ProblemP2994

Erdős Problem 41 on Distinct Triple Sums

Problem statement

Erdős Problem 41: A set $A \subseteq \mathbb{N}$ is said to satisfy the triple distinct sums condition if for any two finite subsets $I, J \subseteq A$ with $|I| = |J| = 3$, the equality $\sum_{i \in I} i = \sum_{j \in J} j$ implies $I = J$. In other words, all sums of three elements from $A$ are distinct aside from trivial coincidences arising from reordering the same three elements. Let $A \subseteq \mathbb{N}$ be an infinite set satisfying the triple distinct sums condition. For each positive integer $N$, let $|A \cap \{1, \ldots, N\}|$ denote the number of elements of $A$ that are at most $N$. Does the lower limit \[ \liminf_{N \to \infty} \frac{|A \cap \{1, \ldots, N\}|}{N^{1/3}} \] equal $0$?

Recorded solution requirements

Give a rigorous proof or counterexample that completely resolves this statement: Erdős Problem 41: A set $A \subseteq \mathbb{N}$ is said to satisfy the triple distinct sums condition if for any two finite subsets $I, J \subseteq A$ with $|I| = |J| = 3$, the equality $\sum_{i \in I} i = \sum_{j \in J} j$ implies $I = J$. In other words, all sums of three elements from $A$ are distinct aside from trivial coincidences arising from reordering the same three elements. Let $A \subseteq \mathbb{N}$ be an infinite set satisfying the triple distinct sums condition. For each positive integer $N$, let $|A \cap \{1, \ldots, N\}|$ denote the number of elements of $A$ that are at most $N$. Does the lower limit \[ \liminf_{N \to \infty} \frac{|A \cap \{1, \ldots, N\}|}{N^{1/3}} \] equal $0$?

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TheoremDB content license: CC0-1.0
Stable alias: erdos-problem-41-attempt-resolution-route
Source: www.erdosproblems.com ↗

Problem packetResearch packetR1239

R1239Recorded attemptOutcome: Open

Resolve the stated acceptance condition

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Authored summary

Give a rigorous proof or counterexample that completely resolves this statement: Erdős Problem 41: A set $A \subseteq \mathbb{N}$ is said to satisfy the triple distinct sums condition if for any two finite subsets $I, J \subseteq A$ with $|I| = |J| = 3$, the equality $\sum_{i \in I} i = \sum_{j \in J} j$ implies $I = J$. In other words, all sums of three elements from $A$ are distinct aside from trivial coincidences arising from reordering the same three elements. Let $A \subseteq \mathbb{N}$ be an infinite set satisfying the triple distinct sums condition. For each positive integer $N$, let $|A \cap \{1, \ldots, N\}|$ denote the number of elements of $A$ that are at most $N$. Does the lower limit \[ \liminf_{N \to \infty} \frac{|A \cap \{1, \ldots, N\}|}{N^{1/3}} \] equal $0$?

The author reports this result. The outcome applies to this attempt's recorded scope.

Attempt outcome: open strategy

Recorded scope: No scope is recorded.

Originating problem: Erdős Problem 41 on Distinct Triple Sums

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Authored record and scope

Authored title: Resolve the stated acceptance condition
Record type: attempt
Stored status: open_strategy
Evidence grade: self_reported

Work and source credit

Recorded action: No action description supplied.
Authored result summary: Read complete authored result summary
Give a rigorous proof or counterexample that completely resolves this statement: Erdős Problem 41: A set $A \subseteq \mathbb{N}$ is said to satisfy the triple distinct sums condition if for any two finite subsets $I, J \subseteq A$ with $|I| = |J| = 3$, the equality $\sum_{i \in I} i = \sum_{j \in J} j$ implies $I = J$. In other words, all sums of three elements from $A$ are distinct aside from trivial coincidences arising from reordering the same three elements. Let $A \subseteq \mathbb{N}$ be an infinite set satisfying the triple distinct sums condition. For each positive integer $N$, let $|A \cap \{1, \ldots, N\}|$ denote the number of elements of $A$ that are at most $N$. Does the lower limit \[ \liminf_{N \to \infty} \frac{|A \cap \{1, \ldots, N\}|}{N^{1/3}} \] equal $0$?
Reported outcome: No separate outcome supplied.
Recorded status: open_strategy
Recorded evidence grade: self_reported
Recorded scope: No explicit scope supplied.

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2Authored explanation

Target the displayed statement directly. Give a rigorous proof or counterexample that completely resolves this statement: Erdős Problem 41: A set $A \subseteq \mathbb{N}$ is said to satisfy the triple distinct sums condition if for any two finite subsets $I, J \subseteq A$ with $|I| = |J| = 3$, the equality $\sum_{i \in I} i = \sum_{j \in J} j$ implies $I = J$. In other words, all sums of three elements from $A$ are distinct aside from trivial coincidences arising from reordering the same three elements. Let $A \subseteq \mathbb{N}$ be an infinite set satisfying the triple distinct sums condition. For each positive integer $N$, let $|A \cap \{1, \ldots, N\}|$ denote the number of elements of $A$ that are at most $N$. Does the lower limit \[ \liminf_{N \to \infty} \frac{|A \cap \{1, \ldots, N\}|}{N^{1/3}} \] equal $0$? Preserve exact hypotheses, source locators, and any finite certificates so later work can distinguish a full resolution from partial progress.

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Agent handoff

Use this record for: For Erdős Problem 41 on Distinct Triple Sums, record erdos-problem-41-attempt-resolution-route (“Resolve the stated acceptance condition”) documents a concrete method, search boundary, or failed route. The record states: Give a rigorous proof or counterexample that completely resolves this statement: Erdős Problem 41: A set $A \subseteq \mathbb{N}$ is said to satisfy the triple distinct sums condition if for any two finite subsets $I, J \subseteq A$ with $|I| = |J| = 3$, the equality $\sum_{i \in I} i = \sum_{j \in J} j$ implies $I = J$.
Evidence boundary: The recorded outcome applies to this attempt's stated scope.
Remaining questions: Unrecorded.
Blockers: Unrecorded.
Reopen when: Unrecorded.
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Verification source: www.erdosproblems.com ↗, Editorial research route recorded 2026-07-31

4How it connects

Addresses

R1239addressesCurrent status and unresolved remainder
claim

Recorded for

R1239recorded forerdos problem 41
problem

Machine-readable record

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  "body": "Target the displayed statement directly. Give a rigorous proof or counterexample that completely resolves this statement: Erdős Problem 41: A set $A \\subseteq \\mathbb{N}$ is said to satisfy the triple distinct sums condition if for any two finite subsets $I, J \\subseteq A$ with $|I| = |J| = 3$, the equality $\\sum_{i \\in I} i = \\sum_{j \\in J} j$ implies $I = J$. In other words, all sums of three elements from $A$ are distinct aside from trivial coincidences arising from reordering the same three elements. Let $A \\subseteq \\mathbb{N}$ be an infinite set satisfying the triple distinct sums condition. For each positive integer $N$, let $|A \\cap \\{1, \\ldots, N\\}|$ denote the number of elements of $A$ that are at most $N$. Does the lower limit\n\\[\n\\liminf_{N \\to \\infty} \\frac{|A \\cap \\{1, \\ldots, N\\}|}{N^{1/3}}\n\\]\nequal $0$? Preserve exact hypotheses, source locators, and any finite certificates so later work can distinguish a full resolution from partial progress.",
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6Provenance

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Project: erdos-problem-41-source-review
Locator: Editorial research route recorded 2026-07-31
License: CC0-1.0
Contributors: TheoremDB maintainers
Source: www.erdosproblems.com ↗
Public record: R1239
Stable alias: erdos-problem-41-attempt-resolution-route
Projection: Reproduction fields are derived from the immutable record.

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Resolve the stated acceptance condition

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