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FLT says that the Diophantine equation $a^n+b^n=c^n$ isn't satisfied by any triplet $(a,b,c)$ where $n\in\mathbb{N}$ and $n>2$.

But what happens if $n\in\mathbb{Z}$ and thus can be negative?

$\textbf{My first thoughts:}$ If we have a negative exponent coupled with a real $a^s$ we can re-write it as $1/a^{-s}.$ So if $n\in\mathbb{Z}$ then the FLT equation takes the rubbish form $$\frac{1}{a^n}+\frac{1}{b^n}=\frac{1}{c^n}$$ which means $$\frac{a^n+b^n}{(ab)^n}=\frac1{c^n}$$ $$\frac{(ab)^n}{a^n+b^n}=\frac{c^n}1$$ $$(ab)^n=(ac)^n+(bc)^n$$ $ab$, $ac$ and $bc$ are all elements of the integer set commonly denoted as $\mathbb{N}$, we thus arrive at a Diophantine equation that is equivalent to the first one we've seen. Therefore the Diophantine equation $a^n+b^n=c^n$ isn't satisfied by any triplet $(a,b,c)$ where $n\in\mathbb{Z}$ and $\mathbf{\color{red}{|n|>2}}$ where $|x|$ is the absolute value of the number $x$.

Is my proof correct?

Furthermore, what happens when the triplet $(a,b,c)\in\mathbb{Z}^3$?

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To my knowledge, Wiles proved that it holds true for $|n|\le2$. An obvious example would be $(a+1)^{-1}+[a(a+1)]^{-1}=a^{-1}$. –  Lucian Mar 4 at 22:03
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What's the question? You just proved it! (but it's not equivalent, it just doesn't hold because of FLT) –  user2345215 Mar 4 at 22:17
    
Read again and please upvote my question :) –  LoveFood Mar 4 at 22:21
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2 Answers 2

Let $m<-2$ and $a^m+b^m=c^m$. Let $n=-m>2$ and multiply the equation by $(abc)^n$. Now $$ (bc)^n + (ac)^n = (ab)^n. $$

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This seems to be exactly what is said in the question after the edit. –  JiK Mar 4 at 22:16
    
What about $n = 2$? –  vonbrand Mar 4 at 23:53
    
@vonbrand When $n=2$ we get the Pythagorean Theorem, to which Euler proved the existence of infinitely many triplets $(a,b,c)\in\mathbb{Z}^3$ that satisfy it. –  LoveFood Mar 5 at 8:43
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If you had $a^{-5}+b^{-5}=c^{-5}$

then you would also have $(\frac{d}{a})^{5}+(\frac{d}{b})^{5}=(\frac{d}{c})^{5}$ for any $d$

but if we choose $d = lcm(a,b,c)$ then we have a solution to $a^5+b^5=c^5$ in integers.

So we can only solve for $-n$, if we can solve for $n$.

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