<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] First, show that this is true for $n=1$:
$2^{4+2}+3^{1+2}=13\cdot7$
Second, assume that this is true for $n$:
$2^{4n+2}+3^{n+2}=13k$
Third, prove that this is true for $n+1$:
$2^{4(n+1)+2}+3^{(n+1)+2}=$
$2^4\cdot2^{4n+2}+3^1\cdot3^{n+2}=$
$16\cdot2^{4n+2}+3\cdot3^{n+2}=$
$(13+3)\cdot2^{4n+2}+3\cdot3^{n+2}=$
$13\cdot2^{4n+2}+3\cdot2^{4n+2}+3\cdot3^{n+2}=$
$13\cdot2^{4n+2}+3\cdot(\color{red}{2^{4n+2}+3^{n+2}})=$
$13\cdot2^{4n+2}+3\cdot\color{red}{13k}=$
$13\cdot2^{4n+2}+13\cdot3k=$
$13\cdot(2^{4n+2}+3k)$

Please note that the assumption is used only in the part marked red.
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