证明8整除n(n+1)(n+2)(n+3)

Question

证明8整除n(n+1)(n+2)(n+3)

Answer 1

证：
n为偶数时,令n=2k (k∈N+)
n(n+1)(n+2)(n+3)
=(2k)(2k+1)(2k+2)(2k+3)
=4k(k+1)(2k+1)(2k+3)
n为奇数时,令n=2k-1 (k∈N+)
n(n+1)(n+2)(n+3)
=(2k-1)(2k-1+1)(2k-1+2)(2k-1+3)
=(2k-1)·(2k)·(2k+1)·(2k+2)
=4k(k+1)(2k-1)(2k+1)
无论n为奇数还是偶数,表达式中均存在因式乘积4k(k+1)
无论k为奇数还是偶数,k,k+1必为一奇一偶,其乘积能被2整除,又两表达式中均包含因数4,因此能被2×4=8整除.
综上,得n(n+1)(n+2)(n+3)能被8整除.