设俩个连续的偶数为2n,2n+2
(2n+2)²-(2n)²
=(2n+2+2n)(2n+2-2n)
=(4n+2)×2
=4(n+1)
4(n+1)能被4整除
∴俩个连续的偶数的平方差是4的倍数