Let p be a prime divisor of the Fermat number $F_n = 2^{2^n} + 1$, and let $a = 2^{2^{n - 2}}(2^{2^{n - 1}} - 1)$. Prove that $\operatorname{ord}_p(a) = 2^{n + 2}$.
I already know how to show that $2^{n + 2} \mid (p - 1)$ (Lucas) using Legendre symbols, but I need to prove this to prove it without using that. Any hints?
Assuming $n \ge 2$ for $2^{n-2}$ to be an integer, then using
$$a = 2^{2^{n - 2}}(2^{2^{n - 1}} - 1), \;\; m = \operatorname{ord}_{p}(a)$$
and that
$$p \mid 2^{2^{n}} + 1 \;\to\; 2^{2^{n}} \equiv -1 \pmod{p} \;\to\;\ 2^{2^{n+1}} \equiv 1 \pmod{p}$$
note we get
$$\begin{equation}\begin{aligned} a^2 & \equiv 2^{2^{n-1}}\left(2^{2^{n}} - 2(2^{2^{n-1}}) + 1\right) \\ & \equiv 2^{2^{n-1}}\left((2^{2^{n}} + 1) - 2(2^{2^{n-1}}) \right) \\ & \equiv 2^{2^{n-1}}\left((-2)2^{2^{n-1}}\right) \\ & \equiv (-2)2^{2^{n}} \\ & \equiv 2 \pmod{p} \end{aligned}\end{equation}$$
Thus, we then have
$$a^{2^{n+2}} \equiv (a^2)^{2^{n+1}} \equiv 2^{2^{n+1}} \equiv 1 \pmod{p}$$
This shows the multiplicative order $m \mid 2^{n+2}$. If $m$ were any smaller factor of $2^{n+2}$, then we'd get $2^{2^{n}} \equiv 1\pmod{p}$, contradicting that actually $2^{2^{n}} \equiv -1\pmod{p}$. Thus, this means that
$$m = \operatorname{ord}_{p}(a) = 2^{n+2}$$
Lemma $\ a = b\,(b^2\!-\!1),\ \color{#0a0}{b^4 \equiv -1}\!\pmod{\!p}\,\Rightarrow\, o_p(a) \!=\! 16k,\,$ if $\,\color{#0a0}b = \color{#c00}2^k,\ k = 2^{n-2}$
Proof $\ \ \ \color{#c00}{a^2} = b^2(\color{#0a0}{b^4\!+\!1}-2b^2) \equiv -2\color{#0a0}{b^4} = \color{#c00}2,\,$ so by $\:\!\rm\color{#0af}{OT}$ = Order Test
$a^{8k}\!\equiv\color{#c00}a^{\color{#c00}2(4k)}\!\equiv\! (\color{#c00}2^k)^4\! \equiv\color{#0a0}{b^4 \equiv-1}$ $\,\underset{\rm\color{#0af}{OT}}\Rightarrow o_p(a)\!=\!16k\,$ (by $\,a^{16k/q}\!\color{#0a0}{\not\equiv\! 1}\:\forall$ prime $\:\!q\!\mid\! 16k\!=\! 16\cdot 2^{n-2})$
Or $\ \color{#0a0}b\equiv\color{#c00}a^{\color{#c00}2k}\,$ so $\,\color{#0a0}{o(b)\!=\!8}\!=\! o(a^{2k})\,$ $\Rightarrow o(a)= \color{#0a0}8(2k)\,$ by Rad OT $ $ (by prime $\,q\mid 2k\Rightarrow 2\mid\color{#0a0}8)$
i.e. $\ \bbox[5px,border:1px solid #c00]{o(\sqrt[\large 2k]b) = 2k\, o(b)}\, $ when all primes $\,q\mid 2k\Rightarrow q\mid o(b)$