Estermann’s evaluation of Gauss Sum

Let p be an odd prime, and consider the quadratic Gauss sum

\displaystyle g_p:=\sum_{n=0}^{p-1}\exp\left(\frac{2\pi i n^2}{p}\right).

This sum admits a second description in terms of the Legendre symbol. Each nonzero quadratic residue modulo p occurs exactly twice among the numbers n^2 , while each nonresidue occurs zero times. Consequently,

\displaystyle g_p=\sum_{m=0}^{p-1}\left(\frac{m}{p}\right)\exp\left(\frac{2\pi i m}{p}\right).

Using the character form we first compute its absolute value:

\displaystyle |g_p|^2=\sum_{x,y\bmod p}\left(\frac{xy}{p}\right)\exp\left(\frac{2\pi i(x-y)}{p}\right).

The terms with y=0 vanish. For y\ne0 , write x=ty . Then

\displaystyle |g_p|^2=\sum_{t\bmod p}\left(\frac{t}{p}\right)\sum_{y=1}^{p-1}\exp\left(\frac{2\pi i(t-1)y}{p}\right).

The inner sum is p-1 when t=1 , and -1 otherwise. Since \displaystyle \sum_{t\bmod p}\left(\frac{t}{p}\right)=0, we obtain \displaystyle |g_p|^2=p. Next we have

\\displaystyle \overline{g_p}=\sum_{x\bmod p}\left(\frac{-x}{p}\right)\exp\left(\frac{2\pi i x}{p}\right)=\left(\frac{-1}{p}\right)g_p.

Therefore \displaystyle p=g_p\overline{g_p}=\left(\frac{-1}{p}\right)g_p^2, and hence

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

This already determines the magnitude of g_p

\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)

Thus the only remaining issue is the sign. The square calculation tells us that there are only two possibilities, but it does not distinguish between them.

Estermann’s idea is to rotate the Gauss sum onto the real axis and then prove that the resulting real number cannot be negative. Define

\displaystyle \eta_p:=\frac12(1-i)(1+i^p).

This factor is chosen so that

\displaystyle \eta_p=\begin{cases}1,&p\equiv1\pmod4,\\ -i,&p\equiv3\pmod4.\end{cases}

Therefore the normalized quantity \displaystyle H_p:=\eta_p g_p is real in both congruence classes. Since the square calculation already shows that g_p has magnitude \sqrt p , we know that \displaystyle H_p\in\{-\sqrt p,\sqrt p\}.

The problem has now become very concrete: it is enough to prove \displaystyle H_p>-\sqrt p. Any strict lower bound above -\sqrt p rules out the negative possibility and forces H_p=\sqrt p .

Using the symmetry of the terms indexed by n and p-n , we have

\displaystyle g_p=1+2\sum_{r=1}^{(p-1)/2}\exp\left(\frac{2\pi i r^2}{p}\right).

Therefore \displaystyle H_p=\eta_p+(1-i)(1+i^p)\sum_{r=1}^{(p-1)/2}\exp\left(\frac{2\pi i r^2}{p}\right).

The expression in the last line has a useful interpretation. The even integers n=2r between 1 and p-1 contribute

\displaystyle \exp\left(\frac{\pi i(2r)^2}{2p}\right)=\exp\left(\frac{2\pi i r^2}{p}\right).

The odd integers can be written as n=p-2r , and they contribute

\displaystyle \exp\left(\frac{\pi i(p-2r)^2}{2p}\right)=i^p\exp\left(\frac{2\pi i r^2}{p}\right).

Thus the two factors 1 and i^p combine into one sum:

\displaystyle H_p=\eta_p+(1-i)\sum_{n=1}^{p-1}\exp\left(\frac{\pi i n^2}{2p}\right).

This is the form that Estermann estimates. Taking real parts, and using the fact that H_p is real, gives

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

The constant term causes no difficulty: {\text{Re}}\eta_p is either 0 or 1 .

The strategy is now visible. For small n , the angle \frac{\pi n^2}{2p} lies in the first quadrant, so the corresponding terms have a definite positive contribution. For larger n , the phases oscillate; one does not try to estimate them term by term, but instead proves that their total contribution is bounded by cancellation.

Set \displaystyle M:=\lfloor\sqrt p\rfloor. When 1\le n\le M , we have \displaystyle 0\le\frac{\pi n^2}{2p}\le\frac{\pi}{2}. On this interval, \displaystyle \cos x+\sin x\ge1. Thus the initial portion of the sum contributes at least

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

This is the source of positivity in the proof: the first approximately \sqrt p terms all lie in a favorable quadrant and point, in aggregate, in a positive direction.

It remains to control the oscillatory tail

\displaystyle T_p:=\sum_{n=M+1}^{p-1}\exp\left(\frac{\pi i n^2}{2p}\right).

A naive bound by the number of terms would be far too large. Instead, Estermann uses a discrete integration-by-parts argument, which exploits the fact that the quadratic phase rotates more and more rapidly.

Define \displaystyle a_n:=\exp\left(\frac{\pi i n(n+1)}{2p}\right),\quad b_n:=\csc\left(\frac{\pi n}{2p}\right).

The key elementary identity is \displaystyle (a_n-a_{n-1})b_n=2i\exp\left(\frac{\pi i n^2}{2p}\right).

Indeed, the difference a_n-a_{n-1} contains the factor

\displaystyle \exp\left(\frac{\pi i n}{2p}\right)-\exp\left(-\frac{\pi i n}{2p}\right)=2i\sin\left(\frac{\pi n}{2p}\right),

and b_n was chosen precisely to cancel this sine factor.

Therefore \displaystyle 2iT_p=\sum_{n=M+1}^{p-1}(a_n-a_{n-1})b_n. Applying summation by parts gives

\displaystyle 2iT_p=a_{p-1}b_{p-1}-a_Mb_{M+1}+\sum_{n=M+1}^{p-2}a_n(b_n-b_{n+1}).

This formula has converted the rapidly oscillating sum into a telescoping expression. The complex numbers a_n all have modulus 1 , while the positive sequence b_n is decreasing for 1\le n\le p . It follows that

\displaystyle 2|T_p|\le b_{p-1}+b_{M+1}+\sum_{n=M+1}^{p-2}(b_n-b_{n+1})=2b_{M+1}.

Hence \displaystyle |T_p|\le\csc\left(\frac{\pi(M+1)}{2p}\right).

Now M+1>\sqrt p . Using the elementary bound \sin x\ge2x/\pi on 0\le x\le\pi/2 , we get

\displaystyle \sin\left(\frac{\pi(M+1)}{2p}\right)\ge\frac{M+1}{p}>\frac{1}{\sqrt p}.

Therefore \displaystyle |T_p|<\sqrt p. This is the crucial cancellation estimate. The tail contains almost p terms, but its total size is only of order \sqrt p . That is exactly the scale needed for the positive initial segment to control the sign.

Combining the initial contribution with the tail estimate, we find

\displaystyle H_p\ge M+{\text{Re}}\bigl((1-i)T_p\bigr).

Since |1-i|=\sqrt2 , \displaystyle {\text{Re}}\bigl((1-i)T_p\bigr)\ge-\sqrt2|T_p|>-\sqrt{2p}. Thus \displaystyle H_p>M-\sqrt{2p}.

Finally, for every odd prime p , \displaystyle \lfloor\sqrt p\rfloor>(\sqrt2-1)\sqrt p. Consequently, \displaystyle M-\sqrt{2p}>-\sqrt p.

We have proved \displaystyle H_p>-\sqrt p. But H_p can only be one of the two values -\sqrt p and \sqrt p . The negative value is therefore impossible, and so \displaystyle H_p=\sqrt p.

Undoing the normalization now gives the classical evaluation

\displaystyle g_p=\begin{cases}\sqrt p,&p\equiv1\pmod4,\\ i\sqrt p,&p\equiv3\pmod4.\end{cases}

The appeal of Estermann’s argument is that it turns a delicate phase question into a robust geometric estimate. The squaring argument reduces the answer to two possible signs. The first \sqrt p summands supply a definite positive bias because their phases lie in the first quadrant. The remaining summands may oscillate in every direction, but summation by parts shows that they cannot collectively cancel enough to overturn that bias. The sign is thus forced by the geometry of the quadratic phases.

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