Is it true that there aren't any three different numbers $x,y,z$ such that $x^3+x \equiv y^3+y \equiv z^3+z \pmod p $? Let $p$ be a prime number. Is it true that there aren't any three different numbers $x,y,z$ such that $$x^3+x \equiv y^3+y \equiv z^3+z \pmod p $$
with  $x -y, y-z, z-x$, each of them cannot be divided by $p$ ?
If not, what are the conditions of $p$ so that the statement is true for prime number $p$ ?
I tried with $p=3,7$ and both of them are correct, so I think that $p \equiv 3 \pmod 4$ may satisfy the statement. 
My other attempt: Assume by contradiction, there exist $x,y,z$ such that $$x^3+x \equiv y^3+y \equiv z^3+z \pmod p$$
with  $x -y, y-z, z-x$, each of them cannot be divided by $p$. Then $$x^2+xy+y^2 \equiv y^2+yz+z^2 \equiv z^2+zx+x^2 \pmod p$$
thus $$x+y+z \equiv 0 \pmod p.$$
Here I am stuck. How can I solve this problem ?
(Sorry for my English)

If $p=n^2+1$ (such as $5,17,37\dots$), then $x=0,\ y=n,\ z=(p-n)$ will solve your equivalence with all three terms congruent to $0$. 
$x^3+x=0;\ y^3+y=n(n^2+1)=np\equiv 0 \mod{p}; z^3+z=(p-n)(p^2-2np+n^2+1)=(p-n)(p^2-2np+p)=p(p-n)(p-2n+1)\equiv 0\mod{p}$
$p$ cannot be of the form $n^2+1$ if $p\equiv 3\mod{4}$.
