# Quadratic Gauss sum for prime powers [closed]

Prove that $G(k;p^\alpha) = pG(k;p^{\alpha-2})$ for $p\not|k$ and $\alpha\geq2$ , where $$G(k;n) = \sum_{r=1}^ne^{\frac{2\pi ikr^2}{n}}$$

This is an exercise from Apostol's Introduction to Analytic Number Theory(Chapter 8 Pg 177).

## closed as off-topic by Saad, blub, José Carlos Santos, YiFan, callculusApr 12 at 22:43

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Denote $\zeta=\exp(2\pi i k/p^{\alpha-2})$ which is a primitive $p^{\alpha-2}$-th root of unity by the condition $p\not\mid k$. Then $\zeta^{1/p^2}$ is a primitive $p^\alpha$-th root of unity so that \begin{align}G(k;p^\alpha)&=\sum_{r=1}^{p^{\alpha}}\zeta^{r^2/p^2}\\&=\sum_{p\not\mid r}^{p^{\alpha}}\zeta^{r^2/p^2}+\sum_{r=1}^{p^{\alpha-1}}\zeta^{r^2}\\&=\sum_{p\not\mid r}^{p^{\alpha}}\zeta^{r^2/p^2}+pG(k;p^{\alpha-2})\end{align} where in the last equality we used that $\zeta^{(r+p^{\alpha-2})^2}=\zeta^{r^2}$. Thus, we are left to show the first sum vanishes, and for this we use that the sum decomposes in sums of the form $$\sum_{l=0}^{p-1}\omega^{(r+lp^{\alpha-1})^2}=\omega^{r^2}\sum_{l=0}^{p-1}\omega^{2lp^{\alpha-1}}$$ where $\omega=\zeta^{1/p^2}$. Now, for $p\ne 2$, $\omega^{2p^{\alpha-1}}$ is a primitive $p$-th root of unity and hence the last sum vanishes. For $p=2$, we do the same but we use instead the sums $$\sum_{l=0}^{3}\omega^{(r+l2^{\alpha-2})^2}=\omega^{r^2}\sum_{l=0}^{3}\omega^{l2^{\alpha-1}}=0.$$
• Thanks for the reply, but I don't understand the last part. How does $\zeta^\frac{1}{p^2}$ being a root of this polynomial make the sum vanish. – Devang Agarwal May 22 '18 at 9:49
• Although, I am still not quite sure why you treated the $p=2$ case separately. – Devang Agarwal May 22 '18 at 10:27
• basically if $p=2$, $\omega^{2p^{\alpha-1}}$ is just $1$ so we cannot apply the trick of the primitive $p$-th root – ArtW May 22 '18 at 11:47