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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.

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Sebastien Palcoux
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    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. – Stefan Kohl Nov 10 '15 at 11:08
  • @StefanKohl: the congruence computation suggests a density of $7/11$ for $k=5$, which heuristics suggests a zero density? – Sebastien Palcoux Nov 10 '15 at 11:20
  • 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? – Eric Naslund Nov 10 '15 at 11:22
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    @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 $n=x^5+y^5+z^5$. – Eric Naslund Nov 10 '15 at 11:34
  • @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. – Stefan Kohl Nov 10 '15 at 11:41
  • Related question posted to m.se, http://math.stackexchange.com/questions/1524797/can-a-sum-of-three-fifth-power-of-integers-be-8 – Gerry Myerson Nov 18 '15 at 06:18

1 Answers1

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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).

joro
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  • For $k$ odd it is the same question, I've allowed the mixed sums for extending to $k$ even. – Sebastien Palcoux Nov 10 '15 at 14:20
  • 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? – Sebastien Palcoux Nov 11 '15 at 05:23
  • 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...>>)? – Sebastien Palcoux Nov 12 '15 at 00:24
  • @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. – joro Nov 12 '15 at 05:20
  • 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. – Sebastien Palcoux Nov 12 '15 at 12:47
  • @SébastienPalcoux Yes, indeed the radical can be very big. But it must be sufficiently small to $T=max(|x|^5,|y|^5,|z|^5)$. Assume $x$ is largest in absolute value then $T=|x|^5$, the radical of $n$ ($x_0$ in your notation is constant), so the whole radical of the product is less than $C |x|^3$, which is significantly smaller than $|x|^5$ – joro Nov 12 '15 at 13:12
  • @SébastienPalcoux For the n-conjecture, the radical must be significantly smaller than $T$, so just increase $n$, avoiding any exceptional set. – joro Nov 12 '15 at 13:15
  • @SébastienPalcoux I mean both conjectures imply the radical can't be too small compared to $|x|^n$. – joro Nov 12 '15 at 13:20
  • Ok I better understand this part, now in what it is << highly unlikely >> that << every single solution must be on proper Zariski-closed subset >> ? I note that your argument works only for the case $k>3$ because $C \vert x \vert^3$ must be << significantly smaller than >> $\vert x \vert^k$. Do you think your Zariski belief also relevant for the case $k=4$ ? – Sebastien Palcoux Nov 12 '15 at 13:50
  • @SébastienPalcoux This is my belief. To avoid exceptional set, use larger $k$ and the n-conjecture. For $k=4$ I believe you can't express numbers with sufficiently small radical, though there are polynomial identities which when specialized give exceptional set. – joro Nov 12 '15 at 13:53
  • Perhaps your answer would get more upvotes if you motivate your Zariski belief by some evidences. – Sebastien Palcoux Nov 12 '15 at 14:28
  • @SébastienPalcoux the Zariski argument is not needed for the n-conjecture, which IMHO is at least equally believed. – joro Nov 12 '15 at 14:54
  • If I'm not mistaken, we can apply the $n$-conjecture only for $k$ large ($k>18$ by this answer), but I ask for small $k$. What I call your "Zariski belief" is not the Vojta's conjecture itself but your sentence << ... which appears highly unlikely to me >>. – Sebastien Palcoux Nov 12 '15 at 15:08
  • @SébastienPalcoux I think $k=10$ is enough since $x^{10} \gg C |x|^9$ and in the n-conjecture the exponent is $3$. – joro Nov 12 '15 at 15:35