# $$n$$th power Difference

Let $$x,y,z$$ be integers satisfying $\left\{\begin{array}{l}x-y=z^n\\ y-z=x^n\\ z-x=y^n\end{array}\right.$ for some integer $$n$$.

Prove that there does not exist non-zero solutions for $$x,y,z$$.

Note by Daniel Liu
4 years, 1 month ago

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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{align} x-y&=1\\ y-z&=1\\ z-x&=1 \end{align} Adding these together shows that $$0=3,$$ which obviously leads to a contradiction.

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

Rewrite the equations. \begin{align} x-y&=z\\ y-z&=x\\ z-x&=y \end{align} You can apply the following logic to any of the three variables. Take a look at $$x$$ and the following two equations. \begin{align} y-z&=x&\text{ }(1)\\ z-x&=y&\text{ }(2) \end{align} 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 $$1.$$

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

- 4 years, 1 month ago

Yeah that's what I did. Without as much laborious casework.

- 4 years, 1 month ago

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.

- 4 years, 1 month ago