First I want to give you some context. Then I will ask my questions.


Consider the equations $ a_1x_1^{l_1} + \dots + a_rx_r^{l_r} = b $ with $a_1, \dots , a_r \in F_m^{*}$, where $F_m$ is a finite field with $m$ elements. $N$ is the number of solutions. Now I know that if $b \neq 0$ :

$ N = m^{r-1} + \sum \chi_1\chi_2\cdots\chi_r(b) \chi_1(a_1^{-1})\chi_2(a_2^{-1})\cdots\chi_r(a_r^{-1})J(\chi_1,\dots,\chi_r)$. The summation is over $r$-tuples of characters $\chi_1, \dots, \chi_r$, where $ \chi_i^{l_i} = \varepsilon $ and $\chi_i \neq \epsilon$ for $i = 1, \dots, r$.

I want to specializing this to $x^2 + y^4 = 1$. Obviously $r=2$. So I want to sum over all $2$-tuples of characters $\chi_1, \chi_2$ where $ \chi_i^{l_i} = \varepsilon $ and $\chi_i \neq \epsilon$ for $i = 1,2$.

So $$ N = m + J(\rho,\chi) + J(\rho,\chi^2) + J(\rho,\chi^3)$$ $$= m - 1 + J(\rho,\chi) + J(\rho,\chi^3) = m - 1 + J(\rho,\chi) + \overline{J(\rho,\chi)} $$ where $\rho$ is a character of order $2$ and $\chi$ is a character of order $4$.

First Question ( Answered, I won ) :

I wanted to use that fact above again, but this time for $F_{m^s}$. At this point I had a discussion with my teacher. He said that the fact above is only for $F_m$, but I don't see a problem because I could say $\hat{m} := m^s$ or do you see a problem?

Second Question:

The number of solutions in $F_{m^s}$ has to be $N = m^s - 1 - (-J(\rho,\chi))^s - (- J(\overline{\rho,\chi}))^s$. But to be honest: I don't see why. I only know that we have to work with the Hasse-Davenport Relation. So we have to use the equation: $ (-g(\chi))^s = -g(\chi^{'})$ (gaussian sums), where $\chi$ is a character of $F_m$ and $\chi^{'}$ is a character of $F_{m^s}$. I think that it won't be necessary to give you a exact form of $\chi^{'}$.


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