For $m,n\in \mathbb{N}$, denote the Kloosterman sum $$S(m,n;c)=\sum_{a\bmod c}e\left( \frac{ma+n \overline{a}}{c}\right),$$where $\overline{a}$ denotes the multiplicative inverse of $a\bmod c$.

Does any expert here know something upon the non-trivial power-saving bound for the mean square estimate of the Kloosterman sum, that is, for $m,n$ fixed, whether or not one has an estimate $$\sum_{c\le x}\frac{|S(m,n;c)|^2}{c}\ll_{m,n}\,\,x^{\theta}\tag{$\ast$}$$ for some $\theta<1$?

As far as we know, the non-trivial bound for the first moment is due to Kuznetzov who showed $$\sum_{c\le x}\frac{S(m,n;c)}{c}\ll_{m,n}\,\,x^{\frac{1}{6}}\,\log ^{\frac{1}{3}}x.$$ The Linnik’s conjecture asserts that $$\sum_{c\le x}\frac{S(m,n;c)}{c}\ll_{m,n}\,\,x^{\varepsilon}$$ for any $\varepsilon>0$. For detailed description on the first moment estimate, one may see Sarnak and Tsimerman's paper: https://www.researchgate.net/publication/225888754_On_Linnik_and_Selberg%27s_Conjecture_About_Sums_of_Kloosterman_Sums

Recently I encounter this tricky sum as in ( $\ast$), for which one need to get a saving, compared with the trivial bound $x^{1+\varepsilon}$. There does not seem to be a reference available in the literature. If any expert here have some strategies or references, please give a guide. Many thanks.

Your any opinions are highly appreciated. Thanks in advance.


You cannot have estimates like (*) for any $\theta<1$. Fouvry and Michel showed that (see Theorem 1.2 there) $$ \sum_{c\le x} |S(m,n;c)|/\sqrt{c} \gg_k \frac{x}{\log x} (\log \log x)^k, $$ for any natural number $k$. By Cauchy-Schwarz one also gets a lower bound for the second moment.

They also note that one can get an upper bound for this quantity, saving a power of log. I would guess that $$ \sum_{c\le x} |S(m,n;c)|^2/c $$ is of the order $x$ (assume $mn \neq 0$).

| cite | improve this answer | |
  • 1
    $\begingroup$ Dear Lucia, thanks for timely explanation, much obliged. $\endgroup$ – Fei Jul 2 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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