Let the set $S_k=\{\pm x^k \pm y^k \pm z^k \ \vert \ x,y,z \in \mathbb{Z} \}$.

Note that the signs are independently positive or negative.

First of all $S_2 = \mathbb{Z}$ because (see the answers of this post):

$2m = (m+1)^2 - m^2 - 1$ and $2m+1 = (m+3)^2 - (m+2)^2 - 4$

It was proved by congruence computation that $n \in S_3$ implies $n \not \equiv 4,5 \pmod{9} $, and the converse was conjectured and checked for $n≤1000$ except $33$, $42$, $74$, $114$, $165$, $390$, $579$, $627$, $633$, $732$, $795$, $906$, $921$, and $975$ (see this paper) (see also here).

*Question*: Is there a similar conjecture and computational results for some small $k>3$?

We find by congruence computation that:

- $n \in S_4$ implies $n \not \equiv 4,5,6 \pmod{8} $
- $n \in S_5$ implies $n \not \equiv 4,5,6,7 \pmod{11} $
- $n \in S_6$ implies

$n \not \equiv 4,5 \pmod{7} $

$n \not \equiv 4,5,6 \pmod{8} $

$n \not \equiv 4,5,6,7 \pmod{9} $

$n \not \equiv 4, 5, 6, 7, 8, 9 \pmod{13} $ - $n \in S_7$ implies

$n \not \equiv 8, 9, 20, 21 \pmod{29} $

$n \not \equiv 10, 16, 17, 26, 27, 33 \pmod{43} $

$n \not \equiv 4, 9, 15, 22, 23, 24, 25, 26, 27, 34, 40, 45 \pmod{49} $

and the converse of each point could be conjectured.

*Remark:* this answer states that assuming the generalized abc conjecture, $S_k$ should have zero density for $k>18$, which prevents, in this case, a conjecture describing this set by finitely many congruences, as above.