Prove that $112$ divides the integral part of $4^m-(2+\sqrt{2})^m$ 
Let $m$ be a positive odd integer not divisible by $3$. Prove that $$112 \mid \left [ 4^m-(2+\sqrt{2})^m\right].$$

I wasn't sure how to simplify the expression since we have the integer part $[x]$ and radicals. Using the Binomial Theorem we have $$(2+\sqrt{2})^m = \binom{m}{0}2^m+\binom{m}{1}2^{m-1}\sqrt{2}+\cdots+\binom{m}{m}(\sqrt{2})^m,$$ but I am not sure how to use it.

Since $2-\sqrt{2}\in\left(0,\frac{2}{3}\right)$ and $(2+\sqrt{2})^m+(2-\sqrt{2})^m$ is an integer number, it is enough to study the sequences given by $a_m=(2+\sqrt{2})^m+(2-\sqrt{2})^m$ and $b_m=4^m$. We have:
$$ a_{m+2} = 4a_{m+1}-2a_{m}$$
hence the situation $\!\!\pmod{7}$ and $\!\!\pmod{16}$ is the following:
$$ \begin{array}{|c|c|c|c|c|c|c|c|}\hline n & 0 & \color{red}{1} & 2 & 3 & 4 & \color{red}{5} & 6 & \color{red}{7}\\ \hline  a_n\pmod{7} & 2 & 4 & 5 & 5 & 3 & 2 & 2 & 4 \\ \hline a_n\pmod{16} & 2 & 4 & 12 & 8 & 8 & 0 & 0 & 0  \\ \hline b_n\pmod{7} & 1 & 4 & 2 & 1 & 4 & 2 & 1 & 4 \\ \hline b_n\pmod{16} & 1 & 4 & 0 & 0 & 0 & 0 & 0 & 0\\ \hline b_n-a_n \pmod{7} & 6 & \color{red}{0} & 4 & 3 & 1 & \color{red}{0} & 6 & \color{red}{0} \\ \hline b_n-a_n\pmod{16} & 15 & \color{red}{0} & 4 & 8 & 8 & \color{red}{0}& 0 & \color{red}{0}    \\ \hline \end{array} $$
The claim now follows by direct inspection of the previous table and the chinese theorem.
Let we state it clearly: $n$ odd and not divisible by three means $n\equiv 1,5\pmod{6}$. So we are just interested in the highligthed columns above. The period of $a_n\pmod{7}$ is six and $a_n\equiv 0\pmod{16}$ for any $n\geq 5$. The period of $b_n\pmod{7}$ is three and $b_n\equiv 0\pmod{16}$ for any $n\geq 2$. So, in order to prove that $b_n-a_n\equiv 0\pmod{112}$ for any $n\equiv 1,5\pmod{6}$, it is enough to prove that it holds for $n\in\{1,5,7\}$.
