Exercise 5.4

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 $\sum_{a=1}^{p-1} (a/p) = 0$.

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} Here $p$ is an odd  prime (the result is false if $p=2$). In the interval $[1,p-1]$, there exist $(p-1)/2$ residues, and $(p-1)/2$ nonresidues (Prop. 5.1.2., Corollary 1), so $\sum_{a=1}^{p-1} (a/p) = 0$.
\end{proof}

\begin{proof}
As an alternative proof, let $S = \sum\limits_{a=1}^{p-1} \legendre{a}{p}$, and  $b$ a nonresidue modulo $p$ : $\legendre{b}{p} = -1$ (such a $b$ exists if $p
eq 2$). As $a \mapsto ab$ is a bijection from $\F_p^*$ to itself, 
$$\legendre{b}{p} S = \sum_{a=1}^{p-1} \legendre{ab}{p} = \sum_{c = 1}^{p-1} \legendre{c}{p}  = S,$$
so $-S = S, S=0$.
\end{proof}

Exercise 5.4

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Exercise 5.4

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