Exercise 3.3.16* ($\sum_{n=1}^p \genfrac{(}{)}{}{}{an+b}{p}= 0$)

Question

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

Prove that if $(a,p) = 1$ and $p$ is an odd prime, then
$$\sum_{n=1}^p \legendre{an+b}{p} = 0.$$

Richard Ganaye · Accepted Answer

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

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

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

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

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

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

Notation: if $\alpha = [a]_p \in \F_p = \Z/p\Z$, we define $\legendre{\alpha}{p} = \legendre{a}{p}$.
\begin{proof} 
Let $\alpha = [a]_p ,\beta = [b]_p$. Since $a \wedge p = 1$, $\alpha 
e 0$, thus
$$\varphi
\left\{
\begin{array}{ccl}
\F_p& 	o & \F_p\
x & \mapsto &\alpha x + \beta
\end{array}
ight.
$$
is bijective. Indeed, let $\psi : \F_p 	o \F_p$ defined by $\psi(y) = \alpha^{-1}(y - \beta)$.

Then $\psi \circ \varphi = \varphi \circ \psi = 1_{F_p}$. This proves that $\varphi$ is bijective (and $\varphi^{-1} = \psi$).

Therefore, the change of variable $y = \varphi(x)$ gives
\begin{align*}
\sum_{n=1}^p \legendre{an+b}{p} &= \sum_{x \in \F_p} \legendre{\alpha x + \beta}{p} \
&= \sum_{y \in \F_p} \legendre{y}{p}\qquad \qquad(y = \varphi(x))\
&=\sum_{y \in \F_p^*} \legendre{y}{p}\
&=\sum_{j = 1}^{p-1} \legendre{j}{p}\
&=0,
\end{align*}
by Problem 5. So
$$\sum_{n=1}^p \legendre{an+b}{p} = 0.$$
\end{proof}

Exercise 3.3.16* ($\sum_{n=1}^p \genfrac{(}{)}{}{}{an+b}{p}= 0$)

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Exercise 3.3.16* ($\sum_{n=1}^p \genfrac{(}{)}{}{}{an+b}{p}= 0$)

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