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Assuming the Fermat Theorem, show that there is no natural number $x$, $y$, and $z$ and $n\geq3$ such that $$\frac{1}{x^n} + \frac{1}{y^n} = \frac{1}{z^n}. $$

So far I think proof by contradiction may be the best route, but I cannot find any place where I can start.

BlackAdder
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mrQWERTY
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2 Answers2

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Suppose to the contrary that $a,b,c$ are natural numbers such that $\frac{1}{a^n}+\frac{1}{b^n}=\frac{1}{c^n}$. Multiply through by $a^nb^nc^n$. We get $(bc)^n +(ac)^n=(ab)^n$. Thus $x=bc$, $y=ac$, $z=ab$ would be a natural number solution of the equation $x^n+y^n=z^n$. This contradicts the fact (Fermat's Last Theorem, proved by Wiles) that for $n\ge 3$, the equation has no solution in natural numbers.

André Nicolas
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  • Hi, thank you very much for your response. Unfortunately, I am still unable to get far. I apologize, this is my first time learning discrete mathematics proofs. Can you please help guide me further? – mrQWERTY Feb 06 '14 at 00:53
  • The hint is rather complete. If we had a solution, we would have (as shown) $(yz)^n+(xz)^n=(xy)^n$. with none of the entries $0$. But Fermat's Last Theorem (the theorem of Wiles) says that if $n\ge 3$, there are no solutions of $u^n+v^n=w^n=0$, with all the entries integers. – André Nicolas Feb 06 '14 at 00:57
  • Oh I see. Thank you very much. – mrQWERTY Feb 06 '14 at 01:01
  • You are welcome. I have modified the hint to a full solution. – André Nicolas Feb 06 '14 at 01:02
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The only property of Fermat's Last Theorem used in the problem is that

FLT with exponent $n$ is a homogeneous equation; if $(a,b,c)$ is a solution then $(ka,kb,kc)$ is a solution, for any $k$.

This makes the existence of solutions in rational numbers equivalent to the existence of integer solutions (because an integer solution is a rational solution, and clearing denominators in a rational solution gives an integer solution).

zyx
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