetic progressions seems to have been solved by Sattler as well, in this paper.

  
  

  Sattler also claims the solution to the last question (on the squares greater than $1$) in both of the papers linked above (see the references). However, I did not find the announced third paper where this proof was supposed to be published. Perhaps Sattler could still be trusted. :-)

  
  

  All in all, I think it is safe to mark this problem as fully solved by R. Sattler.

  Vjeko_Kovac — 14:58 on 17 Aug 2025

  👍 0 📝 0 🤖 0

  • Not sure if it is safe to mark solved. If it is announced at some place and didnt appear, maybe some mistake was observed?!
    If it has been solved, likely due to pointing that it is true, it might be easy to rediscover as well. (in contrast to Fermat's last thr :p ) this theorem seems to suggest that the boring choices work)
    Checking Zbmath, I could only find 5 papers attributed to him and so not the desired ones on square free numbers.

    StijnC — 16:01 on 17 Aug 2025

    👍 0 📝 0 🤖 0

    • Ok, I don't mind treating the last question (on squares greater than $1$) as an open problem. I was swayed by the claim that the annonunced paper "On Erdös' property P1 for the sequence of squares" is "to appear", but no journal was mentioned and also, as you say, it is possible that a mistake was found afterwards.

      
      

      I don't see an easy proof or disproof. In fact, I personally find the question on squares harder than the other (mentioned and solved) cases. Just as you say, Graham's theorem (+ easy tricks) can be used to obtain representations like
      $$\frac {1} {2^2} = \frac {1} {3^2} + \frac {1} {4^2} + \frac {1} {5^2} \
      + \frac {1} {7^2} + \frac {1} {12^2} + \frac {1} {15^2} + \frac {1}
      {20^2} + \frac {1} {28^2} + \frac {1} {35^2} $$
      also for other fractions $1/n^2$, so a potential counterexample cannot be a function that is $-1$ on one number and $1$ on the remaining ones. On the one hand, the multitude of identities with several terms on both sides makes the positive answer somewhat likely. On the other hand, such identities get increasingly more complicated and harder to adjust, so I really don't know; I don't even dare to guess if the claim is true or false.

      Vjeko_Kovac — 16:58 on 17 Aug 2025

      👍 0 📝 0 🤖 0