$n$th power Difference

By adding the equations together, we see that $x^n+y^n+z^n=0,$ or $x^n+y^n=-z^n.$ Obviously, $(0,0,0)$ is a solution. Do cases on other values of $n.$

$\textbf{Case 1:}$ $n=0$

We can rearrange the equations as follows. $\begin{aligned} x-y&=1\\ y-z&=1\\ z-x&=1 \end{aligned}$ Adding these together shows that $0=3,$ which obviously leads to a contradiction.

$\textbf{Case 2:}$ $n=1$

Rewrite the equations. $\begin{aligned} x-y&=z\\ y-z&=x\\ z-x&=y \end{aligned}$ You can apply the following logic to any of the three variables. Take a look at $x$ and the following two equations. $\begin{aligned} y-z&=x&\text{ }(1)\\ z-x&=y&\text{ }(2) \end{aligned}$ Some rearranging shows that $x=y-z$ from $(1)$ and $x=z-y$ from $(2).$ $x=\pm(y-z)$ which is only possible if $x=y-z=0.$ As I said before, you can show this with any variable.

Therefore, if $n=1,$ the only possible solution is $(0,0,0).$

$\textbf{Case 3:}$ $n=2$ or some other positive even.

If $n=2,$ then $x^n,$ $y^n,$ and $z^n$ are all non-negative. This means $-z^n$ is non-positive. The only possible way to add two non-negatives together to get a non-positive is if all those numbers are equal to $0.$ So once again, $(0,0,0)$ is the only solution. The same logic can be applied to any other even $n$.

$\textbf{Case 4:}$ $n=3$ or some other positive odd.

If $x^n+y^n=-z^n,$ then $-(x^n+y^n)=z^n$ and therefore $(-x)^n+(-y)^n=z^n$ Substitute $a=-x$ and $b=-y$ to find that $a^n+b^n=z^n.$ $n$ is greater than $2,$ so by Fermat's Last Theorem, there are no nonzero solutions.

$\textbf{Case 5:}$ $n$ is negative.

Two integers cannot have a difference that is not an integer itself. The case where $x,$ $y,$ and $z$ have solutions in the set $\{-1,1\}$ can be disproven with the proof that $n$ cannot equal $1.$

$\mathbb{Q.E.D.}$

By the law of trichotomy, we must be able to arrange $x,y,z$ in order from least to greatest. Suppose, without loss of generality, that $x \le y \le z$.

Then by the first equation, $z^n \le 0$. The only way this could be true is if $z \le 0$.

Similarly, by the third equation we have $y^n \ge 0$, so $y \ge 0$.

However, now we have $z \le 0 \le y$, so $z \le y$. But we also assumed that $y \le z$. Therefore $y = z$.

Since $0 \le y = z \le 0$, then we must have $y = z = 0$. Therefore, $x = 0$ as well.

Hence the only solution is $(0,0,0)$. You get a similar result if you choose any other ordering of $x,y,z$.

Note that this solution works even if $x,y,z \in \mathbb{R}$, rather than integers.