[#P10] Sum of three cubes problem
Problem. The Diophantine equation \(x^3+y^3+z^3=k\) is solvable in integers \(x,y,z\) for each \(k\in\mathbb{Z}\) satisfying \(k\not\equiv\pm4\pmod 9\).
1Context
A simple congruence gives the only known general obstruction, while finding or excluding representations remains difficult.
2Problem setup
Definition 1 (Cubes modulo 9 are congruent only to 0, 1, or -1, so integers congruent to 4 or 5 modulo 9 are impossible). Cubes modulo 9 are congruent only to 0, 1, or -1, so integers congruent to 4 or 5 modulo 9 are impossible.
Definition 2 (The variables may be negative and need not be distinct). The variables may be negative and need not be distinct.
Remark 1. A simple congruence gives the only known general obstruction, while finding or excluding representations remains difficult.
3What counts as a solution
- Prove representability for every integer outside the two obstructed residue classes, or give an admissible integer and prove that it has no representation by three integer cubes.
1Status
Current status (Dated status and exact unresolved remainder). Unresolved in this packet after the dated source check. Strongest checked result: Booker and Sutherland found explicit representations for 33 and 42, completing every admissible positive k<=100. Universal representability outside residues 4 and 5 modulo 9 remains open. Exact unresolved remainder: Prove representation for every admissible integer, or prove one admissible integer is not representable.[1][2][3]
1Records
Notes and companion material
Original intake status. The cited research paper treats universal representability outside the modulo-9 obstruction as unresolved. The source and public status were checked on 2026-07-31. This is an admin-curated seed record, not an independent exhaustive literature review.
- Solutions can involve variables vastly larger than k. Check current computational tables before claiming a first representation of a particular integer.
Recorded example 1. 6 = 2^3 + (-1)^3 + (-1)^3.
Computational notes
- A search bounded by |x|, |y|, and |z| cannot prove that an unsolved target has no representation beyond that bound.
2See also
How to cite
TheoremDB contributors, “Sum of three cubes problem,” TheoremDB research memory, snapshot of July 31, 2026. https://theoremdb.org/statements/sum-of-three-cubes-problemThis page as plain text: sum-of-three-cubes-problem.md
This problem includes 2 records joined by 2 typed links, sourced from arxiv.org[1], current as of July 31, 2026.
1References
- Packet source. Victor Y. Wang, “Sums of cubes and the Ratios Conjectures”. arXiv:2108.03398 (2021). Victor Y. Wang, arXiv:2108.03398, abstract and introduction. ↗preprint · primary source · arXiv:2108.03398, checked 2026-07-31 · checked 2026-07-31Source use: original summary.The cited research paper treats universal representability outside the modulo-9 obstruction as unresolved. The source and public status were checked on 2026-07-22. This is an admin-curated seed record, not an independent exhaustive literature review.Also cited at Abstract and introduction.Also cited at Editorial research route recorded 2026-07-31.Source used to formulate or check the problem record.Source used to assess the problem's recorded status.Packet-linked current theoretical framing.Source named by the research packet.
- Andrew V. Sutherland, Sums of cubes, project results page, joint work with Andrew R. Booker. math.mit.edu checked 2026-08-01. Displayed identities. ↗website · primary source · checked 2026-08-01Source use: original summary.Primary project record for computed representations.
- Edward Dunne, 42, AMS Beyond Reviews, 11 September 2019. Representation of 42 and status through 100. ↗website · secondary source · checked 2026-08-01Source use: original summary.Explains completion of admissible cases through 100.
An original CC0 restatement prepared by TheoremDB maintainers.