结论是相邻四个偶数的乘积加上16,必然是一个整数的平分.
设这四个偶数分别是:2n、2(n+1)、2(n+2)、2(n+3)
则:
2n*2(n+1)*2(n+2)*2(n+3)+16
=16*[n(n+1)(n+2)(n+3)+1]
=16*{[n(n+3)]*[(n+1)(n+2)]+1}
=16 {n(n+3)[n(n+3)+2]+1}
=16[n(n+3)^2 + 2n(n+3)+1]
=16[n(n+3)+1]^2
= [2n*2(n+3)+4]^2
= [4n(n+3)+4]^2
结论是相邻四个偶数的乘积加上16,必然是一个整数的平分.
设这四个偶数分别是:2n、2(n+1)、2(n+2)、2(n+3)
则:
2n*2(n+1)*2(n+2)*2(n+3)+16
=16*[n(n+1)(n+2)(n+3)+1]
=16*{[n(n+3)]*[(n+1)(n+2)]+1}
=16 {n(n+3)[n(n+3)+2]+1}
=16[n(n+3)^2 + 2n(n+3)+1]
=16[n(n+3)+1]^2
= [2n*2(n+3)+4]^2
= [4n(n+3)+4]^2