Exercise 5.6

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 that the number of solutions to $x^2 - y^2 \equiv a \pmod p$ is given by:
$$
  \sum_{y=0}^{p-1}\left(1+\legendre{y^2 + a}{p}ight).
$$

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}
Let $S =\{(\overline{x},\overline{y}) \in \F_p^2\ \vert \ \overline{x}^2 - \overline{y}^2 = \overline{a}\} $.
From Ex.5.2,
\begin{align*}
\vert S \vert &= \sum_{\overline{y} \in \F_p} \mathrm{Card} \ \{\overline{x} \in \F_p\ \vert \ \overline{x}^2 = \overline{y}^2 + \overline{a}\} \
&= \sum_{y=0}^{p-1}\left(1+\legendre{y^2 + a}{p}ight).
\end{align*}
\end{proof}

Exercise 5.6

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Exercise 5.6

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