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

Write 114, the smallest open case, as a sum of three integer cubes, or find a fourth representation of 3 as a sum of three cubes.

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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,zx, y, z with

x3+y3+z3=114.x^3 + y^3 + z^3 = 114.

Task B. Find integers x,y,zx, y, z with x3+y3+z3=3x^3 + y^3 + z^3 = 3 other than permutations of the three known solutions

(1,1,1),(4,4,−5),(569936821221962380720, −569936821113563493509, −472715493453327032).(1, 1, 1), \qquad (4, 4, -5), \qquad (569936821221962380720,\ -569936821113563493509,\ -472715493453327032).

Background

Every cube is 00 or ±1\pm 1 modulo 9, which rules out k≡±4(mod9)k \equiv \pm 4 \pmod 9; Heath-Brown's density heuristic predicts infinitely many solutions for every other kk [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≡−3(mod9)114 \equiv -3 \pmod 9, so there is no congruence obstruction; any solution has x≡y≡z≡−1(mod3)x \equiv y \equiv z \equiv -1 \pmod 3 (Epoch's write-up).
  • A complete search in September 2019 found no solution of x3+y3+z3=114x^3 + y^3 + z^3 = 114 with min⁡(∣x∣,∣y∣,∣z∣)≤1017\min(|x|, |y|, |z|) \le 10^{17}, and a 2020 search over a more skewed region reaching ∣z∣≤1019|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 0.058log⁡B0.058 \log B solutions of height at most BB for k=114k = 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.

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
  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
  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
  4. J. Grantham and P. G. Walsh, Representing integers as a sum of three cubes, 2022. arXiv:2211.12149

Source

Imported from 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.