Prime factors of ${2^2}^n + 1$ congruence to $2^{n+1}$

Question

Prove that for every natural $n$ all prime factors of $${2^2}^n + 1$$ are congruent to $1$ modulo $2^{n+1}$


I tried to prove it using contradiction, assumming there is a prime factor congruent to $-1$, but couldn't prove anything.

Any help/tips appreciated.

Answer 1

If $p$ divides $2^{2^n}+1$ then $2^{2^n}\equiv-1\pmod p$, and hence $2^{2^{n+1}}=\left(2^{2^n}\right)^2\equiv 1\pmod p$.

If $k$ is the order of $2$ in $\Bbb Z_p^\times$ then $k$ divides $2^{n+1}$.

Since every divisor of $2^{n+1}$ is a power of two and $2^n\neq k$ then $k=2^{n+1}$.

Little Fermat's theorem implies that $k$ divides $p-1$. That is

$$p-1\equiv 0\pmod {2^{n+1}}$$

Answer 2

Here’s just a clue:

Perhaps you can use the fact that that for $n>1$, if $m$ is odd, then $(1+2^nm)^2=1+2^{n+1}m'$, for an odd integer $m'$.