**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](https://mathoverflow.net/a/66740/12905), Cassels in a [1985 paper](http://www.ams.org/journals/mcom/1985-44-169/S0025-5718-1985-0771049-4/S0025-5718-1985-0771049-4.pdf) showed that $w=1$ must have,


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


For general $w$, Heath-Brown in a [1992 paper](http://www.ams.org/journals/mcom/1992-59-200/S0025-5718-1992-1146835-5/S0025-5718-1992-1146835-5.pdf) 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](http://www.uni-math.gwdg.de/jahnel/Arbeiten/Liste/threecubes_20070419.txt) 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?
> 
> 
> 



On cubic reciprocity for $x^3+y^3+z^3 = 996$?
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.


