MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was wondering whether anyone knows how to approach the following two generalizations of the quadratic Gauss sum:

Given integers r,s with gcd(r,s)=1 and integers a,b,N

$F(r,s,N,a,b) = \sum_{w = 0}^{rsa}(-1)^{b w}(\sin\frac{\pi w}{s}) \exp(\pi i w^2\frac{N}{2 rs}) $

$G(r,s,N,a,b) = \sum_{w = 0}^{rsa}(-1)^{b w}(\sin\frac{\pi w}{r})(\sin\frac{\pi w}{s}) \exp(\pi i w^2\frac{N}{2 rs}) $

Note that removing the sine terms and the sign, setting a = 2, N = 4, r = 1 and s = prime gives the classical quadratic Gauss sum.

Some experimentation suggests that

$F(r,s,N,a,b) = 0$ for all integers b,N, r,s if a is even and (r,s) =1 and

$G(r,s,N,a,b) = 0$ for all a,b,N and r,s with (r,s) =1

Is there a good reason for these sums to vanish? Or a clean proof/reference?

Is it possible to evaluate F in the case a = 1? It seems to be non-zero then.

I tried reducing to the original Gauss sum by completing the square but this seems to get quite ugly.

More generally, do such Gauss-like sums have a more natural generalization that turns up somewhere?


share|cite|improve this question
Vinogradov did a lot with generalizations of gauss sums. Nothing like you have there. I read some physics papers where they dealt with sums like yours using finite fouler transform methods. – Charlie Frohman May 13 '10 at 3:17

In $G$, it looks like the term with $w=k$ cancels the term with $w=rsa-k$, at least under certain parity assumptions on the variables. Maybe it's worth having a look at that.

share|cite|improve this answer

I think this should just be computations. Define as usual $e(x) = \exp(2 \pi i x)$. Define $$ f(r,s,N,a,b) = \sum_{w=1}^{rsa} e\left(\frac{N}{4 rs} w^2 + \left(\frac{1}{2s}+\frac{b}{2} \right) w\right) $$ Then $F(a,r,s,N,a,b) = \frac{1}{2i } (f(r,s,N,a,b) - f(r,s,N,a, -b))$, at least if I didn't make any computational mistakes.

Now, start with $a = 1$. Then you can use, the method described in .

Next, one has to understand what happens if one passes from $a$ to $a + 1$. For this compute $$ f(r,s,N,a+1,b) - f(r,s,N,a,b) $$ I guess, the result should be of the form $z f(\hat r,\hat s, \hat N, 1, \hat b)$ with $|z| = 1$.

Note: I began summing at $1$ on purpose, so one sums over the group $\mathbb{Z}_{rsa}$, which seems like the correct choice ...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.