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Define the Gauss sum to be $g_p(a) = \sum_{n=0}^{p-1}\left(\frac{n}{p}\right)\zeta_p^{an}.$

We want to compute $g_p(1)^2$ (the answer is it equals $\pm p$ sign depending $p \pmod 4$). To do this you can evaluate the following:

$\sum_{a=0}^{p-1}g_p(a)g_p(-a).$

I am wondering how anyone could come up with that? I don't have any understanding why that sum would be easier to compute (except that after doing it, it wasn't too hard). Can anyone shed light on this? Are there other ways to evaluate the sum?


And is this related to how the double integral of $\exp(-x^2)$ is used to find the single integral?

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    You might also try to calculate the sum directly, when $p=3,5,7,\ldots$. Such experiments often, but not always, lead to a "light-bulb experience".2012-05-28

1 Answers 1

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perhaps it is an answer to your question: $a\mapsto g_p(a)$ is the (discrete) Fourier transform of $n\mapsto (\frac{n}{p})$ (which is a function on the abelian group $\mathbb{Z}/p\mathbb{Z}$ and the dual group is again $\mathbb{Z}/p\mathbb{Z}$). $\sum_a g_p(a) g_p(-a)=\sum_a |g_p(a)|^2$ is then $p\sum_n (\frac{n}{p})^2=p(p-1)$ by Plancherel's formula.