At least one even number among $\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$ 
For any positive integer $n$, prove that the set
$$\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$$
contains at least one even number.

I tried to prove this question by contradiction, assuming that each element is odd. There exist the positive integers $k_1, k_2, ..., k_{n+1}$ such that
$$2k_1-1<2^n\sqrt{2}<2k_1$$
$$2k_2-1<2^{n+1}\sqrt{2}<2k_2$$
$$...$$
$$2k_{n+1}-1<2^{2n}\sqrt{2}<2k_{n+1}$$
But i can't find a contradiction among these inequalities.

For the sake of contradiction assume that each element in the set is odd. Then, for some $m \geq 1$, we have:
$$2m-1 < 2^n\sqrt{2}<2m$$
and multiplying by $2$:
$$4m-2 < 2^{n+1}\sqrt{2}<4m$$
However, since $\lfloor 2^{n+1}\sqrt{2}\rfloor$ is odd, then
$$4m-1<2^{n+1}\sqrt{2}<4m$$
Repeating the process
$$2^{n+1}m-1<2^{2n}\sqrt{2}<2^{n+1}m\Rightarrow \frac{1}{2^{n+1}}>m-2^{n-1}\sqrt{2}=\frac{m^2-2^{2n-1}}{m+2^{n-1}\sqrt{2}}$$
Also, since $2m>2^n\sqrt{2}\Rightarrow m^2>2^{2n-1}\Rightarrow m^2\geq 2^{2n-1}-1$. Thus:
$$\frac{1}{2^{n+1}}>\frac{m^2-2^{2n-1}}{m+2^{n-1}\sqrt{2}}\geq \frac{1}{m+2^{n-1}\sqrt{2}}$$
Therefore:
$$2^{n+1}<m+2^{n-1}\sqrt{2}<\frac{2^{n}\sqrt{2}+1}{2}+2^{n-1}\sqrt{2}=2^n\sqrt{2}+\frac{1}{2}<2^n\sqrt{2}+\frac{2^n}{2}$$
and thus
$$2<\sqrt{2}+\frac{1}{2}$$
which is a contradiction.
