By Votja P. A.

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**Example text**

Possibilities if 0 ≤ z 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 ].

If t is odd, then the equation ϑi ξ zi + ϑ j ξ z j = 0 is possible only if ϑi = −ϑ j and z i = z j . Therefore, the second half (z k+1 , . . , z 2k ) of each solution should be a permutation of the first half (z 1 , . . , z k ). Thus if we fix the first half, then for the second half we have only O(1) possibilities and precisely k! possibilities if 0 ≤ z 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.

We remark that there are close relations between bounds of such short sums and bounds of the number M1 (m, p) which we define and study in Chapter 13. 5) and also show that this bound is sharp. 5 Bounds of Character Sums for Almost All Moduli So far we have considered estimates which hold for each q from some wide class of parameters. Below, we obtain a quite different result where we estimate character sums for almost all moduli (rather than for all) but instead we obtain a much stronger bound.