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.

  • 1
    $\begingroup$ I think at least the problem for $k = 3$ is hard; however the basic heuristics suggests that for $k > 3$, the density of $S_k$ will be zero. $\endgroup$ – Stefan Kohl Nov 10 '15 at 11:08
  • $\begingroup$ @StefanKohl: the congruence computation suggests a density of $7/11$ for $k=5$, which heuristics suggests a zero density? $\endgroup$ – Sebastien Palcoux Nov 10 '15 at 11:20
  • $\begingroup$ Regarding zero density - can one even show that $x^{n}+y^{m}$ has zero density for any $n\neq m\geq 1$, not both even? $\endgroup$ – Eric Naslund Nov 10 '15 at 11:22
  • 1
    $\begingroup$ @StefanKohl: That heuristic is not accurate as $x,y$ and $z$ can take negative values. Other than using congruence's, it's not clear how to determine whether or not $n$ has infinitely many representations of the form $n=x^5+y^5+z^5$. $\endgroup$ – Eric Naslund Nov 10 '15 at 11:34
  • $\begingroup$ @EricNaslund: My feeling would be that that makes obtaining results more difficult, but that it doesn't really affect the heuristics -- but I haven't thought about this or obtained numerical evidence or whatever, so perhaps you are right. $\endgroup$ – Stefan Kohl Nov 10 '15 at 11:41

EDIT This answers previous revision, quite different question.

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:

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.

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.

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).

| cite | improve this answer | |
  • $\begingroup$ For $k$ odd it is the same question, I've allowed the mixed sums for extending to $k$ even. $\endgroup$ – Sebastien Palcoux Nov 10 '15 at 14:20
  • $\begingroup$ I try to understand your argument for $k=5$. Anyway, if (up to Vojta's conjecture) it is working for the previous revision then it is also working for the current revision because $5$ is odd and for any revision we have $x,y,z \in \mathbb{Z}$. Can you extend your argument to $k=4$ for the current revision? $\endgroup$ – Sebastien Palcoux Nov 11 '15 at 5:23
  • $\begingroup$ For helping me to understand, could you explain why examples like: $3⋅11=0^5+1^5+2^5$ or $6257⋅11=6^5+11^5+(−10)^5$, are not concerned by your argument (you wrote << $2$ can be replaced by any other positive natural...>>)? $\endgroup$ – Sebastien Palcoux Nov 12 '15 at 0:24
  • $\begingroup$ @SébastienPalcoux These don't work. As written, n must be of the form 2^m 11^l (or similar) with the radical of $n$ much smaller than $n$. The radical is the product of the primes factors, without multiplicities. In your examples $n$ is squarefree and $n=rad(n)$, which is allowed by both conjectures. Basically to work, $n$ must be large and $rad(n)=22$, which is constant. $\endgroup$ – joro Nov 12 '15 at 5:20
  • $\begingroup$ Ok so we have $x_0 + x_1 + x_2 + x_3 = 0$ with $x_1 = x^5, x_2 = y^5, x_3 = z^5$, and $x_0 = -n = -2^m11^l$, with $m,l$ large enough as above. The inequality of Vojta's conjecture deals with $\vert x_0 \vert , \vert x_1 \vert, \vert x_2 \vert, \vert x_3 \vert $, and $rad(x_0x_1x_2x_3)$, which means that it does not deal only with $\vert x_0 \vert$ and $rad(x_0)$. So even if $rad(n)$ is much smaller than $n$, $rad(x_0x_1x_2x_3)$ could be very big, and I don't understand the argument in this case. $\endgroup$ – Sebastien Palcoux Nov 12 '15 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.