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 \(x,y,z\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestBy 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 \(1.\)

\[\mathbb{Q.E.D.}\]

Log in to reply

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

Log in to reply

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.

Log in to reply