Hi people, I am trying to prove that
Existing a primitive root mod p (being p a prime),
1^n + 2^n + ... + (p-1)^n is congruent to 0 or to a-1 (mod p) , only depending on the integer n>= 0.
I tried to separate the problem in two parts, when n is odd and when n is even. The first part I think I could prove, but the second.. no clue:
The sum is basically
1^n + 2^n + ... +j^n + .. + (p - j)^n + .. + (p-2)^n + (p-1)^n
i.e. we can separate the sum in two parts, because there is p-1 terms in this serie. For each term 1^n ... (p-1/2)^n , if n is odd, there is a corresponding term when we expand the powers of sums that anulates with the terms of the first part. For example:
If n = 3, we have:
1^3 + 2^3 + 3^3 + ... + (p-3)^3 + (p-2)^3 + (p-1)^3
= 1 + 4 + 9 + ... + (p^3 - 9p^2 + 27p -9) + (p^3 - 6p^2 + 12p - 8) + (p^3 - 3p + 3p -1) where for 1, 4, 9... there is always the negative of them in the second part (resp. -1^3, -2^3, -3^3...). So all the constant terms will be canceled. And the terms which will stay will be all multiples of p, so the sum will be congruent to 0 mod p.
This part is ok, but I could not prove for the n = 2.k i.e. n is even. Any help?
Existing a primitive root mod p (being p a prime),
1^n + 2^n + ... + (p-1)^n is congruent to 0 or to a-1 (mod p) , only depending on the integer n>= 0.
I tried to separate the problem in two parts, when n is odd and when n is even. The first part I think I could prove, but the second.. no clue:
The sum is basically
1^n + 2^n + ... +j^n + .. + (p - j)^n + .. + (p-2)^n + (p-1)^n
i.e. we can separate the sum in two parts, because there is p-1 terms in this serie. For each term 1^n ... (p-1/2)^n , if n is odd, there is a corresponding term when we expand the powers of sums that anulates with the terms of the first part. For example:
If n = 3, we have:
1^3 + 2^3 + 3^3 + ... + (p-3)^3 + (p-2)^3 + (p-1)^3
= 1 + 4 + 9 + ... + (p^3 - 9p^2 + 27p -9) + (p^3 - 6p^2 + 12p - 8) + (p^3 - 3p + 3p -1) where for 1, 4, 9... there is always the negative of them in the second part (resp. -1^3, -2^3, -3^3...). So all the constant terms will be canceled. And the terms which will stay will be all multiples of p, so the sum will be congruent to 0 mod p.
This part is ok, but I could not prove for the n = 2.k i.e. n is even. Any help?
