{
  "id": "1310.6721",
  "version": "v7",
  "published": "2013-10-24T19:36:53.000Z",
  "updated": "2013-11-20T15:37:49.000Z",
  "title": "Cubic congruences and sums involving $\\binom{3k}k$",
  "authors": [
    "Zhi-Hong Sun"
  ],
  "comment": "Theorem 2.5 and Corollary 2.10 are new",
  "categories": [
    "math.NT",
    "math.CO"
  ],
  "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 $\\sum_{k=1}^{[p/3]}\\binom{3k}ka^k\\pmod p$, and real the connection between cubic congruences and the sum $\\sum_{k=1}^{[p/3]}\\binom{3k}ka^k$, where $[x]$ is the greatest integer not exceeding $x$. Suppose that $a_1,a_2,a_3$ are rational p-adic integers, $P=-2a_1^3+9a_1a_2-27a_3$, $Q=(a_1^2-3a_2)^3$ and $PQ(P^2-Q)(P^2-3Q)(P^2-4Q)\\not\\equiv 0\\pmod p$. In this paper we show that the number of solutions of the congruence $x^3+a_1x^2+a_2x+a_3\\equiv 0\\pmod p$ depends only on $\\sum_{k=1}^{[p/3]}\\binom{3k}k(\\frac{4Q-P^2}{27Q})^k\\pmod p$. Let $q$ be a prime of the form $3k+1$ and so $4q=L^2+27M^2$ with $L,M\\in\\Bbb Z$. When $p\\not=q$ and $p\\nmid L$, we establish congruences for $\\sum_{k=1}^{[p/3]}\\binom{3k}k(\\frac{M^2}q)^k$ and $\\sum_{k=1}^{[p/3]}\\binom{3k}k(\\frac{L^2}{27q})^k$ modulo p. As a consequence, we show that $x^3-qx-qM\\equiv 0\\pmod p$ has three solutions if and only if $p$ is a cubic residue of $q$.",
  "revisions": [
    {
      "version": "v7",
      "updated": "2013-11-20T15:37:49.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "11A07",
      "11A15",
      "11B39",
      "05A10"
    ],
    "keywords": [
      "cubic congruences",
      "rational p-adic integer",
      "greatest integer",
      "prime greater",
      "cubic residue"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 0,
      "language": "en",
      "license": "arXiv",
      "status": "editable",
      "adsabs": "2013arXiv1310.6721S"
    }
  }
}