It is known that for even $k$, $\omega$ a primitive $k$th root of unity, the Gauss sum
$$\frac{1}{\sqrt{k}} \sum_{a = 0}^{k-1} \omega^{a^2/2}$$
is an 8th root of unity, related to the signature of an associated quadratic form.
I am interested in the following sum, for odd $k$,
$$\frac{1}{\sqrt{k}} \sum_{a = 0}^{k-1} \omega^{(a-1/2)^2/2}$$
which also appears to be an 8th root of unity. However, I haven't seen this sum in the literature. Do you know how to evaluate it, or if it's known?
Thanks!
For $\omega_k=\exp(2\pi i/k)$ a primitive $k$-th root of unity, we have (see also [1] and [2] here) $$\frac{1}{\sqrt{k}}\sum_{j=0}^{k-1}\omega_k^{j^2}=\begin{cases}1+i,& k\equiv 0\pmod4\\\hfill 1,\hfill&k\equiv 1\pmod4\\\hfill 0,\hfill&k\equiv 2\pmod4\\\hfill i,\hfill&k\equiv 3\pmod4\end{cases}$$ as a special case of the conventional Gauss sum. This allows to evaluate both of your sums: $$\sum_{j=0}^{k-1}\omega_k^{j^2/2}=\sum_{j=0}^{k-1}\omega_{2k}^{j^2}\underset{(*)}{\ =\ }\frac12\sum_{j=0}^{2k-1}\omega_{2k}^{j^2}=\frac{1+i}{\sqrt{2}}\sqrt{k}$$ for even $k$, where $(*)$ follows from $(k+j)^2=k(k+2j)+j^2\equiv j^2\pmod{2k}$, and similarly $$\sum_{j=0}^{k-1}\omega_k^{(j\color{red}{+}1/2)^2/2}=\sum_{j=0}^{k-1}\omega_{8k}^{(2j+1)^2}=\frac14\sum_{j=0}^{4k-1}\omega_{8k}^{(2j+1)^2}=\frac14\left(\sum_{j=0}^{8k-1}\omega_{8k}^{j^2}-\sum_{j=0}^{4k-1}\omega_{8k}^{(2j)^2}\right)=\frac{1+i}{\sqrt{2}}\sqrt{k}$$ for odd $k$, this time $\big(2(k+j)+1\big)^2=4k(k+2j+1)+(2j+1)^2\equiv(2j+1)^2\pmod{8k}$, and the last sum is $2\sum_{j=0}^{2k-1}\omega_{2k}^{j^2}=0$ since $2k\equiv 2\pmod4$.
