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Let $S=\sum_{k=1}^{m}e^{2\pi ik^2/m}$,if $m$ is odd,how to directly calculate the absolute value of $S=\sqrt{m}$.Don't use Gauss sum since here it says "it's easily shown"enter image description here

My try is as follows: $$ \begin{align}S^2&=\left(\sum_{k=1}^{m}e^{2\pi ik^2/m}\right)\left(\sum_{k=1}^{m}e^{-2\pi ik^2/m}\right) \\&=\sum_{k=1}^{m-1}\sum_{d=1}^{m-k}2\cos\left(\frac{2\pi}{m}(2dk+k^2)\right)+m\end{align}$$ we must prove $\sum_{k=1}^{m-1}\sum_{d=1}^{m-k}2\cos(\frac{2\pi}{m}(2dk+k^2))=0$.If m is small, I can directly calculate, but if m is large, how to do so by induction or any other solutions.

Thomas Andrews
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math
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2 Answers2

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We have

\begin{align*} |S|^2 &= \sum_{k=0}^{m-1}\sum_{l=0}^{m-1} \exp\left(\frac{2\pi i}{m}(l^2-k^2)\right) \\ &= \sum_{k=0}^{m-1}\sum_{d=0}^{m-1} \exp\left(\frac{2\pi i}{m}(2kd+d^2)\right) \tag{$l\equiv k+d \pmod{m}$} \\ &= \sum_{d=0}^{m-1} \Biggl[ \sum_{k=0}^{m-1} \exp\left(\frac{4\pi i d}{m} k\right) \Biggr] \exp\left(\frac{2\pi i d^2}{m}\right). \end{align*}

As pointed out by @Cade Reinberger, the geometric sum formula shows that the inner sum only survives with the value $m$ when $d=0$, and hence the desired claim follows.

Sangchul Lee
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Using the fact that $$\sum_{n=1}^L \cos(a+nd) = \csc\left(\frac d2 \right) \sin\left(\frac{dL}{2} \right) \cos\left(a+\frac d2(L+1)\right), $$ the inner sum can be evaluated exactly to reduce it down to $$- 2\sum_{k=1}^{m-1} \cot\left(\frac{2\pi k}{m} \right ) \sin\left(\frac{2\pi k^2}{m} \right)$$ Then it can be easily shown that $$ \cot\left(\frac{2\pi k}{m} \right ) \sin\left(\frac{2\pi k^2}{m} \right) $$ changes sign under the transformation $k \mapsto m-k$, and hence this sum is equal to the negative of itself, or in other words equals $0$.

Vishu
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