Exercise 5.8

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}}

Combining the results of Ex. 5.6 and 5.7 show that:
$$
  \sum_{y=0}^{p-1} \left(\frac{y^2 +a}{p}ight) =
  \begin{cases}
   \  -1 \ \quad \mathrm {if}\ p
mid a \
   \  p-1 \quad \mathrm{if}\, p \mid a
  \end{cases}
$$

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}\} $.

We obtain in Ex 5.6, $\vert S \vert =    \sum\limits_{y=0}^{p-1}\left(1+\legendre{y^2 + a}{p}ight)$, and in Ex. 5.7. , $\vert S \vert  =  p-1$ if $p
mid a$, $\vert S \vert  =  2p-1$ if $p\mid a$.

Therefore $$|S| - p = \sum\limits_{y=0}^{p-1} \legendre{y^2 + a}{p} = 
 \begin{cases}
   \  -1 \ \quad \mathrm {if}\ p
mid a \
   \  p-1 \quad \mathrm{if}\, p \mid a
  \end{cases}
  $$
\end{proof}

Exercise 5.8

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Exercise 5.8

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