Exercise 8.4

Question

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

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Show, if $k \in \F_p, k
eq 0$, that $\sum_t \chi(t(k-t)) = \chi(k^2/2^2)J(\chi,ho)$.

Richard Ganaye · Accepted Answer

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

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
We know from Ex. 8.3 that  $J(\chi,ho) = \sum_t \chi(1-t^2)$, so

\begin{align*}
J(\chi,ho) &= \sum_{t \in \F_p}\chi(1-t)\chi(1+t)\
&=\sum_{u \in \F_p} \chi(u) \chi(2-u)\qquad(u = 1-t)\
&=\chi(2^2) \sum_{u \in \F_p} \chi\left(\frac{u}{2}ight) \chi\left(1-\frac{u}{2}ight)\
&= \chi(2^2) \sum_{v \in \F_p} \chi(v) \chi(1-v)\qquad(u = 2v)\
&= \chi(2^2) \chi(k^{-2}) \sum_{v \in \F_p} \chi(kv) \chi(k-kv)\
&=\chi(2^2/k^2) \sum_{t \in \F_p} \chi(t) \chi(k-t)\qquad(t=kv).
\end{align*}

Conclusion: if $k \in \mathbb{F}^*$, and $\chi$ is a non trivial character, $ho$ the character of order 2,
$$ \sum_{t\in \F_p}\chi(t(k-t)) = \chi(k^2/2^2) J(\chi,ho).$$
\end{proof}

Exercise 8.4

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Exercise 8.4

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