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We now that Fermat's last theorem is true so there are not positive integer solutions to $$x^n+y^n=z^n$$ for $n\in\mathbb{N}$ and $n>2$.

But what about if $n\in\mathbb{R}$ or $n\in\mathbb{R}^+$?

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You may find this useful: Fermat’s Last Theorem for Fractional and Irrational Exponents, College Math. J. 41 (2010), 182-185 –  Byron Schmuland May 18 '13 at 0:32
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3 Answers 3

up vote 39 down vote accepted

Suppose $z> \max(x,y)$ then $x^0+y^0 = 2 > z^0$ but there exists some $N$ such that $x^N+y^N<z^N$. Therefore there exists some $n\in[0,N]$ satisfying $x^n+y^n=z^n$.

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You can generalise this to show that the set of real n for which there exist integers $x, y, z$ with $x^n + y^n = z^n$ is dense in $[0,\infty)$. –  Bruno Le Floch May 17 '13 at 18:51

Take $x=4, y=9$ and $n = 0.5$. You can solve to get $z = 25$. So this works!

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Could you give some insights on how did you find that solution? –  Ambesh May 17 '13 at 14:52
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Use any perfect square for x and y if n=0.5 And so you can also use any perfect cube for x and y if you pick n = 1/3, for example x = 27, y = 64, you find z = 343 –  imranfat May 17 '13 at 15:00
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Do you have an example for n rational and > 2? –  Foon May 17 '13 at 16:39
    
No, I don't unfortunately –  imranfat May 17 '13 at 17:07

$1782^n + 1841^n = 1922^n$ with $n \approx 11.999999995097161$

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Yes. Now I realized I can solve this even with my TI-Nspire; $solve(1782^x+1841^x=1922^x,x)$. –  Ambesh May 18 '13 at 7:15
    
Dan, how is this possible? I thought there were not supposed to be integer solutions with n greater than 2? Isn't that what Fermat is about? I typed it in my calculator and it comes out evenly, is my TI rounding or something? –  imranfat May 18 '13 at 16:52
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Ah, there is rounding, never mind. 1782^12 is even, 1841^12 is odd, but 1922^12 is even. Wiles is indeed a clever man... –  imranfat May 18 '13 at 17:05
    
@imranfat: It's possible because the exponent is a non-integer. It just happens to be really close to one. –  Dan May 18 '13 at 22:45

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