# 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?

• True for the golden ratio, at any rate. But that's unusual because for large $n$ every $\alpha^n$ is close to an integer. – Noam D. Elkies Jul 4 '17 at 22:03
• @René, the question of the distribution of the fractional parts of powers of $3/2$ is a notorious open problem, related to Waring's problem. See, e.g., en.wikipedia.org/wiki/Waring%27s_problem – Gerry Myerson Jul 4 '17 at 23:34
• Slightly more generally than for the golden ratio, if $\alpha$ is a Pisot-Vijayaraghavan number, $\|\alpha^n \| = \inf_{m \in \mathbb N} |\alpha^n - m| \to 0$ exponentially as $n \to \infty$. – Robert Israel Jul 5 '17 at 1:25
• . . . and if it's a Salem number then $\|\alpha^n\|$ has a known distribution and does get arbitrarily (albeit not exponentially) small. en.wikipedia.org/wiki/Salem_number – Noam D. Elkies Jul 5 '17 at 4:41
• Yes, Igor, but any nontrivial information about the fractional part of $(3/2)^n$ seems to be hard to come by. Mahler proved $\|(3/2)^k\|>(3/4)^k$ for $k$ sufficiently large, where $\|x\|$ is the distance from $x$ to the nearest integer (I realize this doesn't speak to your question, it's just an example of the kind of thing that is known). Habsieger, Acta Arith. 106 (2003), no. 3, 299–309, has $\|(3/2)^k\|>.57434^k$ for $k\ge5$. – Gerry Myerson Jul 5 '17 at 5:23

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 $n$.

• The scheme above can be used to produce $\alpha$'s whose powers stay away from integers by at least $\epsilon$ for any given $\epsilon<1/2$. Moreover, by picking a big enough starting $\alpha$, the scheme can produce a tree of counterexamples where each node has more than 2 (or more) branches, therefore proving that there are uncountably many such counterexamples. – Yaakov Baruch Jul 7 '17 at 16:14

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.

• In particular (from the last sentence), I assume that no effective bounds are known.... – Igor Rivin Jul 5 '17 at 17:42
• @IgorRivin I am unable to infer any bound from the references I cited above. (Of course, estimates are easy to get for Pisot-Vijayarghavan numbers - I added a one-line explanation for this). – Luc Guyot Jul 7 '17 at 16:43