# Thread: the sum of a^m

1. ## the sum of a^m

if we have any positive integer m ,
the sum of a^m = -1 if (q-1)|m or 0 otherwise.
Note, the sum is over a belongs to Fq , where Fq is a field.

2. Originally Posted by Mike12
if we have any positive integer m ,
the sum of a^m = -1 if (q-1)|m or 0 otherwise.
Note, the sum is over a belongs to Fq , where Fq is a field.

Hint: 1) $\displaystyle \displaystyle{(q-1)\mid m\Longrightarrow m=t(q-1)\Longrightarrow \sum\limits_{a\in\mathbb{F}_q}a^m=\sum\limits_{a\i n\mathbb{F}_q}(a^{q-1})^t}$

2) $\displaystyle \displaystyle{(q-1)\nmid m\Longrightarrow m=t(q-1)+r\,,\,\,0<r<q-1\Longrightarrow \sum\limits_{a\in\mathbb{F}_q}a^m=\sum\limits_{a\i n\mathbb{F}_q}(a^{q-1})^t\cdot a^s}$ .

But there exists $\displaystyle b\in\mathbb{F}_q,\,\,s.t.\,\,b^s\neq 1$ (why?), so

$\displaystyle \displaystyle{b^s\sum\limits_{a\in\mathbb{F}_q}a^s =\sum\limits_{a\in\mathbb{F}_q}(ba)^m$ , and since we have a group here the last sum runs over

all the elements of $\displaystyle \mathbb{F}_q^*$ and zero, so $\displaystyle \displaystyle{b^s\sum\limits_{a\in\mathbb{F}_q}a^m =\sum\limits_{a\in\mathbb{F}_q}a^m}$ , thus...

Tonio

3. Thanks Tonio,