Induction exercise check-up Prove by induction on $n$ that $13$ divides $2^{4n+2} + 3^{n+2}$ for all natural $n$.
For base case it is divisble by 13, and 
$2^{4n+6} + 3^{n+3}$ must be divisble too.
$16 * 2^{4n+2}+ 3* 3^{n+2}$
If we denote $2^{4n+2} + 3^{n+2}$ as some $13y$, 
we have $3*13y + 13*2^{4n+2}$
Didn't I make a mistake in assuming that $2^{4n+2} + 3^{n+2}$  is divisble by 13? In induction we have to prove that if something is divisible/whatever by something then n+1 is divisible too which I did

Your proof isn't wrong per se, but the wording is a bit confusing (and I would probably end up taking a point or two off if I were grading it). Here's how I would write it:
For $n=0$, $$2^{4\cdot0+2}+3^{0+2} = 2^2 + 3^2 = 13.$$
Now assume that $13|(2^{4n+2}+3^{n+2})$ for some integer $n\geqslant0$. Then $2^{4n+2}+3^{n+2}=13k$ for some integer $k$, and
$$2^{4(n+1)+2}+3^{(n+1)+2} = 2^4\cdot2^{4n+2}+3\cdot3^{n+2}=16(2^{4n+2}+3^{n+2})-13\cdot3^{n+2}=13(16k - 3^{n+2}) $$
You could also factor it into
$$3(2^{4n+2}+3^{n+2}) + 13\cdot2^{4n+2}= 13(3k + 2^{4n+2}).$$
