## Are nontrivial integer solutions known for $x^3+y^3+z^3=3$?

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)
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 I'm not currently able to access Michael Beck, Eric Pine, Wayne Tarrant, Kim Yarbrough Jensen, New integer representations as the sum of three cubes, Math. Comp. 76 (2007), 1683-1690, MathSciNet review: 2299795 to see whether it says anything new about 3. – Gerry Myerson Mar 12 2011 at 4:32 @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. – András Salamon Mar 12 2011 at 13:26

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).

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.

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

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Andreas-Stephan Elsenhans and Jörg Jahnel, preprint link given above by the OP, give solutions for 30, 52, 75. – Will Jagy Mar 11 2011 at 21:39
Note that Heath-Brown et al. checked the range up to $10^8$, compared to Elsenhans and Jahnel's $10^{14}. euler.free.fr/docs/HLR93.pdf – András Salamon Mar 12 2011 at 12:02 There is a known integral solution to$x^3 + y^3 + z^3 = 30$. Use$x = 2220422932$,$y = −283059965$, and$z = −2218888517$. See the bottom of p. 18 at arxiv.org/pdf/math/0005139v1.pdf. – KConrad Dec 9 at 17:06 I was aware of this, thanks anyway, since probably the result required some computations... – Luis H Gallardo Dec 9 at 18:33 As$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$is small and establish the Mordell-Weil rank of the curve. If this is ever positive (for values other than the known solutions the OP mentioned) then one would establish that there were other non-trivial rational solutions, even if these had still not been found. Edit: Rereading the OP's post, I notice they are asking for integer solutions rather than rational solutions, and I recall now that there are rational parametrizations anyway. So perhaps this approach isn't very useful after all. -  It's not hard to find rational solutions, but the problem asks for integral ones. [The simplest non-integral solution is$(x,y,z)=(−6, 10/3, 17/3).\$] – Noam D. Elkies Jun 2 2011 at 18:26