proof verification - Using induction, prove that $({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$.

Question: Using the principle of mathematical induction, prove that for every integer $n\geq 1, ({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$.

My attempt:

Let $P(n): ({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$ ($n\geq 1).$

BASE CASE:

$P(1): ({3^2}^1 -1)$ is divisible by $2^{1+2}$ but not by $2^{1+3}$ , that is, $8$ is divisible by $8$ but not by $16$, which is true.

$\therefore P(1)$ is true

INDUCTION HYPOTHESIS:

Let $P(n)$ be true for $n=k$.

$\therefore P(k)$ is true, that is, $({3^2}^k -1)$ is divisible by $2^{k+2}$ but not by $2^{k+3}$.

$\therefore$ Let ${3^2}^k -1=\lambda.2^{k+2}$

$\implies {3^2}^k=1+\lambda.2^{k+2}$

INDUCTIVE STEP:

$P(k+1):({3^2}^{k+1} -1)$ is divisible by $2^{k+1+2}$ but not by $2^{k+1+3}$

We have to prove that $P(k+1)$ is true

${3^2}^{k+1}-1={3^{{2^k}2}}-1=(1+\lambda.2^{k+2})^2-1=1+2\lambda.2^{k+2}+\lambda^2.2^{2(k+2)}-1=\lambda.2^{k+3}+\lambda^2.2^{k+3+k+1}=2^{k+3}(\lambda+\lambda^2.2^{k+1})$

$\therefore {3^2}^{k+1}-1$ is divisible by $2^{k+3}$

My problem: How do I prove that the number is not divisible by $2^{n+3}$? What should I add to the induction hypothesis part and the inductive step. I am not allowed to use congruence. I am allowed to use the properties of numbers and their division. I thought of doing it by proving that the number leaves a particular remainder on dividing by $2^{n+2}$ but then realized that it would be very cumbersome since the remainder can range from $1$ to $2^{n+3}-1$.
Please help.

Answer

In the inductive step, your aim should be to prove that $3^{2^{k+1}}-1$ is divisible by $2^{k+3}$, not $2^{k+2}$. You should add to the inductive hypothesis that $\lambda$ is odd. To be more precise: you should aim at proving this:$$3^{2^k}-1\text{ can be written as }\lambda.2^{k+2}\text{ for some odd }\lambda.$$If this holds for some $k\in\mathbb N$, then$$3^{2^{k+1}}-1=\left(\lambda+\lambda^2.2^{k+1}\right).2^{k+3}$$and $\lambda+\lambda^2.2^{k+1}$ will be odd too.