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First I want to give you some context. Then I will ask my questions. I think that my questions are easy and fast to answer, so I've decided to put them together in one question here.

Context

Consider the equations $ a_1x_1^{l_1} + \dots + a_rx_r^{l_r} = b $ with $a_1, \dots , a_r \in F^{*}$, where $F$ is a finite field. Let us say that $F$ has $m$ 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 the $N = m + J(\rho,\chi) + J(\rho,\chi^2) + J(\rho,\chi^3)$, where $\rho$ is a character of order $2$ and $\chi$ is a character of order $4$.

Questions

1) Why $\chi^2 = \rho$ ? I need that to say $J(\rho,\chi^2) = -1 $.

2) We know $\chi^4 = \varepsilon$, but why this is enough to say that $\chi^3 = \bar{\chi}$ ?

3) Why $J(\rho,\bar{\chi}) = \overline{J(\rho,\chi)} $ ?

4) Now let us say that $\pi = -J(\rho,\chi)$. We know that $\rho$ takes the values $\pm1$ and $\chi \pm1$, $\pm i$. Why is this enougth to say that $\pi = a+bi$?

5)If my four questions are answered I can say that $N = m - 1 - \pi - \bar{\pi}$ . I know that $ a^2 + b^2 = \pi \bar{\pi} = m $. Why can I say that $N = m - 1 - 2a$ ?

I hope that my question are clear. Sorry that they might be trivial for you. I'm an absolute beginner.

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$\chi^2$ has order $2$ and $\rho$ is the only character of order $2$.

$\chi^3=\chi^{-1}=\overline\chi$.

$\overline{J(\rho,\chi)}=J(\overline\rho,\overline\chi)=J(\rho,\overline\chi)$.

$J(\rho,\chi)$ is the sum of terms all of which have the form $\pm1$ or $\pm i$.

$\pi+\overline\pi=(a+bi)+(a-bi)=2a.$

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  • $\begingroup$ Thank you for your answer. I get it now. Happy new year :). $\endgroup$ – Memories Jan 1 at 18:34
  • $\begingroup$ Sorry for disturbing you again, but can I ask you one final question? My question seems so simple that it wouldnt be necessary to use " ask Question". I showed now $ \rho(a) -1 \equiv 0 (2) $ and $\chi(a) -1 \equiv 0(1+i)$. Combine this to get $(\rho(a) -1)(\chi(b) -1) \equiv 0(2+2i)$. Thus $\sum_{a+b=1} (\rho(a) -1)(\chi(b) -1) \equiv 0(2+2i)$. I have expanded this term and got $\sum_a \rho(a)$ and $\sum_b \chi(b)$. Now I need that these two sum are zero. But I don't see why. $\endgroup$ – Memories Jan 2 at 16:46
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    $\begingroup$ @Memories It's a general fact about characters that the values of a non-trivial character sum to zero. $\endgroup$ – Lord Shark the Unknown Jan 2 at 16:51

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