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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.

J. W. Tanner
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ATB
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1 Answers1

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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 $a|2c$

J. W. Tanner
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  • for 1. how do u factorize $2^{2^{2^n}}-1=(2^{2^{n+1}}-1)(2^{2^{2^n}-2^{n+1}}+2^{2^{2^n}-2\cdot2^{n+1}}+2^{2^{2^n}-3\cdot2^{n+1}}+\cdots+2^{{2^{n+1}}}+1)$ ?? how you can prove that $2^{2^{2^n}}-1$ is equal to the right hand side. – ATB Dec 01 '20 at 17:40
  • you can explain what u mean difference of two squares ?
    • – ATB Dec 01 '20 at 17:40
  • By d.o.t.s., I mean $a^2-b^2=(a+b)(a-b)$ – J. W. Tanner Dec 01 '20 at 17:44