Exercise 4.17

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 fact that $2$ is a primitive root modulo $29$ to find the seven solutions to $x^7 \equiv 1 \pmod {29}$.

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 $x \in \Z$, then $x \equiv 2^k \pmod{29} , k \in \N$.
\begin{align*}
x^7 \equiv 1 \pmod{29} &\iff 2^{7k} \equiv 1 \pmod {29}\
&\iff 28 \mid 7k\
&\iff 4 \mid k
\end{align*}
So the group cyclic  $S$ of the roots of $x^7 - 1$ in $\F_{29}$ are
$$S = \{1,2^4,2^8,2^{12},2^{16},2^{20},2^{24}\},$$
$$S = \{1,16,24,7,25,23,20\}.$$
\end{proof}

Exercise 4.17

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Exercise 4.17

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