Proof about Fermat number without induction

Question

I want to show that $7 \mid (F_{2k + 1} + 2)$ where $k \in \mathbb{N_0}$ and $F_n := 2^{2^n} + 1$. I was able to proof this using induction, but was wondering if there is a more direct approach?

Here is a simple sketch of proof (only induction step):

$$ F_{2(k+1)+1} +2= 2^{2^{2k +3}} +3 = 2^{2^{2k + 3}} + 7q - 2^{2^{2k + 1}} $$

for some $q \in \mathbb{Z}$.

$$ = 16^{2^{2k + 1}} + 7q - 2^{2^{2k + 1}} = 7q + 2^{2^{2k + 1}} (8^{2^{2k + 1}} -1) \\ = 7q + 2^{2^{2k + 1}} ((7 + 1)^{2^{2k + 1}} -1) \\ = 7q + 2^{2^{2k + 1}} ((\sum_{i = 0}^{2^{2k + 1}}7^i1^{2^{2k + 1}-i}) -1) \\ = 7q + 2^{2^{2k + 1}} ((\sum_{i = 1}^{2^{2k + 1}}7^i1^{2^{2k + 1}-i})) $$ This is clearly divisible by $7$.

Answer 1

Hint: prove that $2^k \bmod 7$ cycles through residues $1, 2, 4, 1, 2, 4, \dots$.

Answer 2

$$2^{2^n}\equiv2^{2^n\pmod{\phi(7)}}\pmod7$$

Now we need $2^n\pmod6$ as $\phi(7)=6$

As $(2^n,6)=2$ for $n\ge1$

let us find $2^{n-1}\pmod{6/2}$

As $2\equiv-1\pmod3,2^{n-1}\equiv(-1)^{n-1}\pmod3$

$\implies2^n=2\cdot2^{n-1}\equiv2(-1)^{n-1}\pmod{2\cdot3}$

$\implies2^{2^n}\equiv2^{2(-1)^{n-1}\pmod6}\pmod7$

Answer 3

$$F_{2k+1}+2=7+2^{2^{2k+1}}-2^2=7+4\left(2^{4^k-1}-1\right)\left(2^{4^k-1}+1\right)$$ and $2^{4^k-1}-1$ is divided by $7$ because $4^k-1$ is divided by $3$ and $2^3-1=7$.