# How to prove a generalized Gauss sum formula

First let me write a definition of a generalized Gauss sum.

Let $G(a, c)= \sum_{n=0}^{c-1}\exp (\frac{an^2}{c})$, where $a$ and $c$ are relatively prime integers.

(Here is another question. Is the function $e(x)$ defined in the article equal to $\exp(2\pi i/x)$ or $\exp(\pi i/x)$?)

In the article, a formula is given according to values of $a$ and $c$.

For example, if $a$ is odd and $4|c$, then

$G(a, c)=(1+i)\epsilon_a^{-1} \sqrt{c} \big(\frac{c}{a}\big)$, where $\big(\frac{c}{a}\big)$ is the Jacobi symbol.

I would like to prove it but I don't know how to do it. Could you give me a guide?

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$e(x)$ means neither of the things you write, but $e^{2\pi ix}$. The proof of the formula for Gauss sums is long, and I'm certainly not going to write it out, but many books on algebraic number theory will have it. Maybe you could find such a book in the library. In particular, three very good sources are mentioned at the page to which you have linked. – Gerry Myerson Mar 8 '13 at 0:07
@GerryMyerson Thanks for the definition of $e(x)$. None of the number theory books I have deals with a generalized Gauss sum. Is there any online resource for this? – Snow Mar 8 '13 at 1:12
Probably. If there is, you can find it as easily as I can --- I would just type some keywords into the internet and see what comes back. Concerning books, that's what libraries are for. – Gerry Myerson Mar 8 '13 at 4:25

The quadratic Gauss sum is given by \begin{eqnarray*} G(s;k) = \sum_{x=0}^{k-1} e\left(\frac{sx^2}k\right), \end{eqnarray*} where $\displaystyle e(\alpha) = e^{2\pi \imath \alpha}$, $s$ is any integer coprime to $p$ and $k$ is a positive integer. The generalized Gauss sums is given by \begin{eqnarray*} G(a,b,c) = \sum_{x=0}^{|c|-1} e\left(\frac{ax^2+bx}c\right), \end{eqnarray*} where $ac \neq 0$ and $ac+b$ is even.
One method is to show, for $k$ odd, $|G(s;k)|^2 = k$, and then determining the sign of $G(s;k)$ will be the hard part. From here you can use reduction properties of the quadratic Gauss sum and the Chinese Remainder Theorem to prove the even cases.