[讨论][求助]数论试题

tian27546

luyuanhong

此题解答如下：

elimqiu

(p-1)...(p-a+1) ≡ (-1)...(-(a-1)) = (-1)^(a-1) (a-1)! (mod p)
显然 (a,p) = 1
所以  a ( b (-1)^(a-1) (p-1)...(p-a+1)/a!)≡a b (a-1)!/a! ≡ b a!/a! ≡ b (mod p)

tian27546

谢谢luyuanhong教授