Exercise 11.1.5

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

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{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

As noted in the text, if $f\in \Z[x]$ has degree $n$ and its leading coefficient is not divisible by a prime $p$, then $f(x) \equiv 0 \mod p$ has at most $n$ solutions modulo $p$. Here are two questions that explore what happens when $n=2$ and the modulus is arbitrary.
\be
\item[(a)] How many solutions does the congruence $x^2 - 1 \equiv 0 \mod 8$ have modulo 8?
\item[(b)] Fix an integer $m>1$, and assume that every polynomial of degree 2 in $\Z/m\Z[x]$ has at most two roots in $\Z/m\Z$. Is $m$ prime?
\ee

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers \def\dcro{\mbox{]\hspace{-.15em}]}} ewcommand{\be} {\begin{enumerate}} ewcommand{\ee} {\end{enumerate}} ewcommand{\deb}{\begin{eqnarray*}} ewcommand{\fin}{\end{eqnarray*}} ewcommand{\ssi} {si et seulement si } 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{\U}{\mathbb{U}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\im}{\,\mathrm{Im}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{\Gal}{\mathrm{Gal}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \begin{proof} \be \item[(a)] $\pm1,\pm3$ are the roots in $\Z/8\Z$ of the polynomial $x^2-1 \in (\Z/8\Z)[x]$. \item[(b)] Suppose that $m>1$ is not prime. Then $m$ is composite, so $m = uv$, where $14$, the polynomial $f(x) = vx^2 \in\Z/m\Z[x]$ has at least three distinct roots in $\Z/m\Z$. If $m = 4$, the polynomial $g(x) = 2x^2+2x \in \Z/4\Z[x]$ has 4 roots, namely $0,1,2,3$. Conclusion: If $m>1$ is not prime, there exists some polynomial of degree 2 in $\Z/m\Z[x]$ with more than 2 roots in $\Z/m\Z[x]$. If $m>1$, and if every polynomial of degree 2 in $\Z/m\Z[x]$ has at most two roots in $\Z/m\Z[x]$, then $m$ is prime. \ee \end{proof}

Exercise 11.1.5

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Exercise 11.1.5

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