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.