Infinitely many odd and even integers in a sequence. Consider the sequence $\{a_n\}_{n\in\Bbb N}$ where $a_n = \lfloor n \sqrt2 \rfloor + \lfloor n \sqrt3 \rfloor $. Show that there exists infinitely many odd and even integers in this sequence.
 A: $1 \lt (n+1) \sqrt2 - n\sqrt2 \lt 1.5 $ so $1 \le \lfloor (n+1) \sqrt2 \rfloor- \lfloor n \sqrt2 \rfloor \le 2$
$1.5 \lt (n+1) \sqrt3 - n\sqrt3 \lt 2 $ so $1 \le \lfloor (n+1) \sqrt3 \rfloor- \lfloor n\sqrt3 \rfloor \le 2$
So $a_{n+1}-a_{n}$ must be $2$, $3$ or $4$. 
Similarly $2 \lt (n+2)  \sqrt2 - n \sqrt2 \lt 3 $ so $2 \le \lfloor (n+2) \sqrt2 \rfloor- \lfloor n \sqrt2 \rfloor \le 3$ 
and $3 \lt (n+2)  \sqrt3 - n \sqrt3 \lt 4 $ so $3 \le \lfloor (n+2)  \sqrt3 \rfloor- \lfloor n \sqrt3 \rfloor \le 4$
So $a_{n+2}-a_{n}$ must be $5$, $6$ or $7$. 
To avoid  infinitely many odd and even integers in this sequence we would need $a_{n+2k}-a_{n}=6k$ (i.e. even) for all $n$ greater than some $N$ and all positive $k$.  But that is not possible, as $\sqrt2 +\sqrt3 \gt 3.1$ and so for large enough $k$ (e.g $10$ or more) we have  $a_{n+2k}-a_{n} \gt 6k$.
A: Consider the binary representations of $\sqrt{2}$ and $\sqrt{3}$. Imagine multiplying our numbers by $n=2^k$. Infinitely often, the $k$-th binary digit of $\sqrt{2}$ differs from the $k$-th binary digit of $\sqrt{3}$, else $\sqrt{3}-\sqrt{2}$ would be rational.  But infinitely often the $k$-th binary digits  agree, else $\sqrt{2}+\sqrt{3}$ would be rational.  Thus there are infinitely many transitions from agree to disagree, and also infinitely many transitions from disagree to agree,
At any transition from agree at $k$ to disagree at $k+1$, we get that $\lfloor 2^k\sqrt{2}\rfloor+\lfloor2^k \sqrt{3}\rfloor$ is even. 
At any transition from disagree at $k$ to agree at $k+1$, $\lfloor 2^k\sqrt{2}\rfloor+\lfloor2^k \sqrt{3}\rfloor$ is odd. 
