Induction with divisibility QUESTION: Prove that $16 \mid 19^{4n+1}+17^{3n+1}-4$ for all $n \in \mathbb{N}$
This is what I have so far but I'm not sure where to go from here.
PROOF:
Let $16 \mid 19^{4n+1}+17^{3n+1}-4$ equal $S(n)$.
Base case, let $n=0$.
\begin{align*}
16 &\mid 191+171-4\\
  = 16 & \mid 32\\
  = 2
\end{align*}
Therefore, $S(0)$ is true.
Using induction hypothesis, suppose $19^{4k+1}+17^{3k+1}-4$ is divisible by $16$ for all $n \in \mathbb{N}$.
Claim,
$16 \mid 19^{4k+1}+17^{3k+1}-4$,
that is $19^{4k+1}+17^{3k+1}-4=16m$, whereby $m$ is a multiple of $16$.
The above equation simplifies into,
\begin{align*} 
16 & \mid 19^{4n+5}+17^{3n+4}-4\\
16 & \mid 19^4 \cdot 19^{4k+1} + 17^3 \cdot 17^{3n+1}-4
\end{align*}
photo of my working out

To show $16  \mid 19^4 \cdot 19^{4k+1} + 17^3 \cdot 17^{3n+1}-4$ given $16 \mid 19^{4k+1}+17^{3k+1}-4$, note that
$19^4 \cdot 19^{4k+1} + 17^3 \cdot 17^{3n+1}-4=$
$17^3\color{blue}{(19^{4k+1}+17^{3n+1}-4)} + \color{blue}{(19^4-17^3)}\cdot19^{4n+1}+
\color{blue}{(17^3\cdot4-4)},$
and all of the blue terms are divisible by $16$.
