$ \sum\limits_ {k=0}^{n} \binom {2n+1}{2k+1}2^{3k}$ no es divisible por 5

Question

No tengo ni idea

Demuestre que para cualquier $n$ número natural esta suma $$ \sum\limits_ {k=0}^{n} \binom {2n+1}{2k+1}2^{3k}$$ no es divisible por $5$ .

$ \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} $

gracias de antemano

Answer 1

Usaremos $ \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 $ y $-1- \sqrt2 $ son raíces de $x^2-3x-1 \equiv0\pmod5 $ así que tenemos $$ a_n \equiv3a_ {n-1}+a_{n-2} \pmod5\tag {2} $$ donde $a_0=1$ y $a_1=1$ . Por lo tanto, $a_n$ mod $5$ es $$ 1,1,4,3,3,2,4,4,1,2,2,3,1,1, \dots\tag {3} $$ que se repite, ya que $(2)$ es un recurrencia lineal de segundo orden . Por lo tanto, ninguno de los $a_n$ son $0$ mod $5$ .