Powers of 2 as integer parts of irrational multiples

Suppose $\alpha$ is an irrational number in $(0, 2)$.

Prove that there exist infinitely many positive integers $n$ such that the integer part of $n \alpha$  (i,e $[n \alpha]$) is a power of $2$.

Solution.

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Since $\alpha$ is irrational, so is $\beta = \frac{1}{\alpha}$. The binary expansion of $\beta$ contains the binary digit $1$ infinitely often (otherwise it would be rational).

This means, for infinitely many $m$ we have the following

$$2^{m} \beta = N_m + f_m$$

with $N_m$ is the integer part of $2^m\beta$ and $f_m$ is the fractional part, with $\dfrac{1}{2} < f_m < 1$. We choose $m$ so that $2^m \beta$ has $1$ just after the "binary" point.

Now since $\alpha < 2$, $\beta > \frac{1}{2}$ and thus

$$2^{m}\beta + \beta > N_m + f_m + \frac{1}{2} > N_m + 1$$

And thus we have

$$2^{m} \beta < N_m + 1 < 2^m \beta + \beta$$

Dividing by $\beta$ gives

$$2^{m} < (N_m+1) \alpha < 2^m  + 1$$

i.e 

$$ [(N_m+1) \alpha] = 2^m$$

Since we had infintely many choice for $m$, we are done.