題目有誤吧：證明連續(xù)四個(gè)奇數(shù)的乘積減一能被八整除
設(shè)四個(gè)連續(xù)奇數(shù)為2n-3,2n-1,2n+1,2n+3,n是整數(shù)
(2n-3)(2n-1)(2n+1)(2n+3)-1
=[(2n-3)(2n+3)]*[(2n-1)(2n+1)]-1
=(4n²-9)(4n²-1)-1
=16n^4 -10*4n²+9-1
=16n^4 -5*8n²+8
每項(xiàng)都含有因數(shù)8
所以 ,這個(gè)數(shù)能被8整除