Question

求证,如果n为自然数,则(n^2+n)(n^2+5n+6)+1是完全平方数

Answer 1

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