Find all postive integer $n$,such $(x+y+z)|(x^{2n+1}+y^{2n+1}+z^{2n+1})$ 
Let  $0<x<y<z<p$,where $x,y,z$ are postive intgers,$p$ is prime number, and such 
  $$x^3\equiv y^3\equiv z^3\pmod p$$
  Find all postive integer $n$,such $$(x+y+z)|(x^{n}+y^{n}+z^{n})$$

$n=1$ is clear
$n=2$
I have show that
$$p|x^2+xy+y^2, p|y^2+yz+z^2, p|x^2+xz+z^2$$
and
$$p|x+y+z$$
so use this
$$2\sum_{cyc}(x^2+xy+y^2)-(x+y+z)^2=3(x^2+y^2+z^2)$$
so $n=2$ is right.
also I have show $n=5$ also is right
 A: For the congruence $x^3\equiv a\pmod p$ to have three non-congruent solutions $x,y,z$ it is necessary that there exists a primitive cubic root $\omega\in\Bbb{Z}_p$. This is the case if and only if $p\equiv1\pmod3$. Otherwise cubing will be a bijection from $\Bbb{Z}_p$ to itself.
In that case we have $y\equiv\omega x$ and $z\equiv \omega^2x$ or the other way around. This is because in $\Bbb{Z}_p^*$ we have $(y/x)^3=1$ and $(z/x)^3=1$, and the residue classes $1,\omega,\omega^2$ are the only classes
with cubes $\equiv1$.
Without loss of generality we can assume the first case by replacing $\omega$ with $\omega^2$, if necessary. Anyway, because $\omega\not\equiv 1$ and $\omega^3-1=(\omega-1)(\omega^2+\omega+1)\equiv0\pmod p$, we can conclude that $p\mid 1+\omega+\omega^2$ and consequently
$$
p\mid x(1+\omega+\omega^2)\equiv x+y+z.
$$
The inequalities imply that $0<x+y+z<3p$, so we also know that either $x+y+z=p$ or $x+y+z=2p$.
Claim. $(x+y+z)\mid (x^n+y^n+z^n)$ if and only if $3\nmid n$.
Proof. Let's first show that $p\mid (x^n+y^n+z^n)$ if and only if $3\nmid n$.


*

*If $n\equiv1\pmod 3$, then $y^n\equiv \omega^nx^n\equiv\omega x^n$ and similarly $z^n\equiv\omega^{2n}x^n\equiv \omega^2x^n$. So in this case
$$x^n+z^n+y^n=x^n(1+\omega+\omega^2)$$ is divisible by $p$.

*If $n\equiv2\pmod 3$, then we proceed similarly. Only this time $y^n\equiv \omega^2x^n$ and $z^n\equiv \omega x^n$. Again $p\mid(x^n+y^n+z^n)$.

*If $3\mid n$, then, as a consequence of $\omega^3\equiv1$ we get that $x^n,y^n,z^n$ are all congruent modulo $p$. Because $p>3$ and $p\nmid x^n$, the sum $x^n+y^n+z^n$ won't be divisible by $p$.


It remains to show that if $x+y+z=2p$ then $x^n+y^n+z^n$ is also an even number. But this is easy. Among $x,y,z$ there are either $0$ or $2$ odd numbers, so the same holds for the powers $x^n,y^n,z^n$. So $x^n+y^n+z^n$ is divisible by $2p=x+y+z$ if and only if it is divisible by $p$.
The claim follows.
