Proving that $\sum\limits_{a\in \Bbb{F}_q^\times} a^k=0$ for $k\not\equiv 0\mod q-1$

Question

Let $p$ be a prime number, and $q=p^n$ for some $n$. Consider the field $\Bbb{F}_q$. When $k\not\equiv 0\mod q-1$, then $\sum\limits_{a\in \Bbb{F}_q^\times} a^k=0$.

Is this assertion true? And if it is, could someone give a hint as to why it is?

Answer 1

Yes, it is true. Because $\Bbb F_q^*$ is cyclic (it is well known fact) one has for some generator $$\Bbb F_q^*=\{g,\space g^2,\space g^3,\space g^4\cdots,g^n,\cdots,\space g^{q-1}=1\}$$ and since $k\ne m(q-1)$ for all $m$ there is a $g^n\ne 1\Rightarrow g^{nk}\ne 1$; furthermore $a^k\to g^{nk}a^k$ is a bijection ($g^{nk}a^k=g^{nk}b^k\Rightarrow a^k=b^k $ and for all $b^k$ there is $a$ such that

$a^k=g^{-kn}b^k\iff b^k=g^{nk}a^k$) hence $$\{g^{nk}a^k\}_{a\in\Bbb F^*_q}=\Bbb F_q^*$$ It follows $$\sum_{a\in\Bbb F^*_q}a^k=\sum_{a\in\Bbb F^*_q}g^{nk}a^k\iff\sum_{a\in\Bbb F^*_q}a^k-\sum_{a\in\Bbb F^*_q}g^{nk}a^k=0$$ $$(1-g^{nk})\sum_{a\in\Bbb F^*_q}a^k= 0\Rightarrow\sum_{a\in\Bbb F^*_q}a^k=0$$