Hey guys, I'm concerned with bounding the following sum of gauss sums from above $$\sum_{p\leq x}~{\frac{1}{(p-1)^2}}\sum_{m=1}^{p-1}~\sum_{\chi~(p)}~\sum_{a=1}^{p-1}{~\chi^m(a)e\left(\frac{a}{p}\right)},$$ where $p$ runs through the primes $\leq x$, $\chi$ runs through the multiplicative characters modulo $p$ and $e\left(\frac{a}{p}\right)=\exp\left(\frac{2\pi ia}{p}\right)$. By using orthogonality relations of characters one gets $$\sum_{m=1}^{p-1}~\sum_{\chi~(p)}~\sum_{a=1}^{p-1}{~\chi^m(a)e\left(\frac{a}{p}\right)}=(p-1)\sum_{a=1}^{p-1}~{e\left(\frac{a}{p}\right)\frac{p-1}{ord_pa}},$$ where $ord_pa$ denotes the multiplicative order of $a$ modulo $p$. The right side can be bounded trivially by $$(p-1)\sum_{a=1}^{p-1}~{\frac{p-1}{ord_pa}}=(p-1)^2\sum_{d\mid p-1}{\frac{\varphi(d)}{d}},$$ $\varphi(d)$ denoting Euler's totient function. Using $\varphi(n)\leq n$ one gets the estimate $$\left|\sum_{p\leq x}~{\frac{1}{(p-1)^2}}\sum_{m=1}^{p-1}~\sum_{\chi~(p)}~\sum_{a=1}^{p-1}{~\chi^m(a)e\left(\frac{a}{p}\right)}\right|\leq\sum_{p\leq x}{\tau(p-1)},$$ where $\tau(n)$ is the number of divisors of $n$. The latter sum can be shown to be asymptotically equivalent to a positive constant times $x$. I would like to know if there is a way to show that the sum is $o(x)$.


2 Answers 2


There should be a bunch of cancellation. Here is an idea. You need to relate your sums $\sum_a e(a/p)/ord_p a$ to the sums $\sum_{ord_p a | m} e(a/p)/m$. Now, if $mr = p-1$,

$\sum_{ord_p a | m} e(a/p) = (1/r)\sum_{n=1}^{p-1} e(n^r/p) = O(p^{1/2})$

by the Weil bound. This will deal with the elements of large order, I believe. There is work to do, but this should get you going.


This is a comment rather than an answer, but it is too long. Let $g$ be some generator of the multiplicative group. Then


Rearranging yields $$\sum_{d|p-1} \phi(d) \sum_{k\leq\frac{p-1}{d}}e\left(\frac{g^{dk}}{p}\right) $$ so that the entire sum is $$\sum_{p\leq x}\frac{1}{p-1}\sum_{d|p-1}\phi(d)\sum_{k\leq\frac{p-1}{d}}e\left(\frac{g^{dk}}{p}\right).$$

My hope in posting this is that there are existing bounds on sums of the form $\sum_{k\leq\frac{p-1}{d}}e\left(\frac{g^{dk}}{p}\right)$. It might be strange to deal with, as it is a sum over elements chosen for their multiplicative properties. Essentially, we would need a theorem regarding how these multiplicative elements are distributed among the residue classes, and that it cannot be "too far from uniform".


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .