Prove that the sum is divisible by $7$ 
Prove that if $m$ is an odd positive integer not divisible by $3$, then $$4^m-2\sum_{k=0}^{\frac{m-1}{2}} \binom{m}{2k}2^{m-k}$$ is divisible by $7$.

Since $4^m \equiv 4,2,1,\ldots \pmod{7}$, we take two cases: $m \equiv 1 \pmod{3}$ and $m \equiv 2 \pmod{3}$. In the first case, $m = 1+6k$ and so $4^m \equiv 4 \pmod{7}$ and in the second case, $m = 2+3(1+2k) = 5+6k$ and so $4^m \equiv 2 \pmod{7}$. How do we continue from here?
Note: Solving this question also solves part of this question.

If I am not mistaking, the sequence given is $4^m - u_m$ where
$$ u_m = (2+\sqrt{2})^m + (2-\sqrt{2})^m $$
The characteristic equation is $\lambda^2 - 4 \lambda +2=0$ so we
have 
$$ u_{m+2}=4u_{m+1}-2 u_m \ \ \ \mbox{or} \ \ \
  \left( \begin{matrix} u_{m+2} \\ u_{m+1} \end{matrix} \right)
=
\left( \begin{matrix} 4 & -2  \\ 1 & 0  \end{matrix} \right)
\left( \begin{matrix} u_{m+1} \\ u_{m} \end{matrix} \right) $$
Write $M$ for the 2 by 2 matrix. Then $M^6 \equiv {\bf 1}$ mod 7 and thus, $u_{m+6}\equiv u_m$ mod 7. Since also $4^6\equiv 1$ mod 7 it suffices to verify the claim for $1\leq m \leq 6$. Then it is automatically verified when adding multiples of 6.
