I encounter a tricky sum like the Kloosterman sum

$$\sum_{x \mod qP} e ( \frac{x+\overline{x+P}}{qP} ),$$ where $q$ is a positive integer, $P$ is a prime number satisfying $(q,P)=1$, $x \bmod qP,(x,qP)=1,$ $ \overline{x} $ means $ x\overline{x} \equiv 1\pmod {qP}$.

If this sum can be bounded by $O((qP)^{1/2+\varepsilon}$) by invoking Weil bound for Kloosterman sums? Did anyone ever saw this kind of sum before?Please share some comments. Many thanks.

My confusions are as follows: (a) One may try to split the sum into two sums with the modulos being $q$ and $P$. However we may have the issue that $y+1$ is not co-prime with $q$ when writing $qP=xq+yP$ with $x \bmod P,(x,P)=1,$ and $y \bmod q,(y,q)=1.$ So that it seems that one cannot write $\overline{xq+(y+1)P}\mod {qP}$ as the form $x\cdot A+y\cdot B$ for some integers $A,B$.

(b) For typical modulos, for example, to consider the sum $$\sum_{x \mod c} e ( \frac{x+\overline{x+P}}{c} ),$$ where $c$ is an arbitrary positive integer. If we have the square-root cancellation for this type of sum, just like the Wiel bound?

Your any opinions are highly appreciated. Thanks in advance.

  • 1
    $\begingroup$ in the case $y+1$ not coprime with $q$, the problem is not that we cannot right the inverse in that form - the problem is that the inverse does not exist! What do you want that term in the sum to be then? Zero? $\endgroup$ – Will Sawin Oct 17 '16 at 9:33
  • $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$ – GH from MO Sep 2 '18 at 7:57

As Will Sawin pointed out, it is not clear what you mean by $\overline{x+P}$ when $x+P$ is not coprime to $qP$ (i.e. when $x+P$ is not coprime to $q$). If you omit these terms and you assume that $q$ is square-free, then you do get a Weil-like bound by Proposition 4.6 in Polymath's paper New equidistribution estimates of Zhang type, Algebra & Number Theory 8 (2014), 2067-2199. This is also Proposition 4.6 in the latest arXiv version.

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  • $\begingroup$ @Will Sawin,GH from MO $\endgroup$ – Fei Oct 18 '16 at 11:50
  • $\begingroup$ @Fei: You addressed something to Will and to me in the previous comment, but the message is empty. If you just wanted to say thanks, then please say thanks for clarity. $\endgroup$ – GH from MO Oct 18 '16 at 11:55
  • $\begingroup$ @Will Sawin,GH from MO Great thanks to Will Sawin for pointing out the blur. Notice $\overline{x+P}$, one does not have the case that $y+1$ not coprime with $q$. Also great thanks to GH from MO for guiding the reference. Write $x=x_1q+ x_2 P$ with $x_1 \bmod P, (x_1,P)=1$ and $x_2\bmod q, (x_2,q)=1$. We have [\overline{(x+P)} \equiv \overline{ (P+x_1q)q}q + \overline{ (P+x_2 P)P }P \pmod {qP}$$. Hence by the Chinese Remainder Theorem one may write the sum we concerned about as a product of two sums of the modulos being $q$ and $P$ respectively. $\endgroup$ – Fei Oct 18 '16 at 12:05
  • $\begingroup$ Sorry, I just editting the messege but worongly press the Enter . $\endgroup$ – Fei Oct 18 '16 at 12:09

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