Friedlander–Iwaniec twisted incomplete Kloosterman-sum conjecture

From papers

Let a,q2a,q\geq2 be integers with (a,q)=1(a,q)=1 and qq not a perfect square, and let H,K1H,K\geq1 be reals. Define e(u):=e2πiue(u):=e^{2\pi i u}, let ()\left(\frac{\cdot}{\cdot}\right) denote the Jacobi symbol, and let hˉ\bar h denote the multiplicative inverse of hh modulo qq. Friedlander–Iwaniec twisted incomplete Kloosterman-sum conjecture. For any ϵ>0\epsilon>0,

1hH(h,q)=10k<K(hq)e(ahˉk2q)ϵ(H1/2K1/2+H3/4+K+q1/2HK+q1/2K2)qϵ.\begin{aligned} &\mathop{\sum_{1\leq h\leq H}}_{(h,q)=1}\sum_{0\leq k<K}\left(\frac{h}{q}\right)e\left(\frac{a\bar h k^2}{q}\right)\\ &\ll_\epsilon\left(H^{1/2}K^{1/2}+H^{3/4}+K+q^{-1/2}HK+q^{-1/2}K^2\right)q^\epsilon. \end{aligned}

The source invokes this estimate as an assumption behind stronger bounds for quadratic fractional parts; its resolution status is not supplied.

Progress summary

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Sources & referencesView supporting material

Primary source

Tsz Ho Chan, “Finding Almost Squares”, arXiv:math/0502199 (2005).

Solutions 0

No solutions have been posted yet.