# Character Sums with Exponential Functions and their by Sergei Konyagin, Igor Shparlinski

By Sergei Konyagin, Igor Shparlinski

The subject matter of this e-book is the examine of the distribution of integer powers modulo a major quantity. It offers a variety of new, occasionally really unforeseen, hyperlinks among quantity idea and laptop technological know-how in addition to to different parts of arithmetic. attainable purposes contain (but aren't restricted to) complexity conception, random quantity new release, cryptography, and coding idea. the most process mentioned relies on bounds of exponential sums. as a result, the booklet includes many estimates of such sums, together with new estimates of classical Gaussian sums. It additionally includes many open questions and suggestions for additional examine.

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**Extra resources for Character Sums with Exponential Functions and their Applications **

**Example text**

Q1 } and x2 ∈ {1, . . , q2 } such that x ≡ x1 q2 + x2 q1 (mod q). Therefore, q2 q1 Sn (a, q) = e(a(x1 q2 + x2 q1 )n /q). x1 =1 x2 =1 Note that (x1 q2 + x2 q1 )n ≡ (x1 q2 )n + (x2 q1 )n (mod q). 1) where a1 = aq2n−1 , a2 = aq1n−1 . 1) immediately gives G n (q) ≤ G n (q1 ) G n (q2 ). The inverse inequality G n (q) ≥ G n (q1 ) G n (q2 ) follows from the observation that for any (a1 , q1 ) = 1 and (a2 , q2 ) = 1, there exists a such that a1 ≡ aq2n−1 (mod q1 ) and a2 ≡ aq1n−1 (mod q2 ). 5 below, we need some auxiliary assertions.

For some related estimates of exponential sums, see [45, 52, 84]. 15) is quite strong. In particular, it is non-trivial for |Vq| > Nm(q)1/2 . 153 . . 7 below, which is a generalization of the recent result of [72] concerning the case K = Q). For sets of almost all integer ideals, one obtains an even weaker result. 561 . . , where γ is the Euler constant, the bound |Vq| ≥ Nm(q)τr/(r +1) 3 Bounds of Long Character Sums 23 is stated for almost all integer ideals q. 15) is applicable to groups with at least nine generators.

5) hold. 1), we derive 2 n 4h 2 = N j,t (h) j=1 n ≤n N j,t (h)2 = n Nt (h). 2 For any ε > 0 and sufficiently large p for any non-negative h 1 , h 2 with h 1 h 2 ≤ p, the bound Nt (h 1 h 2 ) ≥ Nt (h 1 ) Nt (h 2 ) p −ε holds. Proof We define the function ψ(k) = 4 max τ 2 (m). m≤k 7 Multiplicative Translations of Subgroups of F∗p 53 Then ψ(k) ≤ k ε for sufficiently large k. We assume that h 1 > 0, h 2 > 0 because otherwise the inequality is trivial. Any solution (u 1 , x1 , y1 ) of the congruence u 1 x1 ≡ y1 (mod p) 0 < |x1 |, |y1 | ≤ h 1 , u 1 ∈ V and any solution (u 2 , x2 , y2 ) of the congruence u 2 x2 ≡ y2 (mod p) 0 < |x2 |, |y2 | ≤ h 2 , u 2 ∈ V defines the solution (u = u 1 u 2 , x = x1 x2 , y = y1 y2 ) of the congruence ux ≡ y (mod p) 0 < |x|, |y| ≤ h 1 h 2 , u ∈ V.