Prove that for all $n \in N$,$3^{2n+2} - 8n - 9$ is divisible by $8$ using the principle of mathematical induction.

(N/A) Let the given statement be $P(n)$,i.e.,$P(n): 3^{2n+2} - 8n - 9$ is divisible by $8$.
Step $1$: For $n = 1$,$3^{2(1)+2} - 8(1) - 9 = 3^4 - 8 - 9 = 81 - 17 = 64$. Since $64$ is divisible by $8$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$,i.e.,$3^{2k+2} - 8k - 9 = 8m$ for some $m \in N$. Thus,$3^{2k+2} = 8m + 8k + 9$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$3^{2(k+1)+2} - 8(k+1) - 9$ is divisible by $8$.
Consider $3^{2k+4} - 8k - 8 - 9 = 3^2 \cdot 3^{2k+2} - 8k - 17$.
Substituting $3^{2k+2} = 8m + 8k + 9$:
$= 9(8m + 8k + 9) - 8k - 17$
$= 72m + 72k + 81 - 8k - 17$
$= 72m + 64k + 64$
$= 8(9m + 8k + 8)$.
Since this is a multiple of $8$,$P(k+1)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.