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Let $F_n$ be a Fermat prime number. Prove that $\gamma_{F_n}(2)=2^{n+1}$.

Here $\gamma_{F_n}(2)$ denotes the order of $2$ modulo $F_n$.

My attempt:

$F_0=2^{2^0}=2^{0+1} \quad\checkmark$

$F_1=5 \neq 2^{1+1} $

I don't know why for $n=1$ this is not true

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By definition, $F_n=2^{2^n}+1$. So, $F_n|2^\gamma-1$ implies $2^\gamma-1\ge 2^{2^n}+1$, which implies $\gamma\ge 2^n+1$.

Moreover, $F_n=2^{2^n}+1|(2^{2^n})^2-1=2^{2^{n+1}}-1$. Thus, $\gamma|2^{n+1}$.

So, we have $\gamma|2^{n+1}$ and $\gamma\ge2^{n}+1$. Thus, $\gamma=2^{n+1}$

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