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, 2021

Algebra Difficulty 5.2 AIME, harder Prove it Taiwan

For each positive integer $n$ , define $V_n = \lfloor 2^n \sqrt{2020} \rfloor + \lfloor 2^n \sqrt{2021} \rfloor$ . Prove that, in the sequence $V_1, V_2, \dots$ , there are infinitely many odd integers, as well as infinitely many even integers.
Remark. $\lfloor x \rfloor$ is the largest integer that does not exceed the real number $x$ .

Solution

Let $a_n = 1_{\{\{2^n\sqrt{2020}\}\} > \frac{1}{2}}$ , $b_n = 1_{\{\{2^n\sqrt{2021}\}\} > \frac{1}{2}}$ . Suppose that after some term $N$ the parity of $V_n$ is always the same, then the parity of $V_{n+1} - 2V_n = a_n + b_n$ must also always be the same. This means $a_n = b_n \forall n \ge N$ , or $a_n = 1 - b_n \forall n \ge N$ . In other words, in binary, $2^N\sqrt{2020}$ and $2^N\sqrt{2021}$ either have the same fractional part, or their fractional parts sum to 1, but in either case, one of $\sqrt{2021} \pm \sqrt{2020}$ must be rational. But this implies that $\sqrt{2020} \times 2021$ is rational, a contradiction, hence the original statement holds.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.