Innumber theory,quadratic Gauss sums are certain finite sums of roots of unity. A quadratic Gauss sum can be interpreted as a linear combination of the values of the complexexponential function with coefficients given by a quadratic character; for a general character, one obtains a more generalGauss sum. These objects are named afterCarl Friedrich Gauss, who studied them extensively and applied them toquadratic,cubic, andbiquadratic reciprocity laws.

For an oddprime numberp and an integera, thequadratic Gauss sumg(a;p) is defined as

${\displaystyle g(a;p)=\sum _{n=0}^{p-1}{\zeta }_p^{a{n}^2},}$

where${\displaystyle {\zeta }_p}$ is aprimitivepth root of unity, for example${\displaystyle {\zeta }_p=\exp (2\pi i/p)}$.Equivalently,

${\displaystyle g(a;p)=\sum _{n=0}^{p-1}(1+\left(\frac{n}{p}\right)){\zeta }_p^{an}.}$

Fora divisible byp, and we have${\displaystyle {\zeta }_p^{a{n}^2}=1}$ and thus

${\displaystyle g(a;p)=p.}$

Fora not divisible byp, we have${\displaystyle \sum _{n=0}^{p-1}{\zeta }_p^{an}=0}$, implying that

${\displaystyle g(a;p)=\sum _{n=0}^{p-1}\left(\frac{n}{p}\right){\zeta }_p^{an}=G(a,\left(\frac{\cdot }{p}\right)),}$

where

${\displaystyle G(a,\chi )=\sum _{n=0}^{p-1}\chi (n){\zeta }_p^{an}}$

is theGauss sum defined for any characterχ modulop.

• The value of the Gauss sum is analgebraic integer in thepthcyclotomic field${\displaystyle \mathbb{Q}({\zeta }_p)}$.
• The evaluation of the Gauss sum for an integera not divisible by a primep > 2 can be reduced to the casea = 1:

${\displaystyle g(a;p)=\left(\frac{a}{p}\right)g(1;p).}$

• The exact value of the Gauss sum fora = 1 is given by the formula:^[1]

Remark

In fact, the identity

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

was easy to prove and led to one of Gauss'sproofs of quadratic reciprocity. However, the determination of thesign of the Gauss sum turned out to be considerably more difficult: Gauss could only establish it after several years' work. Later,Dirichlet,Kronecker,Schur and other mathematicians found different proofs.

Leta,b,c benatural numbers. Thegeneralized quadratic Gauss sumG(a,b,c) is defined by

${\displaystyle G(a,b,c)=\sum _{n=0}^{c-1}{e}^{2\pi i\frac{a{n}^2+bn}{c}}}$.

The classical quadratic Gauss sum is the sumg(a,p) =G(a, 0,p).

Properties

• The Gauss sumG(a,b,c) depends only on theresidue class ofa andb moduloc.
• Gauss sums aremultiplicative, i.e. given natural numbersa,b,c,d withgcd(c,d) = 1 one has

${\displaystyle G(a,b,cd)=G(ac,b,d)G(ad,b,c).}$

This is a direct consequence of theChinese remainder theorem.

• One hasG(a,b,c) = 0 ifgcd(a,c) > 1 except ifgcd(a,c) dividesb in which case one has

${\displaystyle G(a,b,c)=gcd(a,c)\cdot G\left(\frac{a}{gcd(a,c)},\frac{b}{gcd(a,c)},\frac{c}{gcd(a,c)}\right)}$.

Thus in the evaluation of quadratic Gauss sums one may always assumegcd(a,c) = 1.

• Leta,b,c be integers withac ≠ 0 andac +b even. One has the following analogue of thequadratic reciprocity law for (even more general) Gauss sums^[2]

${\displaystyle \sum _{n=0}^{|c|-1}{e}^{\pi i\frac{a{n}^2+bn}{c}}={\left|\frac{c}{a}\right|}^{\frac{1}{2}}{e}^{\pi i\frac{|ac|-b^2}{4ac}}\sum _{n=0}^{|a|-1}{e}^{-\pi i\frac{c{n}^2+bn}{a}}}$.

• Define

for every odd integerm. The values of Gauss sums withb = 0 andgcd(a,c) = 1 are explicitly given by

Here(⁠a/c⁠) is theJacobi symbol. This is the famous formula ofCarl Friedrich Gauss.

• Forb > 0 the Gauss sums can easily be computed bycompleting the square in most cases. This fails however in some cases (for example,c even andb odd), which can be computed relatively easy by other means. For example, ifc is odd andgcd(a,c) = 1 one has

${\displaystyle G(a,b,c)={\epsilon }_c\sqrt{c}\cdot \left(\frac{a}{c}\right)e^{-2\pi i\frac{\psi (a)b^2}{c}},}$

whereψ(a) is some number with4ψ(a)a ≡ 1 (modc). As another example, if 4 dividesc andb is odd and as alwaysgcd(a,c) = 1 thenG(a,b,c) = 0. This can, for example, be proved as follows: because of the multiplicative property of Gauss sums we only have to show thatG(a,b, 2^m) = 0 ifn > 1 anda,b are odd withgcd(a,c) = 1. Ifb is odd thenan² +bn is even for all0 ≤n <c − 1. For everyq, the equationan² +bn +q = 0 has at most two solutions in${\displaystyle \mathbb{Z}}$/2ⁿ${\displaystyle \mathbb{Z}}$. Indeed, if${\displaystyle n_1}$ and${\displaystyle n_2}$ are two solutions of same parity, then${\displaystyle (n_1-n_2)(a(n_1+n_2)+b)=\alpha 2^m}$ for some integer${\displaystyle \alpha }$, but${\displaystyle (a(j_1+j_2)+b)}$ is odd, hence${\displaystyle j_1\equiv j_2(\mathrm{mod}{2}^m)}$.^{[clarification needed]} Because of a counting argumentan² +bn runs through all even residue classes moduloc exactly two times. Thegeometric sum formula then shows thatG(a,b, 2^m) = 0.

• Ifc is an oddsquare-free integer andgcd(a,c) = 1, then

${\displaystyle G(a,0,c)=\sum _{n=0}^{c-1}\left(\frac{n}{c}\right)e^{\frac{2\pi ian}{c}}.}$

Ifc is not squarefree then the right side vanishes while the left side does not. Often the right sum is also called a quadratic Gauss sum.

• Another useful formula

${\displaystyle G\left(n,p^k\right)=p\cdot G\left(n,p^{k-2}\right)}$

holds fork ≥ 2 and an odd prime numberp, and fork ≥ 4 andp = 2.

• Gauss sum
• Gaussian period
• Kummer sum
• Landsberg–Schaar relation

• Ireland; Rosen (1990).A Classical Introduction to Modern Number Theory. Springer-Verlag.ISBN 0-387-97329-X.
• Berndt, Bruce C.; Evans, Ronald J.; Williams, Kenneth S. (1998).Gauss and Jacobi Sums. Wiley and Sons.ISBN 0-471-12807-4.
• Iwaniec, Henryk; Kowalski, Emmanuel (2004).Analytic number theory. American Mathematical Society.ISBN 0-8218-3633-1.

• Cyclotomic fields

• Wikipedia articles needing clarification from July 2022