proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$

Question

I am supposed to proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n$ being a natural number (integer, $n > 0$).

What I have so far:

Basis: $n = 1$ \begin{align} 3^{2 \cdot 1-1} + 2^{2 \cdot 1-1} & = 3^1 + 2^1\\ & = 5 \end{align}

Assumption: $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n = k \in \mathbb{N}$.

$5 \mid (3^{2n-1} + 2^{2n-1}) \implies 3^{2n-1} + 2^{2n-1} = 5m$, $m \in \mathbb{Z}$

Proof: Let $n = k + 1$ \begin{align} 3^{2 \cdot (k+1)-1} + 2^{2 \cdot (k+1)-1} & = 3^{2k+2-1} + 2^{2k+2-1}\\ & = 3^{2k+1} + 2^{2k+1}\\ & = 3^{2k} \cdot 3^1 + 2^{2k} \cdot 2^1\\ & = 3^{2k} \cdot 3 + 2^{2k} \cdot 2 \end{align} And here I got stuck. I don't know how to get from the last line to the Assumption. Either I am overlooking a remodeling rule or I have used a wrong approach.

Anyway, I am stuck and would be thankful for any help.

Answer 1

Let's use, from hypothesis, that $$3^{2k-1}+2^{2k-1}=5p\to 3^{2k-1}=5p-2^{2k-1}$$ so

$$3^{2k+1}+2^{2k+1}=9\cdot3^{2k-1}+4\cdot2^{2k-1}=9\cdot(5p-2^{2k-1})+4\cdot2^{2k-1}=\\ 45p-5\cdot2^{2k-1}=5(9p-2^{2k-1})$$

Answer 2

let $$T_n=3^{2n-1}+2^{2n-1}$$ and $$5|T_n$$ we have to Show that $$5|T_{n+1}$$ indeed we have $$T_{n+1}-T_n=3^{2n+1}+2^{2n+1}-3^{2n-1}-2^{2n-1}=3^{2n-1}\cdot 8+2^{2n-1}\cdot 3=3^{2n-1}(10-2)+2^{2n-1}(5-2)=5(2\cdot3^{2n-1}+2^{2n-1})-2(3^{2n-1}+2^{2n-1})$$

Answer 3

From where you got to:

\begin{align} 3^{2 \cdot (k+1)-1} + 2^{2 \cdot (k+1)-1} & = 3^{2k+2-1} + 2^{2k+2-1}\\ & = 9\cdot3^{2k-1} + 4\cdot 2^{2k-1}\\ & = 5\cdot3^{2k-1} + 4(3^{2k-1}+ 2^{2k-1})\\ & =5\cdot3^{2k-1} + 4\cdot 5m \tag {from hypothesis}\\ & =5(3^{2k-1} + 4m)\\ \end{align}

... so divisible by $5$ as required

Answer 4

The sequence $a_n = 3^{2n-1}+2^{2n-1}$ fulfills the recurrence relation $$ a_{n+2} = 13 a_{n+1} - 36 a_n $$ and since both $a_1$ and $a_2$ are $\equiv 0\pmod{5}$, by the previous recurrence relation every term of the sequence is $\equiv 0\pmod{5}$. Anyway, it is probably easier to notice that if $d$ is a positive odd integer, $(x+y)$ is a divisor of $x^d+y^d$, hence $5=(2+3)$ is for sure a divisor of $2^{2n-1}+3^{2n-1}$.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} 3^{2n + 1} + 2^{2n + 1} & = 9 \times 3^{2n - 1} + 4 \times 2^{2n - 1} = \pars{10 - 1}3^{2n - 1} + \pars{5 - 1}\times 2^{2n - 1} \\[5mm] & = \color{#f00}{5}\pars{2 \times 3^{2n - 1} - 2^{2n - 1}} - \pars{\color{#f00}{3^{2n - 1} + 2^{2n - 1}}} \end{align}