Proving that $64$ divides $3^{2n+2}+56n+55$ by induction Let $n ≥ 0$ be an integer. Prove by induction: 64 divides $3^{2n+2} + 56n + 55$
general expression: $3^{2n+2} + 56n + 55 = 64m$
1st I substitute $P(0)$ and it gives me true:
$9+55 = 64$ (if m = 1 the condition is true)
2nd I assume $n = k$ and I substitute saying $3^{2k+2} + 56k + 55$
and then I don't know what to do to prove the claim.

Assume that 
$$3^{2k+2}+56k+55=64m.$$
Then, use 
$$3^{2(k+1)+2}=3^{2k+4}=3^2\cdot 3^{2k+2}$$
and
$$3^{2k+2}=64m-56k-55.$$

$\begin{align}3^{2(k+1)+2}+56(k+1)+55&=9\cdot 3^{2k+2}+56k+56+55\\&=9(64m-56k-55)+56k+56+55\\&=9\cdot 64m-9\cdot 56k-9\cdot 55+56k+56+55\\&=9\cdot 64m-8\cdot 7\cdot 8k-6\cdot 64\\&=64(9m-7k-6)\end{align}$

