Exercise 5.37

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 if $a$ is negative, then $p \equiv q \pmod {4a}$, $p 
mid a$ implies $(a/p) = (a/q)$.

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}
Write $a = -A, A>0$. As $p \equiv q \pmod {4a}$, we know from Prop. 5.3.3. (b) that $(A/p) = (A/q)$.

Moreover,
\begin{align*}
\legendre{a}{p}&= \legendre{-A}{p} = (-1)^{(p-1)/2} \legendre{A}{p}\
\legendre{a}{q}&= \legendre{-A}{q} = (-1^{(q-1)/2} \legendre{A}{q}
\end{align*}
As  $p \equiv q \pmod {4a}$, $ p = q + 4ak, k\in \Z$, so
$$(-1)^{(p-1)/2} = (-1)^{(q+4ak-1)/2} = (-1)^{(q-1)/2},$$
so $(a/p) = (a/q)$.
\end{proof}

Exercise 5.37

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Exercise 5.37

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