Prove that the equation $x^3+y^3+z^3-(x^2z+y^2x+z^2y)=2$ has no solution in natural numbers I asked myself which primes $p$ can be written as $p=x^3+y^3+z^3-(x^2z+y^2x+z^2y)$ with $x,y,z \in \mathbb{N}$.
 But for $p \neq 2$ we have the solution $x=y=\frac{p-1}{2}$ and $z=\frac{p+1}{2}$. So the only prime for which I can not find a solution is $p=2$. But I can not prove that there is not a solution.
Any ideas?

EDIT: You're right stackExchangeUser; my proof doesn't work. With a similar tack, we can still salvage this:
\begin{align*}
&x^3 + y^3 + z^3 - (x^2 z + y^2 x + z^2 y) \\
= ~ &(x + y + z)^3 - 4(x^2 z + y^2 x + z^2 y) - 3(x^2y + y^2 z + z^2 x) - 6xyz
\end{align*}
So, we are solving,
$$(x + y + z)^3 = 2 + 4(x^2 z + y^2 x + z^2 y) + 3(x^2y + y^2 z + z^2 x) + 6xyz$$
Suppose first the left side is divisible by $2$. Again, at least one of $x, y, z$ must be even. If all of them are even, we see that the left side is $0$ mod $8$, but the right side is $2$ mod $8$. Thus, exactly one must be even. But then, the $3(x^2y + y^2 z + z^2 x)$ term is odd, which makes the right hand side odd, and we get a contradiction again.
Thus, the left side is odd. Similarly, either all of $x, y, z$ are odd, or exactly one is. If exactly one of them is odd, then the right hand side is even, hence all $x, y, z$ are odd.
Finally, considering the original formulation, and the fact that $x^2 \equiv 1$ mod $8$ for all odd $x$, we get,
$$x^3 + y^3 + z^3 - (x^2 z + y^2 x + z^2 y) \equiv x + y + z - (z + x + y) \equiv 0$$
mod $8$, which cannot equal $2$.
