7

Problem

Find the largest $n \in \mathbb{N_+} $ such that $\{ \left(2+\sqrt{2}\right)^n\} < \dfrac{7}{8}$, where $\{x\}$ denotes the fractional part of $x.$

My Solution

First, we can prove that $a_n=(2+\sqrt{2})^n+(2-\sqrt{2})^n$ is an integer sequence. For this purpose, we may apply the mathematical induction. However,in fact,by setting up the characteristic equation $x^2-4x+2=0$，we may confirm that the equality above of $a_n$ really gives the general term formula of the recursion sequence as follows $$a_1=2,a_2=12,a_{n+2}=4a_{n+1}-2a_{n}(n=1,2,\cdots).$$ Now, it's clear that $a_n$ are a series of integers. Moreover, notice that $0<(2-\sqrt{2})^n<1.$ We can obtain $\{(2+\sqrt{2})^n\}=1-(2-\sqrt{2})^n.$ Thus, the problem is to ask us to find the largest $n \in \mathbb{N_+}$ such that
$$\left(\frac{2+\sqrt{2}}{2}\right)^n<8.$$ But the left side increases with the increasing $n$. Hence, we only need to test the critical value.

Since $$2^{3/4}=\sqrt{2 \cdot \sqrt{2}}<\frac{2+\sqrt{2}}{2}<\frac{2+2}{2}=2,$$ we have $\left(\dfrac{2+\sqrt{2}}{2}\right)^4>2^3=8$, and $\left(\dfrac{2+\sqrt{2}}{2}\right)^3<2^3=8.$ As a result, the largest $n$ is $3$.

Please correct me if I'm wrong! And I hope to see another new solution. Thanks!

Community
  • 1
mengdie1982
  • 13,840
  • 1
  • 14
  • 39
  • I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – GNUSupporter 8964民主女神 地下教會 May 18 '18 at 19:20
  • Sorry, I'm a newer here. So I'm not familiar to the rules of the website. Thanks for your reminding. @GNU Supporter – mengdie1982 May 18 '18 at 19:26
  • Your use of AGM to get $2^{3/4}\lt{2+\sqrt2\over2}$ so that you can test the inequality for $n=4$ without doing a bunch of computation is really nice. Well done! – Barry Cipra May 18 '18 at 23:58
  • @BarryCipra Thanks for your comment. – mengdie1982 May 19 '18 at 04:32
  • This approach is great. However, it is a bit confusing when you use the fact that $$\color{#C00}{\frac1{\left(2-\sqrt2\right)^n}=}\left(\frac{2+\sqrt2}2\right)^n\lt8$$ It might improve clarity to include something like the red part above – robjohn May 19 '18 at 12:35
  • However，I think it to be obvious.Since that $(2+\sqrt{2})(2-\sqrt{2})=2^2-(\sqrt{2})^2=2,$ hence $ \dfrac{1}{2-\sqrt{2}}=\dfrac{2+\sqrt{2}}{2}$. – mengdie1982 May 19 '18 at 13:30

1 Answers1

0

Hint.

Note that

$$ (2+\sqrt 2)^n = a_n+\sqrt 2 b_n\\ (2+\sqrt 2)^{n+1} = a_{n+1}+\sqrt 2 b_{n+1} $$

then

$$ a_{n+1} = 2(a_n+b_n)\\ b_{n+1} = a_n + 2 b_n $$

with $a_1 = 2, b_1 = 1$

NOTE

As far as I understood, I am afraid that the problem has no solution because

$$ \{(2+\sqrt 2)^n\} = \{\sqrt 2 b_n\} $$

Suppose now we have a $n = n^*$ such that $\{\sqrt 2 b_n\}\lt\frac{7}{8}$ The sequence of $b_n$ follow without limit an then surely will appear another $n^{**}$ that will achieve that.

Cesareo
  • 33,252