If $p-1 | k \Rightarrow k = (p-1) n$, for some $n \in \mathbb{N}$. Then we $ 1^{(p-1)^n} + 2 ^{(p-1)^n} + \cdots + (p-1)^{(p-1)^n} = \underbrace{1 + 1 + \cdots + 1}_{p-1} = p-1 \equiv -1 \mod p.$ The second part of the question did not come out. Induction: $k=1$, which is true because we $1 + 2 + \cdots + (p-1)= \frac{(p-1)(p)}{2} \equiv 0 \mod p$. Suppose true for $k = n$, then $1^n + 2^n + \cdots + (p-1)^n \equiv 0 \mod p$. How to finish? Thank you.
Asked
Active
Viewed 796 times
0
-
http://math.stackexchange.com/questions/511444/1n-2n-cdots-p-1n-mod-p – N. S. Oct 17 '13 at 13:54
-
http://math.stackexchange.com/questions/226023/congruence-modulo-p – lab bhattacharjee Oct 17 '13 at 14:59
1 Answers
1
The group $\mathbb{F}_p^*$ is cyclic, say generated by $\zeta$ with $\zeta^{p-1}=1$. Then your sum is $\sum_{i=0}^{p-2} (\zeta^k)^i$, which is just a geometric series. If $p-1 | k$, we get $\sum_{i=0}^{p-2} 1 = p-1 = -1$, and otherwise $\zeta^k \neq 1$ so that we get $\dfrac{{(\zeta^k})^{p-1} - 1}{\zeta^k -1} = 0$.
Martin Brandenburg
- 163,620
-
Is the same as $\sum_{i=0}^{p-2} (g^k)^i = \sum_{i=1}^{p-1} (g^k)^i$, hence follows the answer. Thank you. – user79048 Oct 18 '13 at 00:16