Counting the solutions of λ1x1k1 + . . . + λtxtkt ≡ c mod n

JOURNAL OF NUMBER THEORY

卷:187

Article

Li, Songsong¹  Ouyang, Yi¹ 

[1] Univ Sci & Technol China, Sch Math Sci, Wu Wen Tsun Key Lab Math, Hefei 230026, Anhui, Peoples R China

关键词: Congruence equation;    Decomposition and lifting of residue classes;    Squares of units;    Powers of units;   

DOI  :  10.1016/j.jnt.2017.10.017

来源: Elsevier

【 摘 要 】

Given a polynomial Q(x(1), . . . ,x(t)) = lambda(1)x(1)(k1) + . . . + lambda(t)x(t)(kt), for every c is an element of Z and n >= 2, we study the number of solutions N-j(Q; c, n) of the congruence equation Q(x(1), . . . ,x(t)) equivalent to c mod n in (Z/nZ)(t) such that x(i) is an element of (Z/nZ)(x) for i is an element of J subset of I = {1, . . . , t}. We deduce formulas and an algorithm to study N-J(Q; c, p(a)) for p any prime number and a >= 1 any integer. As consequences of our main results, we completely solve: the counting problem of Q(x(i)) = Sigma(i is an element of I) lambda(i)x(i) for any prime p and any subset J of I; the counting problem of Q(x(i)) = Sigma(i is an element of I) lambda(i)x(i)(2) in the case t = 2 for any p and J, and the case t general for any p and J satisfying min{v(p)(lambda(i)) vertical bar i is an element of I} = min{v(p)(lambda(i)) vertical bar i is an element of J}; the counting problem of Q(x(i)) = Sigma(i is an element of I) lambda(i)x(i)(k) in the case t = 2 for any p inverted iota k and any J, and in the case t general for any p inverted iota k and J satisfying min{v(p)(lambda(i)) vertical bar i is an element of I} = min{v(p)(lambda(i)) vertical bar i is an element of J}. (C) 2017 Elsevier Inc. All rights reserved.

【 授权许可】

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