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?

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 $2^N = \lfloor \sqrt{2} n_N \rfloor$.
