Exercise 5.31

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}}

Use the results of Exercises 29 and 30 to show that $(RR) = \frac{1}{4}(p-4-\varepsilon)$, where $\varepsilon = (-1)^{(p-1)/2}$

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}
To summarize the results of the Ex. 5.29 and 5.30,
\begin{align*}
(a) (RR) +(RN) + (NR)+ (NN) &=  p-2\
(b) (RR) + (NN) - (RN) - (NR)  &= -1\
\end{align*}
and
\begin{align*}
(c)(RR)+(RN) &=\frac{1}{2}\left(p - 2 - (-1)^{\frac{p-1}{2}}ight)\
(d)(RR)+(NR) &= \frac{p-1}{2} - 1\
\end{align*}
The sum of (a) and (b) gives
$$(e) (RR) + (NN) = \frac{p-3}{2}.$$
The sum of (c),(d),(e) gives (using (a))
$$2(RR) + p-2 = \frac{p-2}{2} + \frac{p-1}{2} + \frac{p-3}{2} - 1 - \frac{(-1)^{\frac{p-1}{2}}}{2},$$
so
$$2 (RR) = \frac{p-1}{2} + \frac{p-3}{2} - \frac{p-2}{2} - 1 - \frac{(-1)^{\frac{p-1}{2}}}{2} = \frac{p}{2} - 2 - \frac{(-1)^{\frac{p-1}{2}}}{2},$$
$$(RR) = \frac{1}{4} (p-4-\varepsilon),\ \mathrm{where} \ \varepsilon = (-1)^{\frac{p-1}{2}}.$$
\end{proof}

Exercise 5.31

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Exercise 5.31

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