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Question:

If $x,y,z,n$ are natural numbers, $x,y,z,n>1$, with $x^n +y^n=z^n$ then show that $x,y,z$ are all greater than $n$

Here to prove this i would like to use Fermat's last theorem, to show that $n=2$ and then use the property of Pythagorean triplets to show that $x,y,z$ are greater than $n$.Can this method be used in a proof?

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  • $\begingroup$ Sounds fine. By property of pythagorean triple, I take you mean the 'smallest' triplet is 3,4,5 I take? $\endgroup$ – Jack Yoon Jan 26 '15 at 17:22
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    $\begingroup$ This is a rather heavy weapon. It's correct (according to todays state of knoowledge) but I would try to check whether a more elementary reasoning allows to cover the case $n\ge 3$ $\endgroup$ – Thomas Jan 26 '15 at 17:24
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    $\begingroup$ Further to @Thomas point, an advanced result like Fermat's last theorem may well already have used the result you are proving, perhaps via a chain of other earlier results, so you may actually have a circular argument. $\endgroup$ – Joffan Jan 26 '15 at 17:29
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    $\begingroup$ @MattSamuel OK - maybe it's just aesthetics then, or my sense of what it means to prove something. $\endgroup$ – Joffan Jan 26 '15 at 17:43
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    $\begingroup$ Notice that as the bases $x,y,z$ are positive, and may be taken to be relatively prime without loss of generality (including a common factor would increase $x,y,z$ without changing $n$), you can assume $x \lt y \lt z$ and focus efforts on showing $x \gt n$. $\endgroup$ – hardmath Jan 26 '15 at 17:51
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By all means examine Pythagorean triplets for n=2, but it isn't really necessary.

For $n\geq 2$ and taking $x \leq y<z$, note that
$\begin{align} x^n = z^n-y^n &= (z-y)\overbrace{(z^{n-1}+z^{n-2}y+\ldots + zy^{n-2}+y^{n-1})}^{n \text{ terms}} \\ &>ny^{n-1} \geq nx^{n-1}\end{align}$

$\Rightarrow x^n > nx^{n-1}\Rightarrow x>n$ as required.

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  • $\begingroup$ This is the same proof given in my book. The factorization that you have done,is show in the solution though its the first time i'm seeing a factorization technique like this . $\endgroup$ – Ananthakrishna Jan 27 '15 at 10:41
  • $\begingroup$ Once you know that $(z-y)$ is a viable factor of $(z^n-y^n)$ - which it always is - you can divide out to find the remaining factor if necessary. $\endgroup$ – Joffan Jan 27 '15 at 15:39

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