1)相邻两个奇数,令2n+1,2n+3
平方差为(2n+3)² - (2n+1)² = [(2n+3)+(2n+1)][(2n+3)-(2n+1)]
= (4n+4)*2=8(n+1) 一定能被8整除
2) 令三个连续整数分别为 n,n+1,n+2
则 平方和为
n²+(n+1)²+(n+2)²
= n²+n²+2n+1+n²+4n+4
= 3n²+6n+5
= 3(n²+2n+1)+2
被3除余2
3)令奇数2n+1
(2n+1)² -1 = 4n²+4n = 4n(n+1),n和n+1有一个必为偶数,所以
4n(n+1)能被8整除