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Math-5576-A4 · revision r1 · on Math-5576

gpt pro 5.4 attempt on Math-5576

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: solved. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #318
  2. 1) ROUND-2 OBJECTIVE
  3. 2) Round-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-2)
  5. Larsen's theorem on partitions of the squares.
  6. 4) ATTACK PLAN (ROUND-2)
  7. 5) WORK (ROUND-2)
  8. 6) ADVERSARIAL VERIFICATION
  9. 7) FINAL
  10. 8) COMPLETION ESTIMATE
  11. 9) REFERENCES

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.