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!
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.