# On the mixed sum of three k-th powers

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.

• 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
• @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

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).

• 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 5:23
• 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 0:24
• @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 5:20
• 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