منابع مشابه
Power Residues and Nonresidues in Arithmetic Progressions
Let A: be an integer > 2 andp a prime such that vk(p) = (k,p — 1) > 1. Let bn + c(n = 0,1,. ..;b > 2,1 < c < b, (b,p) — (c,p) = 1) be an arithmetic progression. We denote the smallest kth power nonresidue in the progression bn + c by g(p,k,b,c), the smallest quadratic residue in the progression bn + c by r2(p,b,c), and the nth smallest prime kth power nonresidue by g„(p,k), n = 0, 1, 2,_ If C(p...
متن کاملPairs of Consecutive Power Residues
Introduction. Until recently none of the numerous papers on the distribution of quadratic and higher power residues was concerned with questions of the following sort: Let k and m be positive integers. According to a theorem of Brauer (1), for every sufficiently large prime p there exist m consecutive positive integers r, r + l , . . . , r + m — 1, each of which is a &th power residue of p. Let...
متن کاملON QUADRUPLES OF CONSECUTIVE ¿th POWER RESIDUES
In [2] it was shown that A(&, 4) = co for fe^ 1048909 and it was conjectured that A(&, 4) = =° for all k. In this paper we establish this conjecture with the following Theorem. A(&, 4) = ». Proof. It suffices to prove the theorem for values of k which are prime. The proof makes use of the following proposition which is a special case of a result of Kummer [l] (see also [3]). Proposition. Let k ...
متن کاملSimple Arguments on Consecutive Power Residues
By some extremely simple arguments, we point out the following: (i) If n is the least positive kth power non-residue modulo a positive integer m, then the greatest number of consecutive kth power residues mod m is smaller than m/n. (ii) Let OK be the ring of algebraic integers in a quadratic field K = Q( √ d) with d ∈ {−1,−2,−3,−7,−11}. Then, for any irreducible π ∈ OK and positive integer k no...
متن کاملOn the constant in Burgess’ bound for the number of consecutive residues or non-residues
We give an explicit version of a result due to D. Burgess. Let χ be a non-principal Dirichlet character modulo a prime p. We show that the maximum number of consecutive integers for which χ takes on a particular value is less than { πe √ 6 3 + o(1) } p1/4 log p, where the o(1) term is given explicitly.
متن کاملذخیره در منابع من
با ذخیره ی این منبع در منابع من، دسترسی به آن را برای استفاده های بعدی آسان تر کنید
ژورنال
عنوان ژورنال: Mathematics of Computation
سال: 1970
ISSN: 0025-5718
DOI: 10.1090/s0025-5718-1970-0277469-8