Exercise 4.12

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 existence of a primitive root to give another proof of Wilson's theorem $(p - 1)! \equiv -1  \pmod p$.

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} As the result is trivial if $p=2$, we suppose that $p$ is an odd prime.

Let $g$ be a primitive root modulo $p$. Then $\overline{g}$ is a generator of $\F_p^*$.

As $(\overline{g}^{(p-1)/2}) ^2 = \overline{g}^{p-1} = \overline{1}$, and $\overline{g}^{(p-1)/2} 
eq 1$ in the field $\F_p^*$,  then $\overline{g}^{(p-1)/2} = -1$, and $(\overline{1},\overline{g},\overline{g}^{2}, \ldots, \overline{g}^{p-2}) $ is a permutation of $ (\overline{1},\overline{2}, \ldots,\overline{p-1})$, thus
\begin{align*}
\overline{(p-1)!} &= \prod_{k=0}^{p-2} \overline{g}^k\
&= \overline{g}^{\sum_{k=0}^{p-2} k}\
&=\overline{g}^{(p-2)(p-1)/2}\
&=\left(\overline{g}^{(p-1)/2}ight)^{p-2}\
&=(-\overline{1})^{p-2}\
&= - 1.
\end{align*}
Hence $(p-1)! \equiv -1\pmod p$ for each prime $p$.
\end{proof}

Exercise 4.12

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Exercise 4.12

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