$ 3^{2^n }- 1 $ is divisible by $ 2^{n+2} $ Prove that if n is a positive integer, then $ \ \large 3^{2^n }- 1  $ is divisible by  $ \ \large 2^{n+2} $ .
Answer:
For  $ n=1 \ $ we have 
$ \large 3^{2^1}-1=9-1=8 \ \ an d \ \ 2^{1+2}=8 $
So the statement hold for n=1. 
For $ n=2 $ we have 
$ \large 3^{2^2}-1=81-1=80  \ \ and \ \ 2^{2+2}=16 \ $
$Also \ \ \  16 /80 $ . 
Thus the statement hold for $ n=2 $ also. 
Let the statement hold for $ n=m \ $
Then $ a_m=3^{2^m}-1 \ \ is \ \ divisible \ \ by \ \ b_m=2^{m+2} \ $ 
We have to show that $ b_{m+1}=2^{m+3}  \ $ divide $ \ \large a_{m+1}=3^{2^{m+1}}-1 \ $ 
But right here I am unable to solve . If there any help doing this ? 
Else any other method is applicable also.
 A: We can write,
$$(3)^{2^n}-1$$
$$(4-1)^{2^n}-1$$
Using binomial,
$$=\color{red}{\binom{2^n}{0}}-\color{blue}{\binom{2^n}{1}4}+\color{green}{\binom{2^n}{2}4^2}-\cdots-\color{orange}{\binom{2^n}{2^n-1} 4^{2^n-1}}+\color{purple}{\binom{2^n}{2^n}4^{2^n}}-\color{red}{1}$$
$$$$

$$\binom{n}{r}=n (r+1) \binom{n-1}{r+1}$$

$$-\color{blue}{ 2^n 2 \binom{2^n-1}{2}4}+\color{green}{2^n 3\binom{2^n-1}{3}4^2}-\cdots+\color{magenta}{2^n (2^n-1) \binom{2^n-1}{2^n-1}4^{2^n-2}} -\color{orange}{2^n 4^{2^n-1}} +\color{purple}{4^{2^n}}$$
$$$$
$$- \color{red}{2^{n+2}}(-\color{blue}{2\binom{2^n-1}{2}}+\color{green}{3\binom{2^n-1}{3}4}-\cdots+\color{magenta}{(2^n-1)\binom{2^n-1}{2^n-1}4})-\color{orange}{2^n4^{2^n-1}}+\color{purple}{4^{2^n}}$$
$$$$
$$\color{red}{2^{n+2}}(-\color{grey}{\binom{2^n-1}{1}}+\color{blue}{2\binom{2^n-1}{2}}-\color{green}{3\binom{2^n-1}{3}4}-\cdots-\color\magenta{(2^n-1)\binom{2^n-1}{2^n-1}4^{2^n-2}}-\color{grey}{2^n +1})-\color{orange}{2^n4^{2^n-1}}+\color{purple}{4^{2^n}}$$
$$$$
$$\color{red}{2^{n+2}}(f'(4)-\color{grey}{2^n+1})-\color{orange}{2^n4^{2^n-1}}+\color{purple}{4^{2^n}}$$

(f(x) being $=(x-1)^{2^n-1}$)$$$$
  Hence$$f'(4)=(2^n-1)3^{2^n-2}$$

$$\color{red}{2^{n+2}}((2^n-1)3^{2^n-2}+\color{grey}{2^n+1}-\color{orange}{4^{2^n-3}}+\color{purple}{2^{2^n}2^{2^n-n-2}})$$
$$$$
$$\color{red}{2^{n+2}}C$$
$$$$
Hence.......
OR
$$3^{2^{m+1}}-1$$
$$3^{2^m×2}-1$$
$$(3^{2^m}-1)(3^{2m}+1)$$
$$k×2^{m+2} ×(3^{2^m}-1+2)$$

$$2^{m+2} ×k× (k2^{m+2}+2)$$$$2^{m+2} ×k× 2×(k2^{m+1}+1)$$$$2^{m+3} ×\gamma$$

A: You could apply [Lifting the Exponent lemma][1]
[1]: https://brilliant.org/wiki/lifting-the-exponent/ for the case $p=2$ and obtain $v_2(3^{2^n} - 1) = v_2(3-1)+v_2(2^n)+v_2(3+1)-1 = n+2$
A: $$\begin{array}{} &3^{2^m} -1 &=x 2^{m+2} & \text{assumed and checked by } m=1 \text{with $x$ odd}\\ 
 &3^{2^{m+1}} -1 &= 3^{2 \cdot 2^m} -1  &  \text{general step in induction} \\ 
 & &= (3^{2^m})^2 -1\\
 & &= (3^{2^m} -1)(3^{2^m} +1) \\
 & &=x 2^{m+2}(3^{2^m} +1) & \text{ with $x$ odd }\\
 & &=x 2^{m+2}(3 \cdot 3^{2^m-1} +1) \\
 & &=x 2^{m+2}(2 \cdot 3^{2^m-1}+ (3^{2^m-1}+1)) \\
 & &=x 2^{m+2}(2 \cdot 3^{2^m-1}+ (3+1) \cdot y) & \text{by $3^{2w+1}+1$}=(3+1)\cdot y\\
 & &=x 2^{m+2}2 \cdot (3^{2^m-1}+ 2 y) &\text{by factoring out }2\\
 & &=x 2^{m+3} z &\text{with $x$ and $z$ odd}\\
\end{array}$$
$$\begin{array}{l}\implies  2^{m+3} \mid 3^{2^{m+1}}-1 & \phantom {\text{yxcyxcyyxcyxcycyxcyc}}& \text{induction successful, proof complete}
\end{array}$$
A: The Carmichael function $\lambda(2^{n+2})=2^n$ for positive $n$. Thus for any odd number $k$, $k^{2^n}\equiv 1 \bmod  2^{n+2}$ and  thus $2^{n+2}$ divides $k^{2^n}- 1$.
