Giải thích các bước giải:

Gọi $UCLN(2^{2n+1}+2^{n+1}+1; 2^{2n+1}-2^{n+1}+1)=d, d\in N^*$

$\to \begin{cases}2^{2n+1}+2^{n+1}+1\quad\vdots\quad d\\2^{2n+1}-2^{n+1}+1\quad\vdots\quad d\end{cases}$

Trừ vế với vế

$\to 2\cdot 2^{n+1}\quad\vdots\quad d$

$\to 2^{2n+1}\quad\vdots\quad d$

$\to \begin{cases}2^{n+1}+1\quad\vdots\quad d\\-2^{n+1}+1\quad\vdots\quad d\end{cases}$

Cộng vế với vế $\to 2\quad\vdots\quad d$

$\to d\in\{1, 2\}$

Nếu $d=2$

$\to 2^{2n+1}+2^{n+1}+1\quad\vdots\quad 2$ loại vì vế trái lẻ

$\to d=1$

$\to (2^{2n+1}+2^{n+1}+1; 2^{2n+1}-2^{n+1}+1)=1$

Ta có:

$AB=(2^{2n+1}+2^{n+1}+1)(2^{2n+1}-2^{n+1}+1)$

$\to AB=(2^{2n+1}+1)^2-(2^{n+1})^2$

$\to AB=2^{4n+2}+2^{2n+2}+1-2^{2n+2}$

$\to AB=2^{4n+2}+1$

$\to AB=4^{2n+1}+1$

$\to AB\equiv (-1)^{2n+1}+1\equiv -1+1\equiv 0(\mod5)$

$\to AB\quad\vdots\quad 5$

$\to A\quad\vdots\quad 5$ hoặc $B\quad\vdots\quad 5$ vì $(A,B)=1$