Do you want to publish a course? Click here

Powers of Gauss sums in quadratic fields

205   0   0.0 ( 0 )
 Added by Koji Momihara
 Publication date 2020
  fields
and research's language is English
 Authors Koji Momihara




Ask ChatGPT about the research

In the past two decades, many researchers have studied {it index $2$} Gauss sums, where the group generated by the characteristic $p$ of the underling finite field is of index $2$ in the unit group of ${mathbb Z}/m{mathbb Z}$ for the order $m$ of the multiplicative character involved. A complete solution to the problem of evaluating index $2$ Gauss sums was given by Yang and Xia~(2010). In particular, it is known that some nonzero integral powers of the Gauss sums in this case are in quadratic fields. On the other hand, Chowla~(1962), McEliece~(1974), Evans~(1977, 1981) and Aoki~(1997, 2004, 2012) studied {it pure} Gauss sums, some nonzero integral powers of which are in the field of rational numbers. In this paper, we study Gauss sums, some integral powers of which are in quadratic fields. This class of Gauss sums is a generalization of index $2$ Gauss sums and an extension of pure Gauss sums to quadratic fields.



rate research

Read More

In this note, we extend the definition of multiple harmonic sums and apply their stuffle relations to obtain explicit evaluations of the sums $R_n(p,t)=sum olimits_{m=0}^n m^p H_m^t$, where $H_m$ are harmonic numbers. When $tle 4$ these sums were first studied by Spiess around 1990 and, more recently, by Jin and Sun. Our key step first is to find an explicit formula of a special type of the extended multiple harmonic sums. This also enables us to provide a general structural result of the sums $R_n(p,t)$ for all $tge 0$.
Can any element in a sufficiently large finite field be represented as a sum of two $d$th powers in the field? In this article, we recount some of the history of this problem, touching on cyclotomy, Fermats last theorem, and diagonal equations. Then, we offer two proofs, one new and elementary, and the other more classical, based on Fourier analysis and an application of a nontrivial estimate from the theory of finite fields. In context and juxtaposition, each will have its merits.
Let $W$ be a smooth test function with compact support in $(0,infty)$. Conditional on the Generalized Riemann Hypothesis for Hecke $L$-functions over $mathbb{Q}(omega)$, we prove that $$sum_{p equiv 1 pmod{3}} frac{1}{2 sqrt{p}} cdot Big ( sum_{x pmod{p}} e^{2pi i x^3 / p} Big ) W Big ( frac{p}{X} Big ) sim frac{(2pi)^{2/3}}{3 Gamma(tfrac 23)} int_{0}^{infty} W(x) x^{-1/6} dx cdot frac{X^{5/6}}{log X},$$ as $X rightarrow infty$ and $p$ runs over primes. This explains a well-known numerical bias in the distribution of cubic Gauss sums first observed by Kummer in 1846 and confirms (conditionally on the Generalized Riemann Hypothesis) a conjecture of Patterson from 1978. There are two important byproducts of our proof. The first is an explicit level aspect Voronoi summation formula for cubic Gauss sums, extending computations of Patterson and Yoshimoto. Secondly, we show that Heath-Browns cubic large sieve is sharp up to factors of $X^{o(1)}$ under the Generalized Riemann Hypothesis. This disproves the popular belief that the cubic large sieve can be improved. An important ingredient in our proof is a dispersion estimate for cubic Gauss sums. It can be interpreted as a cubic large sieve with correction by a non-trivial asymptotic main term. This estimate relies on the Generalized Riemann Hypothesis, and is one of the fundamental reasons why our result is conditional.
201 - Simon Griffiths 2010
A $k$-sum of a set $Asubseteq mathbb{Z}$ is an integer that may be expressed as a sum of $k$ distinct elements of $A$. How large can the ratio of the number of $(k+1)$-sums to the number of $k$-sums be? Writing $kwedge A$ for the set of $k$-sums of $A$ we prove that [ frac{|(k+1)wedge A|}{|kwedge A|}, le , frac{|A|-k}{k+1} ] whenever $|A|ge (k^{2}+7k)/2$. The inequality is tight -- the above ratio being attained when $A$ is a geometric progression. This answers a question of Ruzsa.
We show that the diophantine equation $n^ell+(n+1)^ell + ...+ (n+k)^ell=(n+k+1)^ell+ ...+ (n+2k)^ell$ has no solutions in positive integers $k,n ge 1$ for all $ell ge 3$.
comments
Fetching comments Fetching comments
Sign in to be able to follow your search criteria
mircosoft-partner

هل ترغب بارسال اشعارات عن اخر التحديثات في شمرا-اكاديميا