A naive diophantine approximation question Let $\alpha$ be a positive real number (bigger than one, and irrational if it matters) (the one I am secretly thinking of is $\varphi,$ the golden ratio). I want to know, given an $\epsilon>0,$ whether there always exist $n, m \in \mathbb{N}$ with $n > 0$, such that $|\alpha^n - m|< \epsilon.$ If yes, is there any way to estimate/find them?
 A: The following seems to be implied by most of the direct comments to OP's question, but I prefer to voice it, loud and clear:


The answer is almost yes. Given $\varepsilon > 0$, we can find such integers $m = m(\alpha, \varepsilon)$ and $n = n(\alpha, \varepsilon)> 0$, for almost every real number $\alpha > 1$, in the sense of Lebesgue measure.


This is so, because the sequence $(\alpha^n)_n$ is equidistributed modulo $1$ in the sense of H. Weyl for almost every $\alpha > 1$, see "Uniform distribution of sequences" by  L. Kuipers et H. Niederreiter (Wiley, 1974).
Equidistribution modulo $1$ implies density of the set 
$\{\alpha^n - \lfloor \alpha^n \rfloor : n \in \mathbb{N}\} \subset \lbrack 0, 1\rbrack$, which in turn implies that the sequence $(\alpha^n - \lfloor\alpha^n\rfloor)_n$ of fractional parts accumulates on $0$.
As already observed above, for the golden ratio $\alpha = \frac{1 + \sqrt{5}}{2}$,  the sequence $(\alpha^n)_n$ is not equidistributed modulo $1$, it is not even dense, since $\alpha^n + (\frac{1 - \sqrt{5}}{2})^n$ is an integer for every $n$, but fractional parts do accumulate on $0$. A similar line of reasoning applies to Pisot-Vijayaraghavan numbers; this is Robert Israel's comment. Indeed, when $\alpha$ is a Pisot-Vijayaraghavan number, say of of degree $d$ with conjugates $\alpha_1 = \alpha, \alpha_2, \dots,\alpha_d$, estimates for the smallest $m(\alpha, \varepsilon)$ and $n(\alpha, \varepsilon)$ are easy to come by since $\sum_{i = 1}^d \alpha_i^k$ is an integer for all $k$ while $\sum_{i > 1}^d \alpha_i^k$ tends exponentially fast towards $0$. (See also Noam D. Elkies'comment for Salem numbers.)
It is not known whether $((\frac{3}{2})^n)_n$ is equidistributed modulo $1$, see Gerry Myerson's comment for references. It is also not known whether $(e^n)_n$ or $(\pi^n)_n$ is equidistributed modulo $1$.
 Actually, no explicit example $\alpha > 1$ is known to be equidistributed modulo $1$. All my remarks are borrowed from a lecture on ergodic theory by Pierre de La Harpe.
A: The answer is no.
Start with $\alpha$ between 3.25 and 3.75; take squares of those 2 to see that one can restrict $\alpha^2$ to being between 11.25 and 11.75; take 3/2 powers of those to see that one can restrict $\alpha^3$ to being between 38.25 and 38.75; take 4/3 powers of those etc.
One gets the idea: since $\alpha>3$ a 0.5 wide interval is magnified to an interval of width >1.5 by moving to the next power. Such interval can always be narrowed to $[n+0.25,n+0.75]$ for some integer $n$.
