Sửa đề (vì tus bảo):`(2 + 1)(2² + 1)(2^4 + 1)(2^8 + 1)(2^16 + 1) = 2^32 - 1`
`⇔ (2 - 1)(2 + 1)(2² + 1)(2^4 + 1)(2^8 + 1)(2^16 + 1) = 2^32 - 1`
`⇔ (2² -1)(2^2 + 1(2^4 + 1)(2^8 + 1)(2^16 + 1) = 3^32 - 1`
`⇔ (2^4 - 1)(2^4 + 1)( 2^8 + 1)(2^16 + 1) = 2^32 - 1`
`⇔ (2^8 - 1)( 2^8 + 1)(2^16 + 1) = 2^32 - 1`
`⇔ (2^16 - 1)(2^16 + 1) = 2^32 - 1`
`⇔ 2^32 - 1 = 2^32 - 1 (đpcm)`
Ta thêm `2 - 1` vào `2` biểu thức