$\sum_{n=0}^{N-1}\exp(-i\frac{\pi}{N}n^2)=\sqrt{N}\exp(-i\frac{\pi}{4})$

Question

The question is: N is an even positive integer, then $\sum_{n=0}^{N-1}\exp(-i\frac{\pi}{N}n^2)=\sqrt{N}\exp(-i\frac{\pi}{4})$. I thought the terms on the left are the solutions set of some polynomial with coefficients in $\mathbb{Q}(\sqrt{N}\exp(-i\frac{\pi}{4}))$, but I have no idea how to proceed. Thank you all the time.

Answer 1

Let $m=2N$. The value of the Gauss sum $\sum_{x=0}^{m-1}e^{-2\pi i x^2/m}$ is well known. Note that $$\sum_{x=0}^{m-1}e^{-2\pi i x^2/m}=\sum_{n=0}^{N-1}\left(e^{-2\pi i n^2/(2N)}+e^{-2\pi i(N+n)^2/(2N)}\right)=2\sum_{n=0}^{N-1}e^{-\pi in^2/N}.$$ So your desired identity follows.