<s>[INST] 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. 
 [/INST] Extract a factor $2\lambda$ from your final bracket. Since you have raised the exponent by $1$ you need to prove divisibility by $2^{k+3}$ but not $2^{k+4}$. You should be able to show that $\lambda$ is odd, as is the remaining factor.
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