If $a^n-b^n$ is integer for all positive integral value of $n$, then $a$, $b$ must also be integers. If  $a^n-b^n$ is integer for all positive integral value of n with a≠b, then a,b must also be integers.  
Source: Number Theory for Mathematical Contests, Problem 201, Page 34.
Let $a=A+c$ and $b=B+d$  where A,B are integers and c,d are non-negative fractions<1.
As a-b is integer, c=d.
$a^2-b^2=(A+c)^2-(B+c)^2=A^2-B^2+2(A-B)c=I_2(say),$  where $I_2$ is an integer
So, $c=\frac{I_2-(A^2-B^2)}{2(A-B)}$ i.e., a rational fraction $=\frac{p}{q}$(say) where (p,q)=1.
When I tried to proceed for the higher values of n, things became too complex for calculation.

assuming $a \neq b$
if $a^n - b^n$ is integer for all $n$, then it is also integer for $n = 1$ and $n = 2$.
From there you should be able to prove that $a$ is integer.
