Cubic congruences and sums involving $\binom{3k}k$

Zhi-Hong Sun

paper

Cubic congruences and sums involving $(k 3 k)$

arXiv:1310.6721

Abstract

Let $p$ be a prime greater than $3$ and let $a$ be a rational p-adic integer. In this paper we try to determine $\sum_{k = 1}^{[p /3]} (k 3 k) a^{k} (mod p)$ , and real the connection between cubic congruences and the sum $\sum_{k = 1}^{[p /3]} (k 3 k) a^{k}$ , where $[x]$ is the greatest integer not exceeding $x$ . Suppose that $a_{1}, a_{2}, a_{3}$ are rational p-adic integers, $P = - 2 a_{1}^{3} + 9 a_{1} a_{2} - 27 a_{3}$ , $Q = (a_{1}^{2} - 3 a_{2})^{3}$ and $P Q (P^{2} - Q) (P^{2} - 3 Q) (P^{2} - 4 Q) \neq \equiv 0 (mod p)$ . In this paper we show that the number of solutions of the congruence $x^{3} + a_{1} x^{2} + a_{2} x + a_{3} \equiv 0 (mod p)$ depends only on $\sum_{k = 1}^{[p /3]} (k 3 k) (\frac{4 Q - P ^{2}}{27 Q})^{k} (mod p)$ . Let $q$ be a prime of the form $3 k + 1$ and so $4 q = L^{2} + 27 M^{2}$ with $L, M \in Z$ . When $p \neq = q$ and $p ∤ L$ , we establish congruences for $\sum_{k = 1}^{[p /3]} (k 3 k) (\frac{M ^{2}}{q})^{k}$ and $\sum_{k = 1}^{[p /3]} (k 3 k) (\frac{L ^{2}}{27 q})^{k}$ modulo p. As a consequence, we show that $x^{3} - q x - q M \equiv 0 (mod p)$ has three solutions if and only if $p$ is a cubic residue of $q$ .

Theorem 2.5 and Corollary 2.10 are new

Cubic congruences and sums involving (k3k​)

Abstract

References in corpus (2)

Cubic congruences and sums involving $(k 3 k)$