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

By Sergei Konyagin, Igor Shparlinski

The topic of this booklet is the learn of the distribution of integer powers modulo a major quantity. It presents quite a few new, occasionally relatively unforeseen, hyperlinks among quantity idea and laptop technological know-how in addition to to different parts of arithmetic. attainable purposes contain (but will not be restricted to) complexity conception, random quantity new release, cryptography, and coding thought. the most procedure mentioned relies on bounds of exponential sums. therefore, the e-book includes many estimates of such sums, together with new estimates of classical Gaussian sums. It additionally comprises many open questions and recommendations for additional study.

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5) x=1 where c(t) > 0 depends only on t. 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.

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). Taking into account that ordl Res(lT, H ) ≥ m, we obtain the statement. We believe that the restriction that the zeros of G are simple is redundant. 4. 3 for two polynomials G(X ), F(X ) ∈ Z[X ] having m common zeros (possibly multiple) modulo l. 2 can be explicitly evaluated (see the comment at the end of [77]). The implied constant in the following theorem may depend on k. 5 For each integer t and prime l ≡ 1 (mod t), we fix some element gt,l of multiplicative order t modulo l.

Proof Because n ≥ 3, we have n ≥ ord p n + 2. Indeed, it is easy to verify that, for x ≥ 4, x ≥ log x + 2 ≥ ord p x + 2 and also that 3 ≥ ord p 3 + 2. Therefore m ≥ n + 1 ≥ ord p n + 3 ≥ γ p (n). 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}.

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