The first one is k | (n k-n), right? This is obtained from Fermat's last theorem. You can read some books on number theory or abstract algebra, or Baidu Encyclopedia: /view/263807.htm? fr=ala0_ 1

Second: according to the knowledge of congruence, 30=2*3*5, as long as it is proved that 2, 3 and 5 can be divisible by n 5-n respectively.

Since n 5-n = n (n- 1) (n+ 1) (n 2+1), n-1is three consecutive numbers.

So 2 and 3 are divisible.

Experiments show that when n = 5k, 5k+ 1, 5k+2, 5k+3, 5k+4,5 can be divisible by n (n-1) (n 2+1).

So the conclusion is established.