Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $\chi$ is a real non-trivial primitive character modulo $q\ge 1$, then how could one show that $$\sum_{n\in \mathbb Z/q\mathbb Z} \chi(n)e\left(\frac nq\right) = \sum_{n\in \mathbb Z/q\mathbb Z} e\left(\frac {n^2}q\right) $$ where $e(z) = \exp(2\pi i z)$? I think one should be able to see this directly, but I simply cannot figure out anything useful. If $n$ is a square mod $q$, then of course $\chi(n) = 1$, so maybe that would be a good place to start... But I've been stuck for quite some time on this.

Some help would be greatly appreciated. Thank you.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

You need to assume that $q$ is prime (you can check for instance that neither of the primitive characters modulo $8$ have this property). When $\chi$ is the quadratic character modulo a prime $q\ge3$, my hint is that you first verify the identity $\chi(n)=\#\{y \pmod q\colon y^2\equiv n\pmod q\}-1$, and then plug this into the left-hand side and see how it equals the right-hand side. (It's slightly easier to see if you make the index of summation on the right-hand side $y$ instead of $n$....)

share|improve this answer
    
Thank you very much for your answer! –  Sam Nov 27 '12 at 6:36

Your Answer

 
discard

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.