Does the equation $a^n+b^n=c^n$ have positive integer solutions? We know that the equation $a^n+b^n=c^n$ has infinitely many positive integer solutions when $n=1,n=2$. Also, by Fermat's last theorem, we know that this equation has no positive integer solution for any integer value of $n$ greater than $2$.
My question is: Does the equation $a^n+b^n=c^n$ have positive integer solutions for non-integer $n>2$?
Say for example, $a^{2.5}+b^{2.5}=c^{2.5}$
 A: Consider the function 
$$f_{a,b,c}(x)=a^x+b^x-c^x$$
with $a=4,b=5,c=6$. Then for $x=2$ the function is positive, while for $x\to\infty$ ($x=3$ is actually enough) the function will be negative. By continuity (intermediate value theorem) the function must at least one zero, i.e. a solution.
A similar argument will lead to various solutions.
A: Well, [1] obviously not for every non-integer $n$ will have integer solutions, $a,b,c$.
But obviously [2] some (many) integer $a,b,c$ will have non-integer $n$ solutions.
[1].  It'd be rather bizarre and strange if none of the $n\in \mathbb Z; n> 2$ will have integer solutions to $a^n + b^n = c^n$ but that every $\x \not \in \mathbb Z; x > 2$ will have integer solutions to $a^x + b^x = c^x$.
And it's obviously not true as there are uncountably many $x$ but only countably many triplets $(a,b,c)$.  [Each $x$ would have a unique  set of triplets as $a^x + b^x \ne a^y + b^y$ if $x\ne y$.]
[2]  If $a< c < b$ where $a^2 + b^2 > c^2$ there will be a value where $a^m + b^m < c^m$ and by intermediate value theorem the is an $x; 2 < m < x$ where $a^x + b^x = c^x$.
....
As for $a^{2.5} + b^{2.5} = c^{2.5}$, I highly suspect has no integer solutions.  
==== old answer====
Suppose $a^k + b^k > c^k$ and $a^m + b^m < c^m$; $k, m \in \mathbb N$ and $k \ge 2$.
That would mean there is a $j\in \mathbb R; k < j < m$ so that $a^j + b^j = c^k$.
So for instance $4^2 + 5^2> 6^2$ but $4^3 +  5^3 = 189 <216 = 6^3$ so there must be a $j; 2 < j < 3$ where $4^j + 5^j = 6^j$.
As for any $x > 1$ we have $(a + b)^x > a^x + b^x$ we find can assure that for and $a < c < b$ we can have $a^x + b^x > c^x$ (although $x$ may be less than $2$) and there will be $y > x$ so that $a^y + b^y < c^y$ and $j; x < j < y$ where $a^j + b^j =c^j$. 
