Number Theory

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

Posted On March 23, 2017 at 12:41 pm by / Comments Off on Character Sums with Exponential Functions and their by Sergei Konyagin, Igor Shparlinski

By Sergei Konyagin, Igor Shparlinski

The subject of this publication is the learn of the distribution of integer powers modulo a major quantity. It presents a number of new, occasionally rather unforeseen, hyperlinks among quantity thought and computing device technological know-how in addition to to different parts of arithmetic. attainable functions contain (but should not restricted to) complexity idea, random quantity iteration, cryptography, and coding concept. the most strategy mentioned is predicated on bounds of exponential sums. for this reason, the e-book comprises many estimates of such sums, together with new estimates of classical Gaussian sums. It additionally comprises 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

P (n)>m≥1 n≥m≥1 Therefore A(n) = {G n ( p m )/ p m(1−1/n) , 1}, max p∈P γ p (n)>m≥1 where P denotes the set of prime numbers. We estimate the part of the product which is taken over primes p | n by using the following trivial inequalities: max p∈P p|n {G n ( p m )/ p m(1−1/n) , 1} γ p (n)>m≥1 ≤ max p∈P p|n { p m/n , 1} γ p (n)>m≥1 p (γ p (n)−1)/n ≤ p∈P p|n = p 2/n ≤ n 3/n , n 1/n p∈P p|n thus max{G n ( p)/ p 1−1/n , 1}. A(n) ≤ n 3/n p∈P gcd( p,n)=1 Now we are going to show that in fact we may consider only a finite product.

5) p∈P, p≤Q n (d) gcd (n, p−1)=d holds. 20) implies that Q n (d) d 3/2 . 20)) can be easily evaluated explicitly. 7 The asymptotic formula A(n) = 1 + O(n −1 τ (n) ln n) holds. Proof Let p ≤ Q n (d) be a prime with gcd( p − 1, n) = d ≥ 2. We put t = ( p − 1)/d and show that if ϕ(t) ≥ 6, then σd ( p) ≤ p 1−1/n provided that n is large enough. 6) implies that d p 2/3 , thus t = O( p 1/3 ). 4), taken with r = 6, we obtain σd ( p) ≤ 1 + d(t − Ct 1/6 p −1/3 ) = p − Cdt 1/6 p −1/3 = p − C( p − 1)t −5/6 p −1/3 ≤ p − cp 7/18 for some absolute constants C, c > 0.

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. Therefore, any partition of the set {1, . . , 2k} on pairs is available. The total number of partitions is (2k − 1)!!.

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