Proof by Induction - Algebra Problem (Steps included but not understood) I do not quite understand this proof, if anyone could explain the steps for me it would be greatly appreciated. It's probably something glaringly obvious I'm not seeing, thanks in advance.
Prove that for every integer $n \ge 0,$ the number $4^{2n+1}+3^{n+2} $ is a multiple of 13.
Proof. We use induction on n, starting with $n=0$
$P(0):4^{2(0)+1}+3^{0+2}=4+9=13=13\cdot1$

Assume $P(k):4^{2k+1}+3^{k+2}=13t$ for some integer $t$. We must prove
$P(k+1): 4^{2(k+1)+1}+3^{(k+1)+2}$ is a multiple of $13$.
We have
$4^{2(k+1)+1}+3^{(k+1)+2}=4^{(2k+1)+2}+3^{(k+2)+1}$
$=4^2(4^{2k+1})+4^2(3^{k+2}-3^{k+2})+3\cdot3^{k+2}$
$=4^2(4^{2k+1}+3^{k+2})+3^{k+2}(-4^2+3)$
$=16\cdot13t+3^{k+2}\cdot(-13)$    (by $P(k)$)
$=13(16t-3^{k+2})$, proven.

We know something about $4^{2k+1}+3^{k+2}$, but in the expression
$$
4^2\cdot 4^{2k+1}+3\cdot3^{k+2}
$$
we have just $4^{2k+1}$. So we insert a term $4^2\cdot 3^{k+2}$ in order to apply the induction hypothesis. We add nothing provided we subtract the same term:
$$
4^2\cdot 4^{2k+1}+3\cdot3^{k+2}=4^2\cdot 4^{2k+1}+4^2\cdot 3^{k+2}-4^2\cdot 3^{k+2}+3\cdot3^{k+2}
$$
and go on with
$$
4^2\cdot 13t-3^{k+2}(16-3)
$$
which is a multiple of $13$.
A different way of doing the same thing is rewriting the induction hypothesis as
$$
4^{2k+1}=13t-3^{k+2}
$$
and substituting it in the expression for $P(k+1)$:
$$
4^2\cdot 4^{2k+1}+3\cdot3^{k+2}=4^2(13t-3^{k+2})+3\cdot3^{k+2}.
$$
