<s>[INST] $\sum\limits_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}$ isn't divisible by 5 I have no idea 

Prove that for any $n$ natural number this sum $$\sum\limits_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}$$ isn't divisible by $5$.

$\begin{array}{l}
 \left( {1 + x} \right)^{2n + 1}  - \left( {1 - x} \right)^{2n + 1}  = \sum\limits_{k = 0}^{2n + 1} {\left( {\begin{array}{*{20}c}
   {2n + 1}  \\
   k  \\
\end{array}} \right)} x^k  - \sum\limits_{k = 0}^{2n + 1} {\left( {\begin{array}{*{20}c}
   {2n + 1}  \\
   k  \\
\end{array}} \right)} \left( { - x} \right)^k  = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c}
   {2n + 1}  \\
   {2k + 1}  \\
\end{array}} \right)} x^{1 + k}  \\ 
 x = 2 \Rightarrow 3^{2n + 1}  + 1 = \sum\limits_{k = 0}^{2n + 1} {\left( {\begin{array}{*{20}c}
   {2n + 1}  \\
   k  \\
\end{array}} \right)} 2^k  - \sum\limits_{k = 0}^{2n + 1} {\left( {\begin{array}{*{20}c}
   {2n + 1}  \\
   k  \\
\end{array}} \right)} \left( { - 2} \right)^k  \\ 
 3^{2n + 1}  + 1 = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c}
   {2n + 1}  \\
   {2k + 1}  \\
\end{array}} \right)} 2^{k + 1}  \\ 
 \end{array}$
$
\begin{array}{l}
 \left\{ \begin{array}{l}
 3^0  \equiv 1\left[ 5 \right] \\ 
 3^1  \equiv 3\left[ 5 \right] \\ 
 3^2  \equiv 4\left[ 5 \right] \\ 
 3^3  \equiv 2\left[ 5 \right] \\ 
 3^4  \equiv 1\left[ 5 \right] \\ 
 \end{array} \right. \Rightarrow \forall m \in\mathbb N:\left\{ \begin{array}{l}
 3^{4m}  \equiv 1\left[ 5 \right] \\ 
 3^{4m + 1}  \equiv 3\left[ 5 \right] \\ 
 3^{4m + 2}  \equiv 4\left[ 5 \right] \\ 
 3^{4m + 3}  \equiv 2\left[ 5 \right] \\ 
 \end{array} \right. \\ 
 3^{2\left( {2m} \right) + 1}  + 1 \equiv 2\left[ 5 \right]and3^{2\left( {2m + 1} \right) + 1}  + 1 \equiv 3\left[ 5 \right] \\ 
 \end{array}
$
thank you in advance
 [/INST] We will use $\mathbb{Z}_5[\sqrt2]$
$$
\begin{align}
a_n
&=\sum_{k=0}^n\binom{2n+1}{2k+1}2^{3k}\\
&=\frac1{4\sqrt2}\left(\sum_{k=0}^{2n+1}\binom{2n+1}{k}\sqrt8^{\,k}-\sum_{k=0}^{2n+1}\binom{2n+1}{k}(-\sqrt8)^k\right)\\[3pt]
&=\frac{\left(1+2\sqrt2\right)^{2n+1}-\left(1-2\sqrt2\right)^{2n+1}}{4\sqrt2}\\[9pt]
&\equiv3(1+\sqrt2)(-1+\sqrt2)^n+3(1-\sqrt2)(-1-\sqrt2)^n\pmod5\tag{1}
\end{align}
$$
$-1+\sqrt2$ and $-1-\sqrt2$ are roots of $x^2-3x-1\equiv0\pmod5$, so we have
$$
a_n\equiv3a_{n-1}+a_{n-2}\pmod5\tag{2}
$$
where $a_0=1$ and $a_1=1$. Therefore, $a_n$ mod $5$ is
$$
1,1,4,3,3,2,4,4,1,2,2,3,1,1,\dots\tag{3}
$$
which repeats, since $(2)$ is a second order linear recurrence. Therefore, none of the $a_n$ are $0$ mod $5$.
</s>