# Sums of three cubes: 114, and a fourth way to write 3

Which integers are sums of three integer cubes? Numbers congruent to 4 or 5 modulo 9 never are, and it is conjectured that every other integer is, in infinitely many ways. Since 33 and 42 were found in 2019, the smallest open case is 114; a second, long-standing target is a fourth representation of 3.

## Problem

**Task A.** Find integers $x, y, z$ with
$$x^3 + y^3 + z^3 = 114.$$

**Task B.** Find integers $x, y, z$ with $x^3 + y^3 + z^3 = 3$ other than permutations of the three known solutions
$$(1, 1, 1), \qquad (4, 4, -5), \qquad (569936821221962380720,\ -569936821113563493509,\ -472715493453327032).$$

## Background

Every cube is $0$ or $\pm 1$ modulo 9, which rules out $k \equiv \pm 4 \pmod 9$; Heath-Brown's density heuristic predicts infinitely many solutions for every other $k$ [3]. Booker found the first representation of 33 in 2019 [1], and Booker and Sutherland then found 42 and the third representation of 3, answering a question of Mordell [2]. These solutions have coordinates far beyond the reach of naive search, so the absence of small solutions is weak evidence that none exist.

## What is known

- $114 \equiv -3 \pmod 9$, so there is no congruence obstruction; any solution has $x \equiv y \equiv z \equiv -1 \pmod 3$ (Epoch's write-up).
- A complete search in September 2019 found no solution of $x^3 + y^3 + z^3 = 114$ with $\min(|x|, |y|, |z|) \le 10^{17}$, and a 2020 search over a more skewed region reaching $|z| \le 10^{19}$ also found none [2], subject to the usual caveat about undetected computing errors.
- An independent number-field approach by Grantham and Walsh was also unsuccessful for 114 [4].
- Heath-Brown's heuristic predicts about $0.058 \log B$ solutions of height at most $B$ for $k = 114$, so solutions may be extremely large (Epoch's write-up, using the constant computed by Booker and Sutherland [2]).
- formal-conjectures states the general conjecture in Lean as `isSumOfThreeCubes_iff_mod_9` in [SumOfThreeCubes.lean](https://github.com/google-deepmind/formal-conjectures/blob/main/FormalConjectures/Wikipedia/SumOfThreeCubes.lean).

## What would count as progress

- An integer triple, checked exactly in microseconds.
- Epoch scores submissions with its own verifier, with a separate prompt for each target; any triple for 114, or any new triple for 3, is accepted.
- A search method whose cost grows more slowly than the search height, or a much sharper arithmetic filter, would matter beyond these two numbers; the public Booker–Sutherland implementation is the baseline.

## References

1. A. R. Booker, Cracking the problem with 33, *Research in Number Theory* 5 (2019), article 26. [doi:10.1007/s40993-019-0162-1](https://doi.org/10.1007/s40993-019-0162-1)
2. A. R. Booker and A. V. Sutherland, On a question of Mordell, *Proc. Natl. Acad. Sci. USA* 118 (2021), e2022377118. [doi:10.1073/pnas.2022377118](https://doi.org/10.1073/pnas.2022377118)
3. D. R. Heath-Brown, The density of zeros of forms for which weak approximation fails, *Math. Comp.* 59 (1992), 613–623. [doi:10.1090/S0025-5718-1992-1146835-5](https://doi.org/10.1090/S0025-5718-1992-1146835-5)
4. J. Grantham and P. G. Walsh, Representing integers as a sum of three cubes, 2022. [arXiv:2211.12149](https://arxiv.org/abs/2211.12149)

## Source

Imported from [FrontierMath: Open Problems, sum-of-three-cubes](https://epoch.ai/frontiermath/open-problems/sum-of-three-cubes), curated by Epoch AI. Open according to the source as of 2026-10-20. The statement above is written in our own words; see the source for the original wording, history and discussion.