Giải thích các bước giải:

Ta có:

$G=3^{32}-1$

$\to G=(3^{16})^2-1$

$\to G=(3^{16}-1)(3^{16}+1)$

$\to G=((3^{8})^2-1)(3^{16}+1)$

$\to G=(3^8-1)(3^8+1)(3^{16}+1)$

$\to G=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$

$\to G=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$\to G=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$\to G=2\cdot 4\cdot (3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$\to G=8\cdot (3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

Vì $3^2+1, 3^4+1, 3^8+1, 3^{16}+1$ chẵn

$\to  (3^2+1)(3^4+1)(3^8+1)(3^{16}+1)\quad\vdots\quad 2\cdot 2\cdot 2\cdot 2$

$\to  (3^2+1)(3^4+1)(3^8+1)(3^{16}+1)\quad\vdots\quad 16$

$\to  8(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)\quad\vdots\quad 16\cdot 8$

$\to G\quad\vdots\quad 128$

$\to đpcm$