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&lt;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.
Denote $I_n=a^n-b^n$. Note that if $(a,b)$ has the property, then so has $(ma,mb)$ for $m\in\mathbb N$ as this just replaces $I_n$ with $m^nI_n$.
We have $I_1\ne 0$ because $a\ne b$ and therefore find that $a=\frac{I_2+I_1^2}{2I_1}\in \mathbb Q$, say $a=\frac uv$ with $u\in\mathbb Z$, $v\in\mathbb N$. Then $w:=vb=u-vI_1$ is an integer, hence $b=\frac wv\in\mathbb Q$. If $v=1$, we are done.
And if $v&gt;1$, let $p$ be a prime dividing $v$. By the observation above about multiples, we may assume wlog. that $v=p$. 
Write $a=A+\frac rp$ with $0&lt;r&lt;p$ and $A\in \mathbb Z$. Wirth $B:=A+I_1$ we obtain $b=B+\frac rp$. From $a\ne b$ we find $A\ne B$ and hence can write $A-B=p^st$ with $s\ge 0$ and $p\not\mid t$.

Pick $m\in\mathbb N$ with $mp\ge s+3$ and $p\not\mid m$. 
Then 
$$\begin{align}(pa)^{mp}&amp;=(pA+r)^{mp}=r^{mp}+mp^2Ar^{mp-1}+\sum_{k=2}^{mp-1}{mp\choose k}p^kA^kr^{mp-k}+p^{mp}A^{mp}\end{align}$$
and hence
$$\begin{align}p^{mp}I_{mp}
&amp;=mp^2r^{mp-1}(A-B)+\sum_{k=2}^{mp-1}{mp\choose k}p^kr^{mp-k}(A^k-B^k)+p^{mp}(A^{mp}-B^{mp})\\
&amp;=mp^{2+s}r^{mp-1}t+p^{3+s}t\sum_{k=2}^{mp-1}\frac{{mp\choose k}}pp^{k-2}r^{mp-k}\frac{A^k-B^k}{A-B}+p^{mp}(A^{mp}-B^{mp})\\\end{align}$$
where everything that is written as a fraction is in fact an integer by well-known divisibilities.
Now we have a contradiction 
because the left hand side and all summands on the right hand side except the first are multiples of $p^{3+s}$.
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.