Show that $a_n = [n \sqrt{2}]$ contains an infinite number of integer powers of $2$ 
Show that the sequence $\{a_n\}_{n \geq 1}$ defined by $a_n = [n \sqrt{2}]$ contains an infinite number of integer powers of $2$. ($[x]$ is the integer part of $x$.)

I tried listing out the first few values, but I didn't see a pattern: $1,2,4,5,7,8,9,11,\ldots.$ Should we do a proof by contradiction?
 A: Let $1/\sqrt{2} = \sum_{j=1}^\infty d_j 2^{-j}$ be the base-2 expansion of $1/\sqrt{2}$, where each $d_j$ is $0$ or $1$.  Since $1/\sqrt{2}$ is irrational, there are infinitely many $0$'s and infinitely many $1$'s.
Let $x_N = \sum_{j=1}^N d_j 2^{-j}$ and $n_N = 1 + 2^N x_N = 1 + \sum_{j=1}^N d_j 2^{N-j}$ which is a positive integer.  If $d_{N+1} = 1$ we have
$$1/\sqrt{2} - 2^{-N-1} < x_N + 2^{-N-1} < 1/\sqrt{2}$$ and then
$$2^N < \sqrt{2}\; n_N = \sqrt{2} (x_N + 1/2^N)2^N < 2^N+\sqrt{2}/2 < 2^N + 1$$
so that $2^N = \lfloor \sqrt{2} n_N \rfloor$.
A: Let $p_k=[\frac {2^k}{\sqrt 2}], q_k=\{\frac {2^k}{\sqrt 2}\}$.  $2^k$ will equal $a_{p_k+1}$ unless $[\sqrt 2(p_k+1)]=2^k+1$, which happens when $\sqrt 2 (p_k+1) \gt 2^k+1$.  We have $\sqrt 2(p_k+1)=2^k-q_k\sqrt 2+\sqrt 2$, so this happens when $q_k \lt \frac {\sqrt 2-1}{\sqrt 2}\approx 0.293$.  Given any $k$ for which $q_k \lt \frac{\sqrt 2-1}{\sqrt 2}$, after some finite number of doublings we will reach a value where $q_{k'} \gt \frac {\sqrt 2-1}{\sqrt 2}$ and $2^{k'}$ will be in the sequence.
