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I. The Diophantine equation,

$$x^3+y^3+z^3 = 3w^3\tag1$$

with $x\geq y \geq z$ and $w=1$ has only two known solutions, namely $1,1,1$ and $4,4,-5$. Are there larger ones? As Noam Elkies points out in this post, Cassels in a 1985 paper showed that $w=1$ must have,

$$x\equiv y\equiv z \bmod 9\tag2$$

For general $w$, Heath-Brown in a 1992 paper showed that,

$$\text{either}\;x\equiv y\equiv z \bmod 9,\; \text{or one of}\; x,y,z\;\text{is}\;9m\tag3$$

Checking the Elsenhans-Jahnel list for $x^3+y^3+z^3=N$ with $N<1000$, one finds large solutions for $w=2,3,4$. (The list only covers $10^{14}$, so if it can be raised higher, maybe a large one can be found for $w=1$ as well.)

II. Trying to find something similar to $(1)$, and if I did my search right, there are at least two other $N$ that obeyed $(2)$, namely $N=996$,

$$x^3+y^3+z^3 = 996\tag4$$

with,

$$x,y,z = 11,2,\,-7$$

$$x,y,z = 2169364505441,\, -631266388780,\, -2151398424325$$

and $N = 2^4\times3^3+3=435$,

$$x^3+y^3+z^3 = 435\tag5$$

with seven known (and rather large) solutions, all of which satisfied $(2)$.

Questions:

  1. Is it true that all $x,y,z$ of $(4)$ and $(5)$ also obey $x\equiv y\equiv z \bmod 9$?
  2. In general, for what $N$ does this condition hold?
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    $\begingroup$ I can't seem to see where Cassels' proof breaks down; doesn't it work for all N with $N \equiv \pm 3 \bmod 9$? $\endgroup$ Dec 8, 2015 at 21:09
  • $\begingroup$ @FranzLemmermeyer: I added $N = 16\times27+3 = 435$ which seems to be a better candidate. $\endgroup$ Dec 9, 2015 at 6:58
  • $\begingroup$ Open question separately. About the equation $x^3+y^3+z^3=qw^3$ ? To find the parameterization as best you can describe all the solutions. $\endgroup$
    – individ
    Dec 17, 2015 at 13:56

1 Answer 1

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Problems of this type are studied in the article

Colliot-Thélène, Wittenberg - Groupe de Brauer et points entiers de deux familles de surfaces cubiques affines.

Here, in remark 5.7, they give an explanation of Cassels' result in terms of a Brauer-Manin obstruction to strong approximation. Other examples of Brauer-Manin obstruction to strong approximation are also included, which in general imply the existence of other "non-obvious impossible congruences".

One has to work very hard to make these congruences explicit. In the Cassels example, they prove it using the fact that $3$ is the only prime of bad reduction. However in your case of $996$, there are other primes of bad reduction, namely $2$ and $83$. So the same method does not apply. Moreover, your ''evidence'' in the $996$ case is very sparse, and I would see no reason why the same Cassels congruence should hold in this case. One would need to carefully work out the Brauer-Manin obstruction in this case and see what one gets.

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  • $\begingroup$ Thanks. I added $N=16\times27+3=435$. Maybe I should have chosen it instead, but I only noticed it later. All its seven known solutions $x,y,z,$ $$4460467, -2755337, -4078175 \\ 5530891, -2058260, -5434196 \\ 34587148, 26017420, -38927093 \\ 11116411450, 7938556819, -12328866524 \\ 509795655496, -981038126, -509795654285 \\ 4625447432587, -2276258604581, -4433868387803 \\ 6523252530778, -5076143048600, -5275053974573 $$ obey the congruence. And this one is less sparse. :) $\endgroup$ Dec 9, 2015 at 7:01

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