By Larry Joel Goldstein

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Let t be the multiplicative order of g modulo p, gcd(g, p) = 1 and let P be integer. If t≥ ln p Ln ln1−ε p , P ≥ min{t, 6 ln p Ln ln p + 1} then P e(g x / p) ≤ P 1 − x=1 26 C(ε) ln1+ε p . 4 Bounds of Short Character Sums 27 We remark that we cannot obtain good upper estimates on S(p, V ) if |Vp| is too small. It is known from [35] that, for any ε > 0, there are infinitely many primes p and subgroups Vp of ∗p such that 2 ln p ln p ≤ Vp ≤ −1 ln ε ln ε−1 and S(p, V ) ≥ |Vp| 1 − 2π ε . 1−ε Finally, we note that, over finite fields of bounded characteristics, the method of [81, 82] allows us to obtain a non-trivial estimate for fairly short sums.

N. Thus we define H p (t) = max{H : ∃ j ∈ {1, . . , n}, ∃M ∈ Z such that M + z ∈ V j , z = 1, . . , H }. To study H p (t), we introduce several more functions. Namely, for j = 1, . . , n, we denote by N j,t (h) the number of u ∈ V j with 1 ≤ |u| ≤ h, that is N j,t (h) = u ∈ V j : 1 ≤ |u| ≤ h , and also put S j (t) = e(v/ p) v∈V j and extend this definition for all j ∈ Z periodically with period n. Relations between H p (t) and N j,t (h), S j (t), are given by the following statement. 1. 5t j=1 49 50 Multiplicative Translations of Sets is satisfied for all k = 1, .

Z k ≤ t − 1 are pairwise distinct. t k + O(t k−1 ) such solutions. 34 Bounds of Character Sums If t is even, then there is one more possibility: ϑi = ϑ j and z i = z j + t/2. Therefore, any partition of the set {1, . . , 2k} on pairs is available. The total number of partitions is (2k − 1)!!. t k + O(t k−1 ) solutions. 3. Let l be a prime number. Assume that m ≥ 1 simple roots of the congruence G(x) ≡ 0 (mod l) with a polynomial G(X ) ∈ Z[X ] are also roots of the congruence F(x) ≡ 0 (mod l) with another polynomial F(X ) ∈ Z[X ].