# Gauss sum variation $\sum_{n=0}^{p-1}\left(\frac{a+bn}{p}\right)\zeta_p^{cn} = ?$

I'm having trouble evaluating this, for $a, b, p$ all pairwise coprime, $p$ an odd prime, $c$ any integer.

$$\sum_{n=0}^{p-1}\left(\frac{a+bn}{p}\right)\zeta_p^{cn}$$

Any help/references would be appreciated! Thanks.

• $n\mapsto m=a+bn$ is a bijection $\mathbb{F}_p \to \mathbb{F}_p$ so it is $$\sum_{n=0}^{p-1}\left(\frac{a+bn}{p}\right)\zeta_p^{cn}=\sum_{m=0}^{p-1} (\frac{m}{p}) \zeta_p^{cb^{-1}(m-a)} = \zeta_p^{-ab^{-1}c} (\frac{cb^{-1}}{p}) \sum_{m=0}^{p-1} (\frac{m}{p}) \zeta_p^m$$ where $\sum_{m=0}^{p-1} (\frac{m}{p}) \zeta_p^m$ is $\sqrt{p}$ or $i \sqrt{p}$ depending on $p \bmod 4$ – reuns Apr 27 '17 at 16:59
• Ah, so simple! Thanks. – HmmmBeee Apr 27 '17 at 17:11

I'll assume $c$ is also coprime to $p$. Then $c\equiv br\pmod p$ for some $r$. The sum is $$\zeta^{-ra}\sum_n\left(\frac{a+bn}p\right)(\zeta^r)^{a+bn} =\zeta^{-ra}\sum_n\left(\frac{m}p\right)(\zeta^r)^m$$ which is a conventional Gauss sum.