Fix an odd prime $p$. Then for a positive integer $a$, I can look at the [quadratic Legendre symbol Gauss sum](http://en.wikipedia.org/wiki/Quadratic_Gauss_sum) 

$$ G\_p(a) = \sum\_{n \,\bmod\, p} \left( \frac{n}{p} \right) e^{2 \pi i a n / p}$$

where I use $\left( \frac{\cdot}{\cdot} \right)$ to be the Legendre Symbol. We know how to calculate it explicitly, and that's all well and good.

Recently, I've had reason to consider a "prime power" Gauss Sum

$$G\_{p^k}(a) = \sum\_{n\,\bmod\,p^k}\left(\frac{n}{p^k}\right)e^{2\pi i a n/p^k}$$

and I've noticed that as long as $k > 1$, $G\_{p^k}(a) = 0$. 

*A sketch of why this is true*: 

1. If $k$ is even, this is very easy to see as the Legendre symbol goes away, so we are summing the $p^k$th roots of unity.
2. If $k$ is odd, the way I see it is a bit less obvious: reduce the Legendre symbol to be $\left( \frac{n}{p} \right)$ so that it's $p$-periodic, write $n = pn' + n''$ fpr $0 \leq n' < p^{k-1}$ and $0 \leq n'' \leq p$. Reorder the sum to sum over the $p^{k-1}$ roots of unity first, which for $k > 1$ will still give $0$.

My question is this: there are many ways of viewing Gauss sums. They can be viewed as discrete Fourier transforms, generators of quadratic extensions in $\mathbb{Q(e^{2\pi i / p})}$, as eigenfunctions, natural character sums, etc. I suspect that there is some 'obvious' or 'clear' reason why these prime power Gauss sums *should* be zero.

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> Is there an obvious or clear reason why $G\_{p^k}(a) = 0$?
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