<s>[INST] 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
 [/INST] HINT:
If $f(n)=2^{4n+2}+3^{n+2}$
$f(m+1)-3f(m)=2^{4(m+1)+2}+3^{(m+1)+2}-3[2^{4m+2}+3^{m+2}]=2^{4m+2}(2^4-3)$ which is divisible by $13$
So, $13|f(m+1)\iff13|f(m)$
Alternatively try with $f(m+1)-2^4f(m)$
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