1.
n∧3-n = n(n^2 -1) = n(n+1)(n-1)
-(1)- n为正整数,则n,n+1,n-1中必有一个3的倍数
-(2)- n为正整数,则n,n+1中必有一个2的倍数
所以n(n+1)(n-1)为6的倍数.
2.
n(n+1)(n+2)(n+3) +1
拆开再合上.
n(n+1)(n+2)(n+3)+1
=n(n+3)(n+1)(n+2)+1
=(n^2+3n)(n^2+3n+2)+1
令n^2+3n＝X
上式= X(X+2)+1
= X^2+2X+1
= (X+1)^2.
= (n^2+3n+1)^2.
所以必是一个完全平方数.
另外,1^3 -1 = 1-1 = 0,0是6的倍数..