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)
 A: Solutions exist for all primes $p\ge5,p\neq7$.
As the OP observed, we have the Vieta relation $x+y+z=0$ as $x,y,z$ are the zeros of the cubic
$$
P(T)=T^3+T+c=(T-x)(T-y)(T-z)
$$
in the field $\Bbb{F}_p$. Here $-c=-xyz$ is the shared value of $x^3+x,y^3+y$ and $z^3+z$ (treated as elements of $\Bbb{F}_p$ turning congruences into equalities).
The relation $z=-x-y$ takes care of the quadratic term, and
we are well placed to take advantage of the degree of freedom to select $c$ any which way we wish. Let's concentrate on the linear term! Expanding $(T-x)(T-y)(T+x+y)$ tells us that
$$
(T-x)(T-y)(T+x+y)=T^3-T(x^2+xy+y^2)-xyz,
$$ 
so we want to be able to choose distinct elements $x,y\in\Bbb{F}_p$ such that $x^2+xy+y^2=-1$.
This is possible whenever $p>3$.
Assume first that $p\equiv1\pmod3$. In this case there is a primitive cubic root of unity $\omega\in\Bbb{F}_p$. It satisfies the equation
$$
\omega^2+\omega+1=0.
$$
And that relation gives us the factorization
$$
a^2+ab+b^2=(a-\omega b)(a-\omega^2b).
$$
So we can select any two numbers $c,d\in\Bbb{F}_p$ such that $cd=-1$. Then the linear system
$$
\left\{\begin{array}{lcl}
a-\omega b&=&c\\
a-\omega^2b&=&d
\end{array}\right.
$$
has a unique solution $(a,b)$. After all, its determinant is $\omega-\omega^2\neq0$.
Then assume that $p\equiv-1\pmod3$. In this case $\omega$ only exists in the extension field $\Bbb{F}_{p^2}$. But, in that case we are dealing with the norm map
$$
N:\Bbb{F}_{p^2}\to\Bbb{F}_p, a-b\omega\mapsto (a-b\omega)(a-b\omega^2)=a^2+ab+b^2.
$$
By elementary properties of finite fields the norm is surjective, and takes each non-zero value in $\Bbb{F}_p$ exactly $p+1$ times. In particular, there are $p+1$ pairs $(a,b)$ such that $a^2+ab+b^2=-1$.

The above argument did not concern the possibility that some of $x,y,z$ may be equal (i.e. $P(T)$ has a multiple root for the resulting $c$). If $x=y$, then $x^2+xy+y^2=3x^2$. If $-1/3$ is a quadratic residue, we need to rule out two possible values of $x$. If $x=-y-x$ then $y=-2x$, and again $3x^2=-1$. Finally, if $y=-y-x$ then $x=-2y$ we need to rule the solutions of $3y^2=-1$.
At most six pairs $(x,y)$ were ruled out. If $p>7$ then in the first case the number of pairs $(c,d)$ such that $cd=-1$ is high enough to leave some solutions. All the cases where we had repetitions among $\{x,y,-x-y\}$ lead to the presence of a square root of $-3\in\Bbb{F}_p$, so the second case of $p\equiv-1\pmod 3$ is not affected.
The claim follows.

It may be worth noting that $p=7$ fails precisely because all the solutions of $a^2+ab+b^2=-1$, namely $(a,b)\in\{(1,3),(3,1),(3,3),(4,4),(4,6),(6,4)\}$ lead to repetitions among $\{a,b,-a-b\}$. None of the six solutions of $cd=-1$ work!
A: This is not true. Consider $x=0$, $y=2$, and $z=3$ mod $5$ (this is a special case of saulspatz counterexample).
A: 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 $p\equiv 3\mod{4}$.
