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Let $F$ be a finite field and $i$ an integer. Calculate the sum of all the elements of $F$, each raised to the $i$th power.

My approach so far:

Let $F=(0,1,\alpha,\alpha^2,...,\alpha^{p^n-1})$, where $p$ is a prime number and let $\sigma=1+\sum_{k=1}^{p^n-1}(α^k)^i$ be the desired sum.

Since $F$ is a ring, each element has an additive inverse. Thus, the trivial case of $i=1$ results in $\sigma=1+[\alpha+(-\alpha)+\alpha^2+(-\alpha^2)+\cdots+\alpha^{p^n-1}+(-\alpha^{p^n-1})]=1$.

We can also calculate the trivial case of $i=0$, where $\sigma=1+p^n-1=p^n$

In the general case, let $i=t (\text{mod } p^n)$.
Then: $i=t+mp^n$, where $t,m$ are integers. We take into account the fact that the order of the multiplicative group of the finite field is $p^n$, so each element raised to the $p^n$ equals $1$. Thus $σ=1+α^t+α^{2t}+\cdots+α^{t(p^n-1)}$. This is the sum of a geometric progression, plus $1$. Thus, $σ=1+α(α^{p^n-1})/α-1$ = $ 1$+($α^{p^n}-α$)/${\alpha-1}$. But according to Lagrange's theorem, for every $\alpha\in F$, the polynomial $x^{p^n}-x$ is zero. It follows that $\sigma=1$.

An important question is raised: Is it true that the characteristic of the finite field is $p$ if its order is $p^n$? If so, the case for $i=0$ leads to $σ=0≠1$ and all cases have been considered?

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    $\begingroup$ The order of the multiplicative group of a finite field $F$ with $p^n$ elements is $p^n-1$. In this case, the characteristic of $F$ is indeed $p$. Also, your case $i=1$ has too many summands, but the idea is right. $\endgroup$ – Thomas Poguntke May 5 '15 at 16:07
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    $\begingroup$ $\alpha^{p^n-1}=1$, so your sum has that term twice. The largest distinct power of a primitive element you have is $\alpha^{p^n-2}$. So your sum contains $1^i$ twice, and the sum is off by one. Similarly, you should look at the congruence $i\equiv t\pmod{p^n-1}$. $\endgroup$ – Jyrki Lahtonen May 5 '15 at 16:08
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    $\begingroup$ If $\text{char } F \neq 2$, each nonzero element has an additive inverse, so for $i = 1$, $\sigma = 0$. $\endgroup$ – Travis Willse May 5 '15 at 16:10
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A different approach. Assume that $i$ is such that $\alpha^i\neq1$. Let $$ \sigma_i=\sum_{x\in\Bbb{F}_{p^n}}x^i $$ be the sum. We get that $$ \alpha^i\sigma_i=\alpha^i\left(\sum_{x\in\Bbb{F}_{p^n}}x^i\right) =\sum_{x\in\Bbb{F}_{p^n}}\alpha^ix^i= \sum_{x\in\Bbb{F}_{p^n}}(\alpha x)^i. $$ Here $\alpha x$ ranges over the field when $x$ does, so the last sum is just $\sigma_i$. Thus $$ \alpha^i\sigma_i=\sigma_i. $$ Given that $\alpha^i\neq1$ I trust you to make the final deductive step. :-)

The condition $\alpha^i\neq 1$ is satisfied, iff $i\not\equiv 0\pmod{p^n-1}$. If $i$ is a multiple of $p^n-1$ your sum consists of several terms equal to $1$, which is another easy case.

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  • $\begingroup$ Certainly a much more compact and refined approach than mine! Thank you! $\endgroup$ – MathematicianByMistake May 5 '15 at 16:17

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