证明：可设这4个连续整数依次为n、n+1、n+2、n+3,则有
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
所以说4个连续整数的积与1的和是一个完全平方数.