Determining the Value of a Gauss Sum. Can we evaluate the exact form of $$g\left(k,n\right)=\sum_{r=0}^{n-1}\exp\left(2\pi i\frac{r^{2}k}{n}\right) $$ for general $k$ and $n$?  For $k=1$, on MathWorld we have that 
$$g\left(1,n\right)=\left\{ \begin{array}{cc}
(1+i)\sqrt{n} & \ \text{when}\ n\equiv0\ \text{mod}\ 4\\
\sqrt{n} & \text{when}\ n\equiv1\ \text{mod}\ 4\\
0 & \text{when}\ n\equiv2\ \text{mod}\ 4\\
i\sqrt{n} & \text{when}\ n\equiv3\ \text{mod}\ 4
\end{array}\right\} .$$
I know how to generalize to all $k$ when $n=p$ is a prime number, but what do we do when $n$ is not prime?  Is there a simple way to rewrite it using whether or not $k$ is a square?  I have a suspicion it should be fairly close to the form above, any help is appreciated.
Thanks,
 A: They do not provide a derivation, but this is actually written up in Wikipedia.
I use the standard notation $e(x) = \exp(2 \pi i x)$.  Assuming $\gcd(k,n) = 1$, we have
$$
\sum_{x \in \mathbb{Z}/n \mathbb{Z}} e\left(\frac{kx^2}{n} \right) = \left\{
\begin{array}{lcl}
\varepsilon_n \left( \frac{k}{n} \right) \sqrt{n} & & n \equiv 1 \pmod{2}\\
0 & &  n \equiv 2\pmod{4}\\
(1 + i) \varepsilon_k^{-1} \left(\frac{n}{k} \right)\sqrt{n} & & k \equiv 1 \pmod{2}, 4 \mid n
\end{array}\right.
$$
where $\left( \cdot \right)$ and for odd $m$,
$$
\varepsilon_m = \left\{ \begin{array}{cc} 
1 & & m \equiv 1 \pmod{4}\\
i & & m \equiv 3 \pmod{4} 
\end{array}\right.
$$
When $\gcd(k,n) > 1$, write $k' = k/\gcd(k,n)$ and $n' = n/\gcd(k,n)$, to see that
$$
\sum_{x \in \mathbb{Z}/n \mathbb{Z}} e\left(\frac{kx^2}{n} \right)= \gcd(k,n) \sum_{x \in \mathbb{Z}/ n' \mathbb{Z}} e\left(\frac{k' x^2}{n'} \right)
$$
since the variable $x$ runs through $\gcd(k,n)$ copies of the above sum.

Added: I looked through Iwaniec-Kowalski and they have a few notes on the sums, but nothing short worth noting here.  These sums are discuss in Chapter 3 (around page 49) of their book.  Anyways, I hope this answer is what you were looking for.
