Bài 8
a, Có 2^32 - 1
= (2^16 - 1).(2^16 + 1)
= (2^8 - 1).(2^8 + 1).(2^16 + 1)
= (2^4 - 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)
= 1.(2^2 + 1).(2^4 + 1).(2^8 + 1).(2^16 + 1)
= (2^2 + 1).(2^4 + 1).(2^8 + 1).(2^16 + 1) ( ĐPCM )
Vậy 2^32 - 1 = (2^2 + 1).(2^4 + 1).(2^8 + 1).(2^16 + 1)
b, Có 100^2 + 103^2 + 105^2 + 94^2
= (98+2)^2 + (101+2)^2 + (107-2)^2 + (96-2)^2
= 98^2 + 2.2.98 + 4 + 101^2 + 2.2.101 + 4 + 107^2 - 2.2.107 + 4 + 96^2 - 2.2.96 + 4
= 98^2 + 101^2 + 107^2 + 96^2 + 4.(98+101-107-96) + 4 + 4 + 4 + 4
= 98^2 + 101^2 + 107^2 + 96^2 + 4.(-4) + 4.4
= 98^2 + 101^2 + 107^2 + 96^2 + 4.(-4+4)
= 98^2 + 101^2 + 107^2 + 96^2 + 4.0
= 98^2 + 101^2 + 107^2 + 96^2
Vậy chứng tỏ 100^2 + 103^2 + 105^2 + 94^2 = 101^2 + 98^2 + 96^2 + 107^2