{
  "id": "1311.6364",
  "version": "v2",
  "published": "2013-11-25T16:59:35.000Z",
  "updated": "2013-12-02T14:51:01.000Z",
  "title": "Quartic residues and sums involving $\\binom{4k}{2k}$",
  "authors": [
    "Zhi-Hong Sun"
  ],
  "comment": "Section 3 is new",
  "categories": [
    "math.NT"
  ],
  "abstract": "Let $p$ be an odd prime and let $m\\not\\equiv 0\\pmod p$ be a rational p-adic integer. In this paper we reveal the connection between quartic residues and the sum $\\sum_{k=0}^{[p/4]}\\binom{4k}{2k}\\frac 1{m^k}$, where $[x]$ is the greatest integer not exceeding $x$. Let $q$ be a prime of the form $4k+1$ and so $q=a^2+b^2$ with $a,b\\in\\Bbb Z$. When $p\\nmid ab(a^2-b^2)q$, we show that for $r=0,1,2,3$, $p^{\\frac{q-1}4}\\equiv (\\frac ab)^r\\pmod q$ if and only if $$\\sum_{k=0}^{[p/4]}\\binom{4k}{2k}\\Big(\\frac{a^2}{16q}\\Big)^k\\equiv (-1)^{\\frac{p^2-1}8a+\\frac{p-1}2\\cdot \\frac{q-1}4}\\Big(\\frac pq\\Big) \\Big(\\frac ab\\Big)^r\\pmod p,$$ where $(\\frac pq)$ is the Legendre symbol. We also establish congruences for $\\sum_{k=0}^{[p/4]}\\binom{4k}{2k}\\frac 1{m^k}\\pmod p$ in the cases $m=17,18,20,32,52,80,272$.",
  "revisions": [
    {
      "version": "v2",
      "updated": "2013-12-02T14:51:01.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "11A07",
      "11A15",
      "11B39",
      "11B65",
      "11E25"
    ],
    "keywords": [
      "quartic residues",
      "rational p-adic integer",
      "legendre symbol",
      "odd prime",
      "greatest integer"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 0,
      "language": "en",
      "license": "arXiv",
      "status": "editable",
      "adsabs": "2013arXiv1311.6364S"
    }
  }
}