Exercise 5.18

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}}

Let $D$ be a square-free integer that is also odd and positive. Show that there is an integer $b$ prime to $D$ such that $(b/D) = -1$.

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 $D = p_1p_2\cdots p_k$, where the $p_i$ are distinct odd primes.

Let $s$ be a nonresidue modulo $p_k$. By the Chinese Remainder Theorem, as $p_i \wedge p_j = 1$ if $i
e j$,  there exists an integer $b$ such that
$$b \equiv 1 \pmod {p_1}, b \equiv 1 \pmod {p_2},\cdots, b \equiv 1 \pmod {p_{k-1}}, b \equiv s \pmod {p_k}.$$
Then $(b/p_i) = 1,\ i=1,2,\ldots k-1$, $(b/p_{k}) = -1$, so $b\wedge p_i = 1$ for all $i = 1,2,\ldots,k$. Then $b\wedge D = b\wedge p_1\cdots p_k = 1$ , and 
$$\legendre{b}{D} = \prod_{i=1}^k \legendre {b}{p_i} = \legendre{b}{p_k} = -1.$$
\end{proof}

Exercise 5.18

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Exercise 5.18

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