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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.

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$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 $a_{n+2k}-a_{n} \gt 6k$.

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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.

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Why do you need the transitions? Isn't it enough to regard the infinite places where it agrees to get even and where it disagrees to get odd? –  Phira Nov 27 '12 at 14:05
    
Sure. I wrote it down the way my visualization went. If you don't mind, later I can shorten things as per your suggestion. –  André Nicolas Nov 27 '12 at 16:36
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