Estermann’s evaluation of Gauss Sum

We show Esterman’s evaluation of the Gauss sum

$\displaystyle g_p=\sum_{n=0}^{p-1} e^{2 \pi in^{2} / p}=\sum_{m=0}^{p-1}\left( \frac{m}{p}\right)e^{2 \pi i m / p}.$

It’s easy to see by squaring that

$\displaystyle g_p^2 = \left( \frac{-1}{p}\right) p$

which implies that

$\displaystyle g_{p}=\pm \sqrt{p} \text { if } p \equiv 1~(\bmod 4) \quad \text { and } g_{p}=\pm i \sqrt{p} \text { if } p \equiv 3 ~(\bmod 4)$

What about the sign?

Both cases together can be written as

$\displaystyle \frac{1}{2}(1-i)\left(1+i^{p}\right) g_{p}=\pm \sqrt{p}$

To show that the sign is positive we will prove that the real part of the above expression ${\frac{1}{2}(1-i)\left(1+i^{p}\right) g_{p}}$ is at least ${-\sqrt p}.$

Start with

$\displaystyle g_p-1= \sum_{n=1}^{p-1} e^{2 \pi in^{2} / p}. = 2\sum_{n=1}^{(p-1)/2} e^{2 \pi in^{2} / p}.$

Multiplying with ${1+i^p,}$ we have

$\displaystyle (1+i^p)\sum_{n=1}^{(p-1)/2} e^{2 \pi in^{2} / p}=\sum_{n=1}^{(p-1)/2} e^{2 \pi in^{2} / p}+ \sum_{n=1}^{(p-1)/2} e^{2 \pi in^{2} / p} e^{\frac{\pi i p}{2}}$

$\displaystyle \implies (1+i^p)\sum_{n=1}^{(p-1)/2} e^{2 \pi in^{2} / p}= \sum_{n=1}^{(p-1)/2} e^{2 \pi in^{2} / p} + \sum_{n=1}^{(p-1)/2} e^{2 \pi i (p/2-n)^{2} / p}$

$\displaystyle \implies (1+i^p)\sum_{n=1}^{(p-1)/2} e^{2 \pi in^{2} / p} = \sum_{n=1}^{p-1} e^{\pi in^{2} / 2p}.$

So we have

$\displaystyle \frac{1}{2}(1-i)\left(1+i^{p}\right) g_{p} = (1-i) \sum_{n=1}^{p-1} e^{\pi in^{2} / 2p}.$

Now consider the real part of the sum $(1-i)\sum_{n=1}^{p-1} e^{\pi in^{2} / 2p}$

which equals

$\displaystyle \sum_{n=1}^{p-1} \left(\cos \frac{\pi n^{2}}{2 p}+\sin \frac{\pi n^{2}}{2 p}\right).$

First the contribution of the terms ${n=1,2, \cdots [\sqrt p]}$ is at least ${[\sqrt p]}$ since each of the term is

$\displaystyle \left(\cos \frac{\pi n^{2}}{2 p}+\sin \frac{\pi n^{2}}{2 p}\right) \ge 1, \quad \text{for } n \le \sqrt p.$

What about the rest of the terms?

We have

$\displaystyle \left(a_{n}-a_{n-1}\right) b_{n}=2 i e^{\pi i n ^{2} /(2 p)},$

where

$\displaystyle a_{n}=e^{\pi i n(n+1) /(2 p)} \text{ and } b_{n}=\csc \left(\frac{\pi n }{2 p}\right).$

Therefore the sum

$\displaystyle 2i\sum_{[\sqrt p]+1}^{p-1} e^{\pi i n ^{2} /(2 p)} =\sum_{[\sqrt p]+1}^{p-1} (a_n-a_{n-1})b_n$
$\displaystyle =\sum_{[\sqrt p]+1}^{p-1} (b_n-b_{n-1})a_n - a_{[\sqrt p]}b_{[\sqrt p]+ 1} +a_{p-1}b_p$

And we have the bound

$\displaystyle \left| 2 \sum_{[\sqrt p]+1}^{p-1} e^{\pi i n ^{2} /(2 p)} \right| \le \sum_{[\sqrt p]+1}^{p-1} (b_n-b_{n-1}) +b_{[\sqrt p]+ 1} + b_p =2b_{[\sqrt p]+ 1} \le 2 \sqrt p$