Prove that $17$ divides $3^{4n} + 4^{3n+2}$ for all natural numbers $n$ by induction. I would like to preface my very first question here by stating that this is not homework. Indeed, I am a mathematics major and computer science minor. However, I am doing extra work on my own to develop my understanding. I am unsure of the informal proof I have created, and my professor is not immediately available, so I am here asking for help.
My understanding goes about as far as multivariable calculus. After this semester, I will be heading into junior level classes. Anyway, that is enough preface.
Here is my informal proof by induction. I've only been writing proofs for about a month, so please be gentle.
$17$ divides $3^{4n} + 4^{3n+2} \ \forall n \in \Bbb{N}$
Proof: (PMI)
 case 1: $n = 1$ 
  $3^{4(1)} + 4^{3(1)+2} = 1,105,$
 $\implies$ $17$ | $1,105$,
 $\implies$ $17m$ = $1,105, \exists m \in \Bbb{N},$
 $\implies m = 65,$
 $\implies$ $17$ | $3^4 + 4^5.$
 case 2: $n = k$
   Assume: $17$ | $3^{4k} + 4^{3k+2}$ given $k \in \Bbb{N}.$
   WTS: $17$ | $3^{4(k+1)} + 4^{3(k+1)+2}$
 $\implies$ $\exists k_1 \in \Bbb{N}$ s.t. $17k_1 = 3^{4k} + 4^{3k+2}.$
 $\implies$ $3^{4(k+1)} + 4^{3(k+1)+2} = 3^{4k+4} + 4^{3k+3+2},$
            $= 3^{4} \cdot 3^{4k} + 4^{2} \cdot 4^{3} \cdot 4^{3k},$
            $= 3^{4}(3^{4k}) + 4^{3}(4^{3k+2}),$
            $=81(3^{4k} + 4^{3k+2}) - 81(4^{3k+2}) + 64(3^{4k} + 4^{3k+2})-64(3^{4k})$,
            $=81(3^{4k} + 4^{3k+2} - 4^{3k+2}) + 64((3^{4k} + 4^{3k+2} - 3^{4k})$,
            $=(81-64)(3^{4k} - 4^{3k+2})$,
            $=17(3^{4k} - 4^{3k+2})$,
            $=17k_1$,
          $\implies 17$ | $3^{4k+4} + 4^{3k+5}$.
$\Box$
If I am way off base here, I would greatly appreciate your solutions (to study) or nudges in the right direction. Thank you.

Assume true for $n=k$.  For $n=k+1$ we have (using $3^4=81$ and $4^3=64$) $81\times 3^{4k}+64\times 4^{3k+2}=17\times 3^{4k}+64\times(3^{4k}+4^{3k+2})$, which is divisible by $17$.
