I'm stuck at the following question: let $d$ divide $2^{2^n} +1 $. Show that $d \equiv 1 \pmod{2^{n+1}}$.
All I've managed to do so far was to notice that $2^{n+1} \mid 2^{2^n}$ for every natural $n$, so if $dl = 2^{2^n} + 1$, we have $dl \equiv 1 \pmod{2^{n+1}}$ - which merely states that $d$ is invertible, which is trivial since it is odd. This is homework, are there any hints?
Suppose prime $\,p\mid d.\ $ ${\rm mod}\ p\!:\ 2^{\large 2^{\Large n}}\!\!\equiv -1\overset{\rm square}\Longrightarrow 2^{\large 2^{\Large n+1}}\!\equiv 1\, $ so Order Test $\Rightarrow 2\,$ has $\color{#0a0}{{\rm order}\ 2^{\large n+1}},\,$ so $\,2^{\:\!\large\color{#c00}{p-1}}\!\equiv 1\Rightarrow \color{#0a0}{2^{\:\!\large n+1}}\!\mid \color{#c00}{p\!-\!1}\,$ by the Order Theorem. Thus since each prime $\,p_i\,$ dividing $\,d\,$ satisfies $\,\color{#c00}{p_i\equiv \bf 1}\pmod{\!2^{\large n+1}}\,$ so too their product $\,d = \prod \color{#c00}{p_i}^{\!\large k_i}\!\equiv\prod \color{#c00}{\bf 1}^{\large k_i}\!\equiv {\color{#c00}{\bf 1}}\pmod{\!2^{\large n+1}}$.