<s>[INST] 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.
 [/INST] 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 $$T_{n+1}-T_n=3^{2n+1}+2^{2n+1}-3^{2n-1}-2^{2n-1}=3^{2n-1}\cdot 8+2^{2n-1}\cdot 3=3^{2n-1}(10-2)+2^{2n-1}(5-2)=5(2\cdot3^{2n-1}+2^{2n-1})-2(3^{2n-1}+2^{2n-1})$$
</s>