On the mixed sum of three k-th powers

Question

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.

Answer 1

EDIT This answers previous revision, quite different question.

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Let $k = 5$. I think a single representation outside your forbidden congruences would violate "Vojta's more general abc conjecture".

In A more general abc conjecture, p. 7 Paul Vojta conjectures:

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If $x_0,\ldots x_{n-1}$ are nonzero coprime integers satisfing $x_0 + \cdots x_{n-1}=0$

$$ \max\{|x_0|,\ldots |x_{n-1}|\} \le C \prod_{p\mid x_0 \cdots x_{n-1}}p^{1+\epsilon}\qquad (1) $$

for all $x_0 , \ldots, x_{n-1}$ as above outside a proper Zariski-closed subset.

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Let $n=2^m 11^l$, where $m,l$ are coprime to $5$ and $m,l$ are coprime.

$m,l$ can be arbitrary large. The radical of $n$ is constant and we have $n=2^m 11^l=x^5+y^5+z^5$.

By Vojta's conjecture, as $m,l$ vary as above, every single solution must be on proper Zariski-closed subset, which appears highly unlikely to me.

To ensure coprimality, clear the gcd.

$2$ can be replaced by any other positive natural and $n$ large, but with small radical satisfying your congruences will in general work, unless clearing the gcd solves it.

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Similar argument for larger $k$ and $n$ sufficiently larger than its radical would contradict also the n-conjecture.

The n-conjecture is a generalization of abc and basically says that the if $a_1 + \ldots + a_n=0$, no proper subsum vanishes and $a_i$ are coprime, then the radical of $a_1\cdots a_n$ can't be too small (without exceptional set).