猜想n(n+1)(n+2)(n+3)+1是完全平方数
证明:
n(n+1)(n+2)(n+3)+1
=[n(n+3)][(n+1)(n+2)]+1
=[(n^2+3n)(n^2+3n+2)]+1
=(n^2+3n)^2+2(n^2+3n)+1
=(n^2+3n+1)^2
所以成立。
(n+1)(n+2)(n+3)(n+4)+1
=(n+1)(n+4)(n+2)(n+3)+1
=(n²+5n+4)(n²+5n+4+2)+1
=(n²+5n+4)²+2(n²+5n+4)+1
=(n²+5n+4+1)²
=(n²+5n+5)²
n(n+1)(n+2)(n+3)+1
=[n(n+3)][(n+1)(n+2)]+1
=(n^2+3n)(n^2+3n+2)+1
=(n^2+3n)^2+2(n^2+3n)+1
=[(n^2+3n)+1]^2
=(n^2+3n+1)^2
同理
(n+1)(n+2)(n+3)(n+4)+1=(n^2+5n+5)^2
n(n+1)(n+2)(n+3)+1
= n(n+3)(n+1)(n+2) + 1
=(n^2+3n)(n^2+3n+2) + 1
=(n^2+3n)^2+2(n^2+3n) + 1
=(n^2+3n+1)^2
n(n+1)(n+2)(n+3)+1=(n²+3n+1)²
每个数的平方都是中间两个数的乘积减一
比如:3*4*5*6+1=361=19
4*5-1=19