Quadratic Gauss sums were first formulated and studied by Carl Friedrich Gauss himself, motivated partly by his desire to prove the law of quadratic reciprocity in different ways. They are sums of the form

where

They are called *quadratic* Gauss sums because under certain circumstances they can be expressed in terms of Legendre symbols, which are quadratic characters modulo a prime number. I will explore some properties of Gauss sums associated with Dirichlet characters in a later note. In this short note I want to record a proof I have worked out of a multiplicative property of quadratic Gauss sums which says that

(1)

when . A standard way to prove this in the literature is to begin by proving a more general multiplicative property which says

when and . One then obtains the previous result by setting . I found it instructive to go through the process of proving (1) directly, however, without going through the general case first, and I want to record the steps involved in doing this here. I’ll also give an illustrative example of (1) at the end.

It is necessary to prove two results first, and then these can be used to prove (1).

**Result 1**: We can write where the summation is taken over any complete set of residues mod , because whenever (mod ).

** Proof**: If (mod ) then for some integer , so

since is an th root of unity.

Any set of integers is a complete set of residues modulo provided no two of them are congruent modulo . Each integer in a complete set of residues modulo will be congruent to one, and only one, integer in the set . By the result obtained above, therefore, it makes no difference whether we sum over the integers , or over the integers in any complete set of residues modulo . Therefore

where the summation extends over any complete set of residues, as claimed. QED

**Result 2**: Let and be integers such that and let and run through complete sets of residues mod and respectively. Then runs through a complete set of residues mod and

(mod )

** Proof**: Let

be a complete set of residues mod , and let

be a complete set of residues mod . Let

where , . Then the set

contains elements corresponding to the possible different ways of combining an element from with an element from . The claim to be proved is then that the set is a complete set of residues mod .

To see this, suppose and are two distinct elements from , and suppose they are congruent mod . I will show that this leads to a contradiction. We have

(mod )

(mod )

(mod )

(mod )

and (mod )

(since ). These last two congruences imply

(mod ) and (mod )

which cannot both be true since and are distinct elements of and so at least one of the pairs and must consist of distinct elements from the corresponding complete set of residues mod or respectively. This is a contradiction, so it follows that any two distinct elements of must be incongruent, so is a complete set of residues mod .

We have

(mod )

as claimed. QED

We can now use these two results to prove the multiplicative property in (1) above. By Result 1, we can sum over any complete sets of residues. Therefore using the sets , and as defined in Result 2, we have

Therefore

where the penultimate equality follows from the fact that

(mod ) as is clear from the expression for in Result 2. QED

**Example**: Evaluating each term of the equation when and we get

and so . END