Exercise 5.5

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}}

Prove that $\sum_{x=1}^{p-1} ((ax + b)/p) = 0$ provided that $p 
mid a$.

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}}

There is a mistake in the sentence : we must read

{\it  Prove that $\sum_{x=0}^{p-1} ((ax + b)/p) = 0$ provided that $p 
mid a$.}

For instance, 
$$\sum_{x=1}^{5-1} \legendre{x+1}{5} = \legendre{2}{5} +  \legendre{3}{5}+  \legendre{4}{5} = -1
eq 0.$$
\begin{proof} 
From exercise 5.3, as $\legendre{0}{p} = 0$, we know that
$$\sum_{\overline{x} \in \F_p} \legendre{x}{p} = \sum _{x=0}^{p-1} \legendre{x}{p} = \sum_{x=1}^{p-1} \legendre{x}{p} =0.$$

(This sum is well defined, since $\legendre{x}{p}$ depends only of $\overline{x}$ : $x\equiv x' \pmod p \Rightarrow \legendre{x}{p} =\legendre{x'}{p}$.)

As $\overline{a} 
eq \overline{0}$ in $\F_p$,
$
f :
\left\{
\begin{array}{ccc}
  \F_p &	o   &\F_p   \
  x & \mapsto   & \overline{a} x+\overline{b}
 \end{array}
ight.
$ is a bijection. Thus
\begin{align*}
\sum_{x=0}^{p-1} \legendre{ax+b}{p} & = \sum_{x \in \F_p} \legendre{f(x)}{p}\
&=\sum_{y \in \F_p} \legendre{y}{p}\qquad (y = f(x))\
&= 0
\end{align*}
\end{proof}

Exercise 5.5

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Exercise 5.5

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