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Trying to solve this I find out the following problem in which it is not necessary the condition $x^3=y^3=z^3$ in some $\mathbb F_p$:

Prove there are infinitely many pairwise coprime triples of distinct natural numbers, $(x,y,z)$, such that:

$$\frac{x^3+y^3+z^3}{x+y+z}\in \mathbb N$$

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  • $\begingroup$ Let $x=0$. It becomes trivial. $\endgroup$ – Patrick Stevens Oct 8 '15 at 17:03
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    $\begingroup$ @Ataulfo You should have said so first :) $\endgroup$ – Patrick Stevens Oct 8 '15 at 17:09
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    $\begingroup$ Please reedit your question to make it clear. $\endgroup$ – YoTengoUnLCD Oct 8 '15 at 17:09
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    $\begingroup$ If you're discarding most of the solutions we've given you, why don't you just ask for all the solutions? We've repeatedly proved that there are infinitely many solutions. $\endgroup$ – YoTengoUnLCD Oct 8 '15 at 17:15
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    $\begingroup$ A reason to downvote/disqualify your question is that you were completely unclear with your use of the term 'non-trivial'. My thoughts as to why you didn't clear this up immediately when asked is that you didn't even think about what you meant by that. I hope my edit shows that. $\endgroup$ – YoTengoUnLCD Oct 8 '15 at 20:20
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Yes, there are infinitely many pairwise coprime solutions $x,y,z$.

$(x,y,z)=(m,n,m+n)$ with $\gcd(m,n)=1$ and $m$ even.

Motivation: $$x^3+y^3+z^3=(x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz$$

Therefore:

$$x+y+z\mid x^3+y^3+z^3\iff x+y+z\mid 3xyz$$

If $(x,y,z)=(m,n,m+n)$, then $$\iff 2(m+n)\mid 3mn(m+n)\iff 2\mid mn$$

Since you want pairwise coprime solutions, we can let $\gcd(m,n)=1$.

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  • $\begingroup$ Completely different of mine. But this works. For interest of other users I'll give my answer below.Regards. $\endgroup$ – Piquito Oct 8 '15 at 18:25
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Let $x=y=z$ then:

$$ \frac {x^3+y^3+z^2}{x+y+z}=\frac {3x^3}{3x}=x^2\in \Bbb N, \forall x\in\Bbb Z-\{0\} $$

Furthermore, let $z=0,y=1$, then:

$$ \frac {x^3+y^3+z^2}{x+y+z}=\frac {x^3+1}{x+1}=x^2-x+1\in\Bbb N, \forall x\in\Bbb N-\{1\} $$

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    $\begingroup$ See the edit please. I mean non trivial solutions. $\endgroup$ – Piquito Oct 8 '15 at 17:06
  • $\begingroup$ Done :~). ${}{}$ $\endgroup$ – YoTengoUnLCD Oct 8 '15 at 17:07
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I want to solve a more general question. I give here an infinite family of solutions without details for the case $n=3$, just for complement the nice answer of @user236182.

$$(x,y,z)=(t^2-2t,\space2t-3,\space t^2-3t+3)$$

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If $(x,y,z)$ is a solution then so is $(px, py, pz)$ for any integer $p$. So even if $x$, $y$, and $z$ have to be distinct there are infinitely many solutions (once you find the first one).

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  • $\begingroup$ Without common factors,please. Non trivial infinitely many $\endgroup$ – Piquito Oct 8 '15 at 17:11
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    $\begingroup$ @Ataulfo It's your responsibility as the person asking the question to define "non-trivial": the term can mean many things in different contexts, so just calling them non-trivial isn't enough. $\endgroup$ – Erick Wong Oct 8 '15 at 17:20
  • $\begingroup$ I think it is a question of common sense to see what is trivial. For example $x=y=z$ or when $xyz=0$ or $(kx,ky,kz)$. The given example in my edition is obviously non-trivial. $\endgroup$ – Piquito Oct 8 '15 at 17:39

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