Prove $3^{2^n}-1$ divisible by $2^{n+2}$. Please, help

Question

loser66 · Answer

I tried and it is a mess!!! :)

loser66 · Answer

Likely we need prove 
$3^{2^{n+1}}-1$ divided by $2^{n+1}$ given that  $3^{2^n}-1$ divided by $2^n$

it is an ugly step.

loser66 · Answer

I tried: 
$2^n$ is an even number, hence, it can be written as 2m. Then
$3^{2m}-1= 9^m-1=(9-1)(9^{m-1}+9^{m-2}+\cdots +1)$

$9-1) =8 \div 4$, hence the original one divided by $4=2^2$

The leftover is to prove it is divided by $2^n$ but how??

holsteremission · Answer

Assume the statement holds for $n=k$, then use that fact to show it holds for $n=k+1$.
$$3^{2^{k+1}}-1=\left(3^{2^k}\right)^2-1=\left(3^{2^k}-1\right)\left(3^{2^k}+1\right)$$By the hypothesis, you know that $3^{2^k}-1$ is divisible by $2^{k+2}$, so all you need to do is show that $3^{2^k}+1$ is divisible by $2$.

This could probably also be shown inductively, but I don't see a need for that unless you want to be as explicit as possible in showing that every power of $3$ is odd.

loser66 · Answer

Thanks a ton @HolsterEmission  I got it. It is elegant logic when you turn $3^{2^{k+1}}$ to $(3^{2^k})^2$