Exercise 3.3.5 ($\sum\limits_{j=1}^{p-1} \genfrac{(}{)}{}{}{j}{p} \equiv 0 \pmod p$)

Question

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
Prove that $\sum\limits_{j=1}^{p-1} \legendre{j}{p} \equiv 0 \pmod p$, $p$ an odd prime.

Richard Ganaye · Accepted Answer

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

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

\begin{proof} Since the there are $(p-1)/ 2$ non residues, there exists some integer $a$ such that $\legendre{a}{p} 
e 1$.
If $\alpha  = [a]_p\in \F_p^* = (\Z/p\Z)^*$, we define $\legendre{\alpha}{p} = \legendre{a}{p}$. This makes sense because $\legendre{a}{p}$ depends only of the class of $a$ modulo $p$.

Put 
$$S = \sum_{j=1}^{p-1} \legendre{j}{p}.$$
Then, for $\alpha = [a]_p$,
\begin{align*}
\legendre{a}{p} S &= \legendre{a}{p} \sum_{j=1}^{p-1} \legendre{j}{p}\
&= \sum_{j=1}^{p-1} \legendre{aj}{p}\
&= \sum_{x \in \F_p^*} \legendre{\alpha x}{p}\
& = \sum_{y \in \F_p^*} \legendre{y}{p}, & (y = \alpha x)
\end{align*}
because the map (change of variable)
$$\varphi
\left\{
\begin{array}{ccl}
\F_p^* & 	o & \F_p^*\
x & \mapsto & y = \alpha x
\end{array}
ight.
$$
is a bijection.

Since 
$$ \sum_{y \in \F_p^*} \legendre{y}{p} =  \sum_{x \in \F_p^*} \legendre{x}{p} = \sum_{j=1}^{p-1} \legendre{j}{p} = S,$$
we obtain
$$\legendre{a}{p} S = S,$$
so $(\legendre{a}{p} - 1) S = 0$, where $\legendre{a}{p} 
e 1$, therefore
$$S =  \sum_{j=1}^{p-1} \legendre{j}{p} = 0.$$

(Alternatively, since there are as many residues that non residues in $[\![1, p-1]\!]$, there are as many terms equal to $1$ than terms equal to  $-1$ in the sum $\sum_{j=1}^{p-1} \legendre{j}{p} $, hence $\sum\limits_{j=1}^{p-1} \legendre{j}{p} = 0$.)
\end{proof}

Exercise 3.3.5 ($\sum\limits_{j=1}^{p-1} \genfrac{(}{)}{}{}{j}{p} \equiv 0 \pmod p$)

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Exercise 3.3.5 ($\sum\limits_{j=1}^{p-1} \genfrac{(}{)}{}{}{j}{p} \equiv 0 \pmod p$)

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