Question

求证一个数论问题1^i+2^i+……+(p-1)^i≡0(modp),其中2≤i≤p-2

Answer 1

1^n+2^n+…+(p-1)^n= (1^n + (p-1)^n) + (2^n + (p-2)^n) + ...+ (((p-1)/2)^n + ((p+1)/2)^n)

= (1^n - 1^n) + (2^n - 2^n) + (3^n - 3^n) + ...+ (((p-1)/2)^n - ((p-1)/2)^n)

= 0 (mod p)