The Diophantine equation $$x^3+y^3+z^3=3$$ has four easy integer solutions: $(1,1,1)$ and the three permutations of $(4,4,-5)$. Elsenhans and Jahnel wrote in 2007 that these were all the solutions known at that time.

Are any other solutions known?

By a conjecture of Tyszka, it would follow that if this equation had finitely many roots, then each component of a solution tuple would be at most $2^{2^{12}/3} \lt 2^{1365.34}$ in absolute value. (To see this, it is enough to express the equation using a Diophantine system in 13 variables in the form considered by Tyszka.) This leaves a large gap, since Elsenhans and Jahnel only considered solutions with components up to $10^{14} \approx 2^{46.5}$ in absolute value. It is also not obvious whether Tyszka's conjecture is true.

OEIS sequence A173515 refers to equations of the form $x^3+y^3=z^3-n$, for $n$ a positive integer, as "Fermat near-misses". Infinite families of solutions are known for $n=\pm 1$, including one constructed by Ramanujan from generating functions (see Rowland's survey).

  • Andreas-Stephan Elsenhans and Jörg Jahnel, New sums of three cubes, Math. Comp. 78 (2009), 1227–1230. DOI: 10.1090/S0025-5718-08-02168-6. (preprint)
  • Apoloniusz Tyszka, A conjecture on integer arithmetic, Newsletter of the European Mathematical Society (75), March 2010, 56–57. (issue)
  • Eric S. Rowland, Known Families of Integer Solutions of $x^3+y^3+z^3=n$, 2005. (manuscript)
  • 1
    $\begingroup$ @Gerry Myerson: Beck et al. do not seem to have dealt with 3, nor do they seem to have gone above $10^{12}$ in any case. So Elsenhans and Jahnel dominate these results. $\endgroup$ Mar 12, 2011 at 13:26
  • 2
    $\begingroup$ For those who will not read further, Tyszka's conjecture is false (as he states below.) $\endgroup$ Dec 7, 2015 at 14:52
  • 2
    $\begingroup$ Andrew Sutherland and Andrew Booker, fresh from finding the first known integer solution to $x^3+y^3+z^3=42$ in September 2019, are apparently hoping to next find a third solution to $x^3+y^3+z^3=3$. $\endgroup$
    – David Roberts
    Sep 9, 2019 at 13:35
  • 18
    $\begingroup$ And they finally find out the first non-trivial solution!!! $569936821221962380720^3 + (-569936821113563493509)^3 + (-472715493453327032)^3$ $\endgroup$
    – Y. Zhao
    Sep 17, 2019 at 21:43
  • 9
    $\begingroup$ It occurs to me that these sorts of questions would be excellent challenge questions to pose to any psychics who claim to be in contact with super-intelligent aliens, since the solutions are already expected to be produced by computer search in a few years but would be instantly verifiable evidence of some extraordinary computational or intellectual resource if produced sooner. $\endgroup$
    – Terry Tao
    Jan 15, 2021 at 22:21

5 Answers 5


Just noticed this question. I agree with L.H.Gallardo that the problem is old (see e.g. Problem D5 in UPINT = Unsolved Problems in Number Theory by R.K.Guy), but not that it is hopeless: the usual heuristics suggest that the number of solutions with $\max(|x|,|y|,|z|) \leq H$ should be asymptotic to a multiple of $\log H$, so further solutions should eventually emerge (though it may indeed be hopeless to prove anything close to the $\log H$ heuristic).

See also my article

Rational points near curves and small nonzero $|x^3-y^2|$ via lattice reduction, Lecture Notes in Computer Science 1838 (proceedings of ANTS-4, 2000; W.Bosma, ed.), 33-63 = math.NT/0005139 on the arXiv.

Among other things it gives an algorithm for finding all solutions of $|x^3 + y^3 + z^3| \ll H$ with $\max(|x|,|y|,|z|) \leq H$ that should run (and in practice does run) in time $\widetilde{O}(H)$; since we expect the number of solutions to be asymptotically proportional to $H$, this means we find the solutions in little more time than it takes to write them down.

D.J.Bernstein has implemented the algorithm efficiently, and reports on the results of his and others' extensive computations at http://cr.yp.to/threecubes.html .

EDIT: for the specific problem $x^3+y^3+z^3=3$, Cassels showed that any solution must satisfy $x\equiv y\equiv z \bmod 9$ in this brief article:

A Note on the Diophantine Equation $x^3+y^3+z^3=3$, Math. of Computation 44 #169 (Jan.1985), 265-266.

This uses cubic reciprocity, and is stronger than what one can obtain from congruence conditions. See also Heath-Brown's paper "The Density of Zeros of Forms for which Weak Approximation Fails" (Math. of Computation 59 #200 (Oct.1992), 613-623), where he gives corresponding conditions for the homogeneous equation $x^3 + y^3 + z^3 = 3w^3$ and also $x^3 + y^3 + z^3 = 2w^3$, and reports that

In a letter to the author, Professor Colliot-Thélène has shown that the above congruence restrictions are exactly those implied by the Brauer-Manin obstruction. Moreover, for the general equation $x^3 + y^3 + z^3 = kw^3$, with a noncube integer $k$, there is always a nontrivial obstruction, eliminating two-thirds of the adèlic points.

  • 2
    $\begingroup$ Welcome to MO, Prof Elkies! $\endgroup$ Jun 3, 2011 at 0:41
  • 1
    $\begingroup$ Thanks, though it's not quite my début here: I answered a chess question (mathoverflow.net/questions/63423/checkmate-in-omega-moves) here a month or so ago. $\endgroup$ Jun 3, 2011 at 17:07
  • $\begingroup$ By computations Cassels's result can be improved to (say, up to permutations) $x \equiv y \equiv 4 \pmod{27}$, $z \equiv -5 \pmod{27}$. $\endgroup$ Jun 8, 2011 at 17:53
  • $\begingroup$ Are you sure? The congruence mod 27 is not satisfied by the smallest solution $x=y=z=1$. $\endgroup$ Jun 8, 2011 at 18:25
  • $\begingroup$ You are right, probably something wrong. $\endgroup$ Jun 8, 2011 at 18:28

My conjecture is false, see

Apoloniusz Tyszka, All functions $g:\mathbb{N}\to\mathbb{N}$ which have a single-fold Diophantine representation are dominated by a limit-computable function $f\colon \mathbb{N}\setminus \{0\}\to\mathbb{N}$ which is implemented in MuPAD and whose computability is an open problem, Computation, cryptography, and network security (eds. N. J. Daras and M. Th. Rassias), Springer, 2015, pp. 577–590, doi:10.1007/978-3-319-18275-9_24, arXiv:1309.2682


Of course the problem is old and probably there is no hope to be resolved.

The following papers of Vaserstein are of interest:

MR1196532 (93k:11090) Payne, G.(1-PAS); Vaserstein, L.(1-PAS) Sums of three cubes. The arithmetic of function fields (Columbus, OH, 1991), 443–454, Ohio State Univ. Math. Res. Inst. Publ., 2, de Gruyter, Berlin, 1992. 11P05

MR1284068 (95g:11128) Conn, W.(1-PAS); Vaserstein, L. N.(1-PAS) On sums of three integral cubes. (English summary) The Rademacher legacy to mathematics (University Park, PA, 1992), 285–294, Contemp. Math., 166, Amer. Math. Soc., Providence, RI, 1994. 11Y50 (11D25) PDF Clipboard Series Chapter Make Link

Over the last 40 years there have been various computational efforts to search for integer solutions to the equation $x^3+y^3+z^3 = t$ for small integers $t$. This paper describes a search that found solutions for $t = 39$ and $t = 84$, as well as a number of other solutions

for small $t$ that are of interest for various reasons. The authors used a symbolic computation package on workstations, and used different search techniques for different regions of interest. They argue that their data supports the conjecture that solutions should exist for all $t$ satisfying the easy necessary condition that $t$ not be congruent to $\pm 4$ modulo 9; the only such $t$ less than 100 for which no solutions are known are now $30,33,42,52,74,75$. The algorithms of this paper are tuned to providing solutions for an interval of possible $t$, whereas a recent algorithm due to Heath-Brown is faster for a fixed value of $t$, although it requires significant precomputation whose complexity depends on the class number of ${\bf Q}(\root 3 \of t)$. An implementation of that algorithm by D. R. Heath-Brown, W. M. Lioen and H. J. J. te Riele [Math. Comp. 61 (1993), no. 203, 235--244; MR1202610 (94f:11132)] also discovered some of the solutions found in the article under review.

  • 4
    $\begingroup$ Andreas-Stephan Elsenhans and Jörg Jahnel, preprint link given above by the OP, give solutions for 30, 52, 75. $\endgroup$
    – Will Jagy
    Mar 11, 2011 at 21:39
  • $\begingroup$ Note that Heath-Brown et al. checked the range up to $10^8$, compared to Elsenhans and Jahnel's $10^{14}. euler.free.fr/docs/HLR93.pdf $\endgroup$ Mar 12, 2011 at 12:02
  • 1
    $\begingroup$ There is a known integral solution to $x^3 + y^3 + z^3 = 30$. Use $x = 2220422932$, $y = −283059965$, and $z = −2218888517$. See the bottom of p. 18 at arxiv.org/pdf/math/0005139v1.pdf. $\endgroup$
    – KConrad
    Dec 9, 2012 at 17:06
  • $\begingroup$ I was aware of this, thanks anyway, since probably the result required some computations... $\endgroup$ Dec 9, 2012 at 18:33

The solution given in the comment of Y. Zhao:

$569936821221962380720^3 + (-569936821113563493509)^3 + (-472715493453327032)^3 = 3$.

Note that each of these numbers is larger than the limit of $10^{14}$ at which Elsenhans and Jahnel stopped their search in 2007. The smallest has 18 digits, while the other two have 21.


As $x^3 + y^3 = c$ for any given (suitable) c is an elliptic curve, perhaps a reasonable strategy would be to try various integers $f$, $g$ for which $c := f^3 - 3 g^3$ is small and establish the Mordell-Weil rank of the curve.

If this is ever positive (for values other than the known solutions the OP mentioned) then one would establish that there were other non-trivial rational solutions, even if these had still not been found.

Edit: Rereading the OP's post, I notice they are asking for integer solutions rather than rational solutions, and I recall now that there are rational parametrizations anyway. So perhaps this approach isn't very useful after all.

  • 2
    $\begingroup$ It's not hard to find rational solutions, but the problem asks for integral ones. [The simplest non-integral solution is $(x,y,z)=(−6, 10/3, 17/3).$] $\endgroup$ Jun 2, 2011 at 18:26

Not the answer you're looking for? Browse other questions tagged or ask your own question.