Number Theory: Prove for all $1\leq n \in \mathbb{N}$ $F_n \mid 2^{F_n}-2$ with $F_n$ is Fermat numbers

Question

Number Theory: Prove for all $1\leq n \in \mathbb{N}$ $F_n \mid 2^{F_n}-2$ with $F_n$ is Fermat numbers

Follow the following sections They will guide you in proof

1. Use polynomial division to prove that $2^{2^{n+1}}-1\mid2^{F_n-1}-1$.

2. Use polynomial division to prove that $F_n\mid2^{2^{n+1}}-1$.

3. Draw the conclusion from these two sections.

Attempt:

if $2^{2^{n+1}}-1$ divide $2^{F_n-1}-1$ and $F_n$ divide $2^{2^{n+1}}-1$ so $F_n$ divide $2^{F_n-1}-1$

and i have no idea how to make it happen.

Answer 1

Hints:

1. $2^{2^{2^n}}-1=(2^{2^{n+1}}-1)(2^{2^{2^n}-2^{n+1}}+2^{2^{2^n}-2\cdot2^{n+1}}+2^{2^{2^n}-3\cdot2^{n+1}}+\cdots+2^{{2^{n+1}}}+1)$

2. $2^{2^{n+1}}-1$ is the difference of two squares

3. if $a|b$ and $b|c$ then $a|2c$