设4个连续自然数为n,n+1,n+2,n+3.
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,4个连续自然数的积加1必是一个完全平方数