For positive integers $ n$, $ f_n = \lfloor2^n\sqrt {2008}\rfloor + \lfloor2^n\sqrt {2009}\rfloor$. Prove there are infinitely many odd numbers and infinitely many even numbers in the sequence $ f_1,f_2,\ldots$.
Problem
Source: China Girls Math Olympiad 2008, Problem 8
Tags: search, number theory unsolved, number theory
26.08.2008 23:35
Similar to this: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2010018844&t=154073
27.08.2008 01:03
cosinator wrote: Similar to this: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2010018844&t=154073 But harder...
27.08.2008 01:12
It boils down to simply proving $ \sqrt {2009}\pm\sqrt {2008}$ is irrational, no?
27.08.2008 01:45
It seems like that according to the link provided. ><
02.09.2008 16:04
cosinator wrote: It boils down to simply proving $ \sqrt {2009}\pm\sqrt {2008}$ is irrational, no? yup... It is rather strange that they still propose problems which can be solved by these standard tricks....I've seen this method when I was in 9th grade and since that time I have solved so much problems using this trick,but I must admit that this method sometimes does not work...if this trick doesn't work I strongly suggest to use dirichlet theorem about irrational numbers...
04.02.2013 06:35
I think there is a very nice solution by considering this base 2. If the sequence was all odd after a while, then that would mean the nth digit of (2008)^.5 is always different from the nth digit of (2009)^.5 when working base 2. therefore (2008)^.5+(2009)^.5 must be rational as can be seen from it having all 1s after a certain point in the base 2 expansion, but its not difficult to show this is false. If the sequence was even after a certain points, the nth digits would be the same and we would only have to show (2009)^.5-(2008)^.5 is not rational which is also not difficult.
08.08.2019 17:55
For each $n \in \mathbb{N}$, let $a_n = 1$ if $\{2^n \sqrt{2008}\} > \frac12$ and $0$ else. Similarly define $b_n$ for $\sqrt{2009}.$ Assume, for contradiction, that $f_1, f_2, \cdots$ had the same parity after a while. Then, $f_{n+1} - 2f_n = a_n + b_n$ also has invariant parity after a while. Therefore, we have that for big enough $n \ge N$, either $a_n = b_n$ or $a_n = 1-b_n$. This implies that $2^N\sqrt{2008}$ and $2^N \sqrt{2009}$ either have the same fractional part or add up to an integer. Hence, it suffices to show that $\sqrt{2008} \pm \sqrt{2009}$ is not a rational number. Since they are both algebraic integers, we just need to show that they are not integers. However this is easy to see since $\sqrt{2008} + \sqrt{2009} \in (89, 90)$ and $\sqrt{2009} - \sqrt{2008} \in (0, 1).$ $\square$
05.06.2022 15:22
Let the binary representation of $\sqrt{2008}$ be $\overline{\ldots a_0.a_1a_2\ldots}_2$ and the binary representation of $\sqrt{2009}$ 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 $\sqrt{2009}-\sqrt{2008}$ has in binary form eventually terminates, so it's rational. But this isn't true by squaring, as $2009 \cdot 2008$ 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 $\sqrt{2009}+\sqrt{2008}$ 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. $\blacksquare$
19.07.2022 18:50
Let $\{\sqrt{2008}\} = 0.\overline{a_1a_2a_3\cdots a_k\cdots}$ and $\{\sqrt{2009}\} = 0.\overline{b_1b_2b_3\cdots b_k\cdots}$ in their binary representations. If there are only finitely many odd $f_i$, for all $k>N$ we have $b_k + a_k = 1$, so $$\sqrt{2008} + \sqrt{2009} = \frac{n+0.\overline{111\cdots}}{2^N} = \frac{n+1}{2^N}$$for some positive integer $n$ is rational, contradiction. If there are only finitely many even $f_i$, for all $k>N$ we have $b_k = a_k$, so $$\sqrt{2009} - \sqrt{2008} = \frac n{2^N}$$for some positive integer $n$ as the repeating parts cancel out. This is rational, which is again a contradiction.