$\sum\limits_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}$ isn't divisible by 5

Question

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

Answer 1

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 $5$.