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:
- Is it true that all $x,y,z$ of $(4)$ and $(5)$ also obey $x\equiv y\equiv z \bmod 9$?
- In general, for what $N$ does this condition hold?
Best Answer
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.