Giải thích các bước giải:
a.Ta có:
$2^{32}-1$
$=(2^{16})^2-1$
$=(2^{16}-1)(2^{16}+1)$
$=((2^{8})^2-1)(2^{16}+1)$
$=(2^{8}-1)(2^8+1)(2^{16}+1)$
$=((2^{4})^2-1)(2^8+1)(2^{16}+1)$
$=(2^{4}-1)(2^4+1)(2^8+1)(2^{16}+1)$
$=((2^{2})^2-1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^{2}-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
b.Ta có:
$100^2+103^2+105^2+94^2=101^2+98^2+96^2+107^2$
$\leftrightarrow (103^2-101^2)+(100^2-98^2)=(107^2-105^2)+(96^2-94^2)$
$\leftrightarrow (103-101)(103+101)+(100-98)(100+98)=(107-105)(107+105)+(96-94)(96+94)$
$\leftrightarrow 2\cdot 204+2\cdot 198=2\cdot 212+2\cdot 190$
$\leftrightarrow204+198=212+ 190$
$\leftrightarrow 402=402$ đúng
$\to đpcm$
