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\(n\)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
2 years, 5 months 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 \(1.\)

\[\mathbb{Q.E.D.}\] Trevor B. · 2 years, 5 months ago

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@Trevor B. Yeah that's what I did. Without as much laborious casework. Finn Hulse · 2 years, 4 months ago

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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. Ariel Gershon · 2 years, 5 months ago

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