Question on Fermats Last Theorem

Question

$a^n + b^n = c^n$, for any integer value of n greater than two where a,b,c are positive integers.

Since this is too hard for me to solve, I tried to change the question a little. I believe Fermat put the rule "for any integer value of n greater than two" because the question would have solutions where n is any number. But even then with n being any number (except 1 and 2) I couldn't imagine of a solution. So I will ask it here if anyone knows to solve it. But if there arent any solutions that anyone knows of why did Fermat state and believe that it only has no solutions when n is an integar greater than 2.

So here is my question :

Is there solutions to $a^n + b^n = c^n$, where n is any value (positive or negative) but not 1 or 2 and a,b,c are positive integers

Answer 1

If $a$, $b$, $c$ are any nonnegative numbers, you question boils down to the existence of solutions $x$ to $$u^x+v^x=1$$ where $u=a/c$ and $v=b/c$.

For $x=0$, $u^x+v^x=2>1$, so to show the existence of a solution, you only need to show there are $x$ with $u^x+v^x<1$ and apply the mean value theorem.

If $u,v<1$ you only need to pick $x$ large and positive. If $u,v>1$ you pick $x$ large and negative.

I'll leave it to you to investigate the remaining cases.

Answer 2

If $n=1$, there are many solutions, right? Just pick and two positive integers for $a$ and $b$ and setting $c=a+b$ satisfies the equation. Even for $n=2$, there are infinitely many Pythagorean triples.

Answer 3

There are solutions for $n=-1$ and $n=-2$. For instance $10^{-1}+15^{-1}=6^{-1}$; and $15^{-2}+20^{-2}=12^{-2}$.