Download e-book for iPad: Character sums with exponential functions and their by Sergei Konyagin, Igor Shparlinski

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By Sergei Konyagin, Igor Shparlinski

ISBN-10: 0511040369

ISBN-13: 9780511040368

ISBN-10: 0521642639

ISBN-13: 9780521642637

The subject matter of this e-book is the research of the distribution of integer powers modulo a first-rate quantity. It offers various new, occasionally really unforeseen, hyperlinks among quantity idea and computing device technology in addition to to different parts of arithmetic. attainable purposes comprise (but will not be constrained to) complexity conception, random quantity iteration, cryptography, and coding idea. the most strategy mentioned relies on bounds of exponential sums. therefore, the booklet comprises many estimates of such sums, together with new estimates of classical Gaussian sums. It additionally includes many open questions and recommendations for additional learn.

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3, we obtain p m−1 Sn (a, p ) = p m−1 e(a( py) / p ) = m n e(ay n / p m−n ) = p n−1 Sn (a, p m−n ), m y=1 y=1 and we are done. 1 shows that it is enough to consider Gaussian sums with a prime power denominator, that is A(n) = sup{G n ( p k )/ p k(1−1/n) , 1}. 5 shows that sup{G n ( p k )/ p k(1−1/n) , 1} = max {G n ( p m )/ p m(1−1/n) , 1}. 4 shows that max {G n ( p m )/ p m(1−1/n) , 1} = max {G n ( p m )/ p m(1−1/n) , 1}. γ 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.

By Theorem 3 of [39], α can be lifted to l-adic roots of G(X ). Therefore over the ring of integer l-adic integers Zl we have G = H Q, where H (X ), Q(X ) ∈ Zl [X ], and H is the monic polynomial of degree m whose roots are l-adic liftings of those roots of G. We write F = H P + R where P(X ), R(X ) ∈ Zl [X ] and deg R < m. From the condition of the lemma, we see that R has m roots modulo l, so it is identical to zero modulo l. Therefore, R(X ) = lT (X ) where T (X ) ∈ Zl [X ]. Now we have Res(F, G) = Res(H P + lT, H Q) = Res(H P + lT, H ) Res(H P + lT, Q) = Res(lT, H ) Res(H P + lT, Q).

4) we have S = 0 again. p−1 j=0 e(an 0 j x n 0 −1 / p) = 0 6 Bounds of Gaussian Sums 41 Now we are prepared to prove the following two assertions giving a recurrent formula and initial values for Gaussian sums. 4. Let p be a prime. We put 2, 3 + ord p n, γ p (n) = if ord p n = 0; if ord p n > 0. Then, for any integer m with γ p (n) ≤ m ≤ n, G n ( p m ) = p m−1 . 3, p m−1 Sn (a, p ) m = e(ax / p ) = n 1≤x≤ p m (x, p)>1 e(a( py)n / p m ) m y=1 p m−1 e(ay n p n−m ) = p m−1 , = y=1 and we are done.

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Character sums with exponential functions and their applications by Sergei Konyagin, Igor Shparlinski


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