a,b,k为大于2的正整数a^k mod (k+1)=n;b^k mod (k+1)=m; 证明 n*m mod (k+

题目条件：
a^k = n (mod k+1)
b^k = m (mod k+1)
m*n = 1 (mod k+1)
所以(ab)^k = 1 (mod k+1) (1)
记k+1的欧拉函数为ψ(k+1),那么在(1,ψ(k+1))内,有且仅有
a^ψ(k+1) = 1 (mod k+1)
b^ψ(k+1) = 1 (mod k+1)
相乘得(ab)^ψ(k+1) = 1 (mod k+1) (2)
由于k >=ψ(k+1)
由(1)(2)可以得到k = p * ψ(k+1)
所以m = a^k = (a^ψ(k+1))^p = 1 (mod k+1)
n = b^k = (b^ψ(k+1))^p = 1 (mod k+1)