CGMO 2008

usjl wrote: 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 $v_1,v_2,\ldots,$ there are infinitely many odd integers, as well as infinitely many even integers. Yeah, similar to @above, We will use the fact that $\sqrt{2021},\sqrt{2020}$ are irrational. And so,if we represent $\sqrt{2020}=a_0.a_1 a_2......$ and $\sqrt{2021}=b_0.b_1 b_2......$ And if ${a_i}+{b_i}=2k+1$ then $V_i$ would be even And if ${a_i}+{b_i}=2k $ then $V_n$ would be odd. And since there are infinitely many such ${a_i},{b_i}$ satisfing the above so infinitely many n exist. @below,sorry for writing two wrong proofs here.i knew this proof however I was desperate to do it in a other way.I hope this one's fine.

@above $\lfloor 2^4\sqrt{2020}\rfloor = 719 \equiv 1 \not \equiv 0 \pmod 2$, and there are many other counterexamples to your claims. Also I think $\lfloor 2^n\sqrt{2020}\rfloor$ modulo $2$ does not follow a nice pattern. (Neither does $\lfloor 2^n\sqrt{2021}\rfloor$) @above Lol, still your both claims are totally wrong and there exist many counterexamples If you want to see those, here are some of them: $V_{17} = 11783364 \equiv 0 \pmod{2} , V_{18} = 23566730 \equiv 0 \pmod{2} , V_{19} = 47133460 \equiv 0 \pmod{2} , V_{20} = 94266922 \equiv 0 \pmod{2} , V_{21} = 188533846 \equiv 0 \pmod{2}$. If you try to find a pattern modulo $2$ in this sequence, you will never succeed. Also it is clear that your proof cannot be fixed since any pattern argument will not work. @above No problem! But your ideas were very natural and I liked it.

Note that $2020$ and $2021$ have no relevance whatsoever. The problem holds for integers $a,b$ such that $\sqrt a+\sqrt b\notin\mathbb{Z}.$ Assume that for all $n>N,$ $V_n\equiv c\bmod{2}.$ Then, for all $n>N, V_n-2V_{n-1}\equiv c\bmod{2}.$ However, the latter rewrites as \[2^n\big(\sqrt{a}+\sqrt{b}\big)-\big\{2^n\sqrt a\big\}-\big\{2^n\sqrt b\big\}-2\bigg(2^{n-1}\big(\sqrt{a}+\sqrt{b}\big)-\big\{2^{n-1}\sqrt a\big\}-\big\{2^{n-1}\sqrt b\big\}\bigg)\]which is equal to $2\big(\big\{2^{n-1}\sqrt a\big\}+\big\{2^{n-1}\sqrt b\big\}\big)-\big\{2^n\sqrt a\big\}-\big\{2^n\sqrt b\big\}.$ For simplicity, let $\big\{2^k\sqrt a\big\}=\varepsilon_k$ and $\big\{2^k\sqrt b\big\}=\delta_k.$ If $\varepsilon_{n-1}<1/2$ then $\varepsilon_n=2\cdot\varepsilon_{n-1}.$ Otherwise, $\varepsilon_n=2\cdot\varepsilon_{n-1}-1.$ The same clearly holds for $\delta_n$ and $\delta_{n-1}$ too. Now, since for all $n>N, \ 2(\varepsilon_{n-1}+\delta_{n-1})-\varepsilon_n-\delta_n\equiv c\bmod 2$ then for all $n>N, \varepsilon_n$ and $\delta_n$ are either both in one of $[0,1/2)$ and $[1/2,1)$ or are in different intervals. However, by Kronecker's lemma, we can find integer solutions to both of the following systems \[\begin{cases}\varepsilon_n<1/2 \\ \delta_n<1/2\end{cases}\begin{cases}\varepsilon_n<1/2 \\ \delta_n\geq 1/2\end{cases}\]which clearly contradicts our observation.

Let the binary representation of $\sqrt{2020}$ be $\overline{\ldots a_0.a_1a_2\ldots}_2$ and the binary representation of $\sqrt{2021}$ be $\overline{\ldots b_0.b_1b_2\ldots}_2$. It is easy to see that $V_i$ is even if and only if $a_i=b_i$. Suppose that there are finitely many odd $V_i$, so every sufficiently large $i$ satisfies $a_i=b_i$. Then $\sqrt{2021}-\sqrt{2020}$ has in binary form eventually terminates, so it's rational. But this isn't true by squaring, as $2021 \cdot 2020$ isn't a square. Likewise, if there are finitely many even $V_i$, every sufficiently large $i$ satisfies $a_i\neq b_i \implies a_i+b_i=1$ since $a_i,b_i \in \{0,1\}$. Then $\sqrt{2021}+\sqrt{2020}$ has a binary expansion that's eventually all $1$'s, which can be "re-expressed" as terminating (since $1_2=0.\overline{1}_2$), so it's rational. But this isn't true by squaring either. $\blacksquare$

My solution. Since $\sqrt{2020}, \sqrt{2021}$ are irrational, their binary representation is unique. Note that $V_n$ is just the sum of the binary representation of $\sqrt{2020}$ and $\sqrt{2021}$ before the $n$-th digit after the radix point. In particular, write $\sqrt{2020} = *.a_1a_2a_3...$ and $\sqrt{2021} = *.b_1b_2b_3...$ in their binary representation, then $V_n$ is odd/even iff $a_n + b_n$ is. To see that there are infinite odd $V_n$s, we assume on the contrary that eventually when $n>N_0$, $a_n-b_n$ are even i.e. $0$. This implies $\sqrt{2020} - \sqrt{2021}$ is rational, as one can verify that $V_{N_0}$ with the middle "+" substituted with "-" is one binary representation of $2^{N_0}(\sqrt{2020} - \sqrt{2021})$, which is finite, a contradiction. To see that there are infinite even $V_n$s, we assume that eventually when $n > N_1$, $a_n+b_n$ are odd i.e. $1$. Then $2^{N_1}(\sqrt{2021}+\sqrt{2020})$ has binary representation $V_{N_1} + 1$; finite, contradiction.