A little bit different quadratic Gauss sum Could anyone give me a direction on how to demonstrate that
$$
\sum\limits_{k = 0}^{N-1} e^{-i\frac{\pi}{N}(k+C)^2} = \sqrt{N} e^{-i\frac{\pi}{4}},
$$
if $N\in \mathbb{N}$ is even, for any $C \in \mathbb{Z}$.
Unfortunately, the left side of the above equation cannot be exactly defined as a quadratic Gauss sum since it does not have the term 2 within the argument of $e$ and the summation coefficient is shifted by $C$.
 A: Let $N=2n$. Note the following:
$$
\sum_{k=0}^{\color{red}4n-1}e^{-i\frac{2\pi}{4n}(k+C)^2}=\sum_{k=0}^{4n-1}e^{-i\frac{2\pi}{4n}k^2}=(1-i)\sqrt{4n}=\color{red}2e^{-i\frac\pi4}\sqrt{N}
$$
Here we used that $e^{-i\frac{2\pi}{4n}}$ is a primitive $4n$-th root of unity and a classical Gauß sum identity (I used Lang's Algebraic Number Theory as a reference for that part). The former part is given since while $k$ runs through a system of representatives of of $\mathbb Z/4n\mathbb Z$ so does $k+C$ for all $C\in\mathbb Z$.
Now it only remains to show that the sum on the LHS is twice the sum from $k=0$ to $k=2n-1$. For that part notice that
$$
((2n+k)+C)^2\equiv (k+C)^2\mod 4n
$$
which a short computation reveals. Therefore,
$$
\sum_{k=0}^{4n-1}e^{-i\frac{2\pi}{4n}(k+C)^2}=\sum_{k=0}^{2n-1}e^{-i\frac{2\pi}{4n}(k+C)^2}+\sum_{k=2n}^{4n-1}e^{-i\frac{2\pi}{4n}(k+C)^2}=2\sum_{k=0}^{2n-1}e^{-i\frac{2\pi}{4n}(k+C)^2}
$$
which concludes the proof (well, up to the knowledge of the "classical Gauß sum identity").


"Classical Gauß sum identity": Let $b\in\mathbb Z$, $b>0$ and $b\equiv0\mod4$. Then
$$
\sum_{k=0}^{b-1}e^{\frac{2\pi i}b x^2}=(1+i)\sqrt b
$$

