Prove that : $2^{2^{n}}+1\mid 2^{x_{n}}-2$ with $n=1,2,3...$

Question

Question :

Let $n>0$ a natural number Use the following inequality $2^{n}≥n+1$ to prove that :

$2^{2^{n}}+1\mid 2^{x_{n}}-2$ where :

$x_{n}=2^{2^{n}}+1$

My attempt :

I think use induction :

$n=1$ then $x_{n}=5$ so $30\mid 5$ correct

Now for $n+1$ we will prove that :

$x_{n+1}\mid 2^{x_{n+1}}-2$.

I don't know how prove it using $2^{n}≥n+1$.

If any one know other method please drop here

Thanks!

Answer 1

$2^{x_n}-2 = 2(2^{x_n-1}-1) = 2(2^{2^{2^n}}-1) = 2(2^{2^{2^n-1}}+1)(2^{2^{2^n-1}}-1)$ , by induction we get $ 2^{x_n}-2 = 2(2^{2^{2^n-1}}+1)(2^{2^{2^n-2}}+1) \cdots (2^{2^n}+1) \cdots(2^{2^0}+1)(2^{2^0}-1) $. Hence $2^{2^n}+1$ divides $2^{x_n}-2$