Exercise 2.9.2 (Number of solutions modulo $p$)

Question

Suppose that $f(x) = ax^2 +bx +c$, and that $D = b^2-4ac$. Show that if $p $  is an odd prime, $p
mid a, p \mid D$, then $f(x) \equiv 0 \pmod p$  has exactly one solution. Show that if $p $  is an odd prime, $p 
mid a, p 
mid D $, then the congruence $f(x) \equiv 0 \pmod p$  has either $0 $  or $2 $  solutions, and that if $x $  is a solution then $f'(x) 
ot \equiv 0 \pmod p$.

Richard Ganaye · Accepted Answer

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} Since $p$ is odd, $2$ has an inverse modulo $p$, which we write $2^{-1}$, and since $p 
mid a$, $a$ has an inverse modulo $p$, which we write $a^{-1}$ (i.e. $a a^{-1} \equiv 1 \pmod p$).

Then 
$$f(x)=a x^2 + bx +c \equiv  a\left[(x + 2^{-1}a^{-1}b)^2 - 2^{-2} a^{-2} Dight] \pmod p.$$
(We may write this more readably $f(x) \equiv  a\left[\left( x + \frac{b}{2a}ight)^2 - \frac{D}{4a^2} ight] \pmod p$).

Since $p$ is prime, and $p
mid a$,
\begin{align}
f(x) \equiv 0 \pmod p \iff (x + 2^{-1}a^{-1}b)^2 \equiv 2^{-2} a^{-2} D \pmod p.
\end{align}

\begin{enumerate}
\item[$\bullet$] If $p \mid D$, 
\begin{align*}
f(x) \equiv 0 \pmod p &\iff (x + 2^{-1}a^{-1}b)^2 \equiv 0\
&\iff x \equiv - 2^{-1}a^{-1}b).
\end{align*}
Therefore the equation $f(x) \equiv 0 \pmod p$ has exactly one solution modulo $p$, which is $ -2^{-1}a^{-1}b$ (this is the root of $f'(x) = 2ax +b$ modulo $p$).
\item[$\bullet$] If $p 
mid D$ and $D$ is not a square modulo $p$ (i.e. $\legendre{D}{p} = -1$), then the equation (1) has no solution.
\item[$\bullet$] If $p 
mid D$ and $D$ is a square modulo $p$ (i.e. $\legendre{D}{p} = 1$), then $D \equiv \delta^2 \pmod p$ for some integer $\delta$, and
$$f(x) \equiv 0 \pmod p \iff ( x + 2^{-1}a^{-1}(b -\delta))( x + 2^{-1}a^{-1}(b +\delta)) \equiv 0 \pmod p.$$
Therefore the equation $f(x) \equiv 0 \pmod p$ has two solutions modulo $p$, which are $2^{-1}a^{-1}(-b - \delta), 2^{-1}a^{-1}(-b + \delta)$. Since $p 
mid D$, then $p 
mid \delta$, thus these solutions are distinct modulo $p$.

If $x \equiv 2^{-1}a^{-1}(-b \pm \delta) \pmod p$ is such a solution, $f'(x) = 2ax +b = \pm \delta 
ot \equiv 0 \pmod p$, since $p 
mid \delta$.
\end{proof}

\end{enumerate}

Exercise 2.9.2 (Number of solutions modulo $p$)

Answers

Comments

Exercise 2.9.2 (Number of solutions modulo $p$)

Answers

Comments

Add answer