Factoring out Kloosterman Sum

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

$$S(n,n;c)=S(n\bar{q},n\bar{q};r)T(n\bar{r},n\bar{r};q)$$

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

$$S(n,n;c)=S(n\bar{q},n\bar{q};r)S(n\bar{r},n\bar{r};q)$$

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 $$(d,q)=1, d and I don't quite see why this is true. Does it have something to do with $$n$$ being square free or is this generally true?

• Where does he say this? – Peter Humphries Feb 22 at 11:41
• Page 78 of Topics in classical automorphic forms while he is estimating Fourier coefficients of cusp forms. – J.Shim Feb 22 at 18:38
• 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. – Peter Humphries Feb 23 at 13:45