proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ 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.
 A: 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
A: Let's use, from hypothesis, that $$3^{2k-1}+2^{2k-1}=5p\to 3^{2k-1}=5p-2^{2k-1}$$ so
$$3^{2k+1}+2^{2k+1}=9\cdot3^{2k-1}+4\cdot2^{2k-1}=9\cdot(5p-2^{2k-1})+4\cdot2^{2k-1}=\\
45p-5\cdot2^{2k-1}=5(9p-2^{2k-1})$$
A: 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 $$T_{n+1}-T_n=3^{2n+1}+2^{2n+1}-3^{2n-1}-2^{2n-1}=3^{2n-1}\cdot 8+2^{2n-1}\cdot 3=3^{2n-1}(10-2)+2^{2n-1}(5-2)=5(2\cdot3^{2n-1}+2^{2n-1})-2(3^{2n-1}+2^{2n-1})$$
A: 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 $2^{2n-1}+3^{2n-1}$.
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\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}
