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Let $\omega = e^{\frac{2\pi i}{d}}$, where $d$ is prime. Also let $0 < a < d$ and $0 < b < d$ (note that $a$ and $b$ are non-zero). Is there a nice formula for:
$S = \sum\limits_{j=0}^{d-1} \omega^{aj + bj(j-1)/2}$
Ideally, I'd like to find $\phi$ such that $S = r e^{i \phi}$. I already know that $r = \sqrt{d}$, but I don't know how to express $\phi$ as a function of $a$, $b$ and $d$.
Assume that $d > 2$, otherwise answer is obvious.
Let's deal with exponent (from this moment, all divisions are modulo $d$): $$ aj + b\frac{j(j-1)}2 = \frac b2 \left(j^2 + \left(\frac{2a}b-1\right)j\right) = \frac b2 \left(j + \frac ab - \frac 12\right)^2 - \frac{(2a-b)^2}{8b} $$ When $j$ runs complete residue system, so does $j + \frac ab - \frac 12$. Then $$ S = \omega^{-(2a-b)^2 /(8b)}\sum_{j=0}^{d-1} \omega^{b/2 \cdot j^2}, $$ and sum rewrites as $$ \sum_{h=0}^{d-1} \left( 1 + \left(\frac hd \right)\right)\omega^{b/2 \cdot h} = \sum_{h=0}^{d-1} \omega^{b/2 \cdot h} + \sum_{h=0}^{d-1} \left(\frac hd \right)\omega^{b/2 \cdot h}, $$ where $\left( \frac \cdot d \right)$ denotes Legendre symbol. Left sum vanishes and right is the well-known Gauss sum, which equals to $\left(\frac {b/2}d\right)\sqrt d$ when $d \equiv 1 \pmod 4$ or $\left(\frac {b/2}d\right)i\sqrt d$ when $d \equiv 3 \pmod 4$.
