证明：
(n^2+n)(n^2+5n+6)+1
=n(n+1)(n+2)(n+3)+1
=[n(n+3)]*[(n+1)*(n+2)]
=(n^2+3n)*(n^2+3n+2)+1
=(n^2+3n)^2+2*(n^2+3n)+1
=(n^2+3n+1)^2
所以(n^2+n)(n^2+5n+6)+1是完全平方数