I'm reading Iwaniec's book and he says Kloosterman sum factors into


where $n$ is square free and $c=rq$ such that $(q,n)=(q,r)=1$(i.e. $q$ is the largest factor of $c$ coprime to $n$) and $T$ is the Salie sum. I know that


since $(q,r)=1, c=qr$ but then his claim means $S(n\bar{r},n\bar{r};q)=T(n\bar{r},n\bar{r};q)$ which means $\displaystyle\left(\frac{d}{q}\right)=1$ for all $(d,q)=1, d<q$ and I don't quite see why this is true. Does it have something to do with $n$ being square free or is this generally true?

  • $\begingroup$ Where does he say this? $\endgroup$ – Peter Humphries Feb 22 at 11:41
  • $\begingroup$ Page 78 of Topics in classical automorphic forms while he is estimating Fourier coefficients of cusp forms. $\endgroup$ – J.Shim Feb 22 at 18:38
  • $\begingroup$ Be careful. The Kloosterman sum $S(m,n;c)$ is not the usual Kloosterman sum $\sum_{d \in (\mathbb{Z}/c\mathbb{Z})^{\times}} e\left(\frac{md + n\overline{d}}{c}\right)$; there is a $\vartheta$-multiplier inserted. $\endgroup$ – Peter Humphries Feb 23 at 13:45

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