<s>[INST] Proof about Fermat number without induction 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 $7$.
 [/INST] $$F_{2k+1}+2=7+2^{2^{2k+1}}-2^2=7+4\left(2^{4^k-1}-1\right)\left(2^{4^k-1}+1\right)$$ and $2^{4^k-1}-1$ is divided by $7$ because $4^k-1$ is divided by $3$ and $2^3-1=7$.
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