Exercise 3.19

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}}

If $p$ is an odd prime, show that $1$ and $-1$ are the only solutions of $x^2 \equiv 1 \pmod {p^a}$.

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}
$$x^2-1\pmod{p^a} \iff p^a \mid (x-1)(x+1).$$
Let $d = (x-1)\wedge(x+1)$. Then $d = 1$ or $d=2$.

$\bullet$ If $d=1$, then $x$ is even (if not, $x-1$ and $x+1$ are even, and $2\mid d$).
As $p^a \mid (x-1)(x+1)$ and $(x-1) \wedge (x+1) = 1$, then $p^a \mid  x-1$, or $p^a \mid x +1$, that is $$x\equiv \pm 1 \pmod {p^a}.$$

$\bullet$ If $d=2$, then $x$ is odd, and
$$p^a \mid 4 \frac{x-1}{2}\frac{x+1}{2}.$$
As $p$ is an odd prime, $p\wedge 4 = 1$, so $p \mid  \frac{x-1}{2}\frac{x+1}{2}$, where  $\frac{x-1}{2}\wedge \frac{x+1}{2} = 1$, hence $p^a \mid  \frac{x-1}{2} \mid x-1$ or $p^a \mid  \frac{x+1}{2} \mid x+1$, thus $$x\equiv \pm 1 \pmod {p^a}.$$

Conclusion: $\{-\overline{1},\overline{1}\}$ is the set of roots of $x^2-\overline{1}$ in $\Z/p^a\Z$.
\end{proof}

Exercise 3.19

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Exercise 3.19

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