Exercise 8.16

Question

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}{\mathrm{N}}

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Suppose that $p\equiv 1 \pmod 4$ and that $\chi$ is a character of order 4. Let $N$ be the number of solutions to $x^4+y^4 = 1$ in $\F_p$. Show that $N = p+1-\delta_4(-1) 4 + 2\, \mathrm{Re}\, J(\chi,\chi) + 4 \, \mathrm{Re}\, J(\chi,ho)$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

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\begin{proof}
Let $\chi$ be a character of order 4 : such a character exists since $p\equiv 1 \pmod 4$. Then

\begin{align*}
N(x^4+y^4=1) &=\sum\limits_{a+b=1} N(x^4=a)N(y^4=b)\
&=\sum\limits_{a+b=1} \sum\limits_{i=0}^3 \chi^i(a) \sum\limits_{j=0}^3 \chi^j(b)\
&= \sum\limits_{i=0}^3 \sum\limits_{j=0}^3\sum\limits_{a+b=1} \chi^i(a)  \chi^j(b)\
&= \sum\limits_{i=0}^3 \sum\limits_{j=0}^3 J(\chi^i,\chi^j)\
&= p-\chi(-1)-\chi^2(-1) - \chi^3(-1) \
&+ J(\chi,\chi) + J(\chi,\chi^2) +J(\chi^2,\chi)\
&+J(\chi^2,\chi^3)+J(\chi^3,\chi^2)+J(\chi^3,\chi^3),
\end{align*}
since from Theorem 1, we have $J(\varepsilon,\varepsilon) = p, J(\varepsilon,\chi^j) = 0$ for $j = 1,2,3$, and $J(\chi^i, \chi^{4-i}) = -\chi^i(-1)$.

Moreover
 $$-[\chi(-1)+\chi^2(-1)+\chi^3(-1)] = 1-[1+\chi(-1)+\chi^2(-1)+\chi^3(-1)],$$
 
 and
 $$
\left\{
\begin{array}{ccll}
  1+\chi(-1)+\chi^2(-1)+\chi^3(-1) = \frac{1-\chi^4(-1)}{1-\chi(-1)} & = 0&\ \mathrm{if}\ \chi(-1) 
eq 1    \
  &     =4&\ \mathrm{if}\ \chi(-1) =1.
\end{array}
ight.
$$

Let $g$ a generator of $\F_p^*$. Recall that  $\chi(g) = e^{qi\pi/2}$ with $q$ odd, so $\chi : a = g^k \mapsto e^{i qk\pi /2} = i^{qk}$, thus
 $$\chi(a) = 1 \iff \chi(g^k)=1 \iff  i^{qk} = 1 \iff 4 \mid k \iff a = b^4, b \in \F^*.$$
 $\delta_4$ is defined by  $\delta_4(a) = 1$ if $a$ is a fourth power, 0 if not. Then  
 $$-[\chi(-1)+\chi^2(-1)+\chi^3(-1)] = 1 - \delta_4(-1)4.$$
 Moreover $J(\chi,\chi)+ J(\chi^3,\chi^3) = 2 \,\mathrm{Re}\,( J (\chi,\chi))$, and
 $$J(\chi,\chi^2)+J(\chi^3,\chi^2) + J(\chi^2,\chi)+J(\chi^2,\chi^3) = 2 \,\mathrm{Re}\,(J(\chi,\chi^2)) + 2\,\mathrm{Re}\,(J(\chi^2,\chi))= 4\,\mathrm{Re}\,(J(\chi,\chi^2)).$$
$\chi$ is of order 4, so $ho = \chi^2$ is the unique character of order 2, the Legendre's character.

In conclusion,
 $$N(x^4+y^4=1) = p+1- \delta_4(-1)4+ 2 \,\mathrm{Re}\,( J (\chi,\chi))+4\,\mathrm{Re}\,(J(\chi,ho)).$$
\end{proof}

Exercise 8.16

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Exercise 8.16

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