TITLE: My conjecture on almost integers.
QUESTION [5 upvotes]: Here when I was studying almost integers , I made the following conjecture - Let $x$ be a natural number  then For sufficiently large $n$ (Natural number) Let   $$\Omega=(\sqrt x+\lfloor \sqrt x \rfloor)^n$$ then $\Omega$ is an almost integer . The value of $n$ depends upon the difference between the number $x$ and its nearest perfect square which is smaller than it.Can anyone prove this conjecture.
Moreover, I can provide examples like 
$(\sqrt 5+2)^{25}=4721424167835364.0000000000000002$
$(\sqrt 27+5 )^{15}=1338273218579200.000000000024486$

REPLY [13 votes]: Well, we have that $$(\sqrt{x}+\lfloor\sqrt{x}\rfloor)^{n} + (-\sqrt{x}+\lfloor\sqrt{x}\rfloor)^{n} $$
is an integer (by the Binomial Theorem), but $(-\sqrt{x}+\lfloor\sqrt{x}\rfloor)^{n}\to 0$ if $\sqrt{x}$ was not already an integer.