PlanetMath: derivation of Gauss sum up to a sign

The Gauss sum can be easily evaluated up to a sign by squaring the original series

$\displaystyle g_1^2(\chi)$	$\displaystyle = \sum_{s \in \mathbb{Z}/p\mathbb{Z}} \left(\frac{s}{p}\right) e^... ...p} \sum_{t \in \mathbb{Z}/p\mathbb{Z}} \left(\frac{t}{p}\right) e^{2 \pi i t/p}$
	$\displaystyle = \sum_{s,t \in \mathbb{Z}/p\mathbb{Z}} \left(\frac{st}{p}\right) e^{2 \pi i(s+t)/p}$
and summing over a new variable $n=s^{-1} t\pmod p$
	$\displaystyle = \sum_{s,n \in \mathbb{Z}/p\mathbb{Z}} \left(\frac{n}{p}\right) e^{2 \pi i(s+n s)}$
	$\displaystyle = \sum_{n \in \mathbb{Z}/p\mathbb{Z}} \left(\frac{n}{p}\right) \sum_{s \in \mathbb{Z}/p\mathbb{Z}} e^{2 \pi i s(n+1)}$
	$\displaystyle = \sum_{n \in \mathbb{Z}/p\mathbb{Z}} \left(\frac{n}{p}\right) (q[n\equiv -1 \pmod p]-1)$
	$\displaystyle = p\left(\frac{-1}{p}\right)-\sum_{n \in \mathbb{Z}/p\mathbb{Z}} \left(\frac{n}{p}\right)$
	$\displaystyle = p\left(\frac{-1}{p}\right)=\begin{cases}\hphantom{+{}}p,&{\rm if\ }p\equiv 1 \pmod 4 ,\\ -p,&{\rm if\ }p\equiv 3 \pmod 4.\end{cases}$

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