$\frac{x^3+y^3+z^3}{x+y+z}\in \mathbb N$ has infinitely many non-trivial solutions 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:
$$\frac{x^3+y^3+z^3}{x+y+z}\in \mathbb N$$
 A: 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:
$$
\frac {x^3+y^3+z^2}{x+y+z}=\frac {x^3+1}{x+1}=x^2-x+1\in\Bbb N, \forall x\in\Bbb N-\{1\}
$$
A: 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 $\gcd(m,n)=1$.
A: 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.
$$(x,y,z)=(t^2-2t,\space2t-3,\space t^2-3t+3)$$
A: 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).
