Numerele Fermat sunt prime sau pseudoprime in baza 2

[Filip Chindea]

Fie \( k \in \mathbb{N} \) iar \( F_k = 2^{2^k} + 1 \). Sa se arate ca \( F_k | 2^{F_k} - 2 \).

[Marius Mainea]

\( 2^{F_k}-2=2(2^{F_k-1}-1)=2(2^{2^{2^k}}-1)=2(2^{2^{2^k-1}}-1)(2^{2^{2^k-1}}+1)=\mathcal{M}(2^{2^{2^k-1}}-1)=\mathcal{M}(2^{2^{2^k-2}}-1)=...=\mathcal{M}(2^{2^{k+1}}-1)=\mathcal{M}(2^{2^k}-1)(2^{2^k}+1)=\mathcal{M}F_k \)