<s>[INST] Strategy for solving $7\vert2^{n+2}+3^{2n+1}$ by induction. So I have to show the following to be true using induction

$7\mid 2^{n+2}+3^{2n+1}$

This is easily checked with the case $n=0$ because $7 \mid 7$, but I assuming this holds for$n=k :$
$$7\mid 2^{k+2}+3^{2k+1}$$
I fail to see how I can use this for $n=(k+1):$
$$\begin{align*}\begin{split}
&7 \mid \left(2^{k+3}+3^{2k+3}\right) &\Longleftrightarrow\\
&7 \mid \left(2\times2^{k+2}+3^2\times3^{2k+1}\right)
\end{split}\end{align*}
$$
I'm assuming my strategy is flawed, but it's a while since I've done these and I can't remember what should be done.
 [/INST] You checked it for $n=0$, so now you have to prove implication:
$$ \left(\forall n \in \mathbb{N}\right)\left(\underbrace{7 \mid 2^{n+2} + 3^{2n+1}}_{\mathrm{Assumption}} \Longrightarrow
\underbrace{7 \mid 2^{(n+1)+2} + 3^{2(n+1)+1}}_{\mathrm{Thesis}}\right)$$
Simple thesis transformation:
$$\begin{align*}\begin{split}
2^{(n+1)+2} + 3^{2(n+1)+1} &= 2^1 \cdot 2^{n+2} + 3^2 \cdot 3^{2n+1}\\
                           &= 2 \cdot 2^{n+2} + 9 \cdot 3^{2n+1}\\
                           &= 2 \cdot 2^{n+2} + (2+7) \cdot 3^{2n+1}\\
                           &= \left(2 \cdot 2^{n+2} + 2\cdot 3^{2n+1}\right) +
7 \cdot 3^{2n+1}\\
                           &=2\cdot\left(2^{n+2} + 3^{2n+1} \right) + 7 \cdot
3^{2n+1}
\end{split}\end{align*}$$
From assumption $(\exists k \in \mathbb{Z})(2^{n+2} + 3^{2n+1} =7k)$, so
$$\begin{align*}\begin{split}
2^{(n+1)+2} + 3^{2(n+1)+1} &= 2\cdot\left(2^{n+2} + 3^{2n+1} \right) + 7 \cdot
3^{2n+1}\\
                           &= 2 \cdot 7k + 7 \cdot 3^{2n+1} \\
                           &= 7 \cdot \left(2\cdot k + 3^{2n+1}\right)
\end{split}\end{align*}$$
$$7 \mid \left(7 \cdot \left(2\cdot k + 3^{2n+1}\right) \right)
\Longleftrightarrow
7 \mid \left(2^{(n+1)+2} + 3^{2(n+1)+1}\right)$$
Which proves the inductive thesis, and we obtain $\left(\forall n \in \mathbb{N}\right)\left(7 \mid  2^{n+2} + 3^{2n+1} \right)$.
</s>