Dem
B.I. La afirmacion se cumple para n=1 $2^{2}-1=3$ y $3|3$
H.I. Suponemos que $3|2^{2^n}-1$ Esto es $2^{2^n}-1=3q$
P.d. $3|2^{2^(n+1)}-1$
(1)\begin{equation} 2^{2^(n+1)}-1 = 2^{2n+2}-1 \end{equation}
(2)
\begin{equation} 2^{2n+2}-1 = 2^{2n} 2^{2}-1 \end{equation}
(3)
\begin{equation} 2^{2n} 2^{2}-1 = 4(2^{2n})- 1 \end{equation}
(4)
\begin{equation} 4(2^{2n})- 1 = (3+1)(2^{2n})- 1 \end{equation}
(5)
\begin{equation} (3+1)(2^{2n})- 1 = 3(2^{2n}) + 2^{2n}-1 \end{equation}
Por H.I. $2^{2^n}-1=3q$
(6)\begin{equation} 3(2^{2n}) + 2^{2n}-1 = 3(2^{2n}) + 3q = 3(2^{2n}+q) \end{equation}
Haciendo $p=(2^{2n}+q)$ tenemos que $2^{2^(n+1)}-1 = 3p$
Por la definición de divisibilidad $3|2^{2^(n+1)}-1$