{
  "id": "1604.05019",
  "version": "v1",
  "published": "2016-04-18T07:22:38.000Z",
  "updated": "2016-04-18T07:22:38.000Z",
  "title": "Proof of a recent conjecture of Z.-W. Sun",
  "authors": [
    "Song Guo",
    "Victor J. W. Guo"
  ],
  "comment": "4 pages",
  "categories": [
    "math.NT",
    "math.CO"
  ],
  "abstract": "The polynomials $d_n(x)$ are defined by \\begin{align*} d_n(x) &= \\sum_{k=0}^n{n\\choose k}{x\\choose k}2^k. \\end{align*} We prove that, for any prime $p$, the following congruences hold modulo $p$: \\begin{align*} \\sum_{k=0}^{p-1}\\frac{{2k\\choose k}}{4^k} d_k\\left(-\\frac{1}{4}\\right)^2 &\\equiv \\begin{cases} 2(-1)^{\\frac{p-1}{4}}x,&\\text{if $p=x^2+y^2$ with $x\\equiv 1\\pmod{4}$,} 0,&\\text{if $p\\equiv 3\\pmod{4}$,} \\end{cases} [5pt] \\sum_{k=0}^{p-1}\\frac{{2k\\choose k}}{4^k} d_k\\left(-\\frac{1}{6}\\right)^2 &\\equiv 0, \\quad\\text{if $p>3$,} [5pt] \\sum_{k=0}^{p-1}\\frac{{2k\\choose k}}{4^k} d_k\\left(\\frac{1}{4}\\right)^2 &\\equiv \\begin{cases} 0,&\\text{if $p\\equiv 1\\pmod{4}$,} (-1)^{\\frac{p+1}{4}}{\\frac{p-1}{2}\\choose \\frac{p-3}{4}},&\\text{if $p\\equiv 3\\pmod{4}$.} \\end{cases} \\sum_{k=0}^{p-1}\\frac{{2k\\choose k}}{4^k} d_k\\left(\\frac{1}{6}\\right)^2 &\\equiv 0, \\quad\\text{if $p>5$.} \\end{align*} The $p\\equiv 3\\pmod{4}$ case of the first one confirms a conjecture of Z.-W. Sun, while the second one confirms a special case of another conjecture of Z.-W. Sun.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2016-04-18T07:22:38.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "11A07",
      "11B65",
      "05A10"
    ],
    "keywords": [
      "conjecture",
      "congruences hold modulo",
      "special case",
      "polynomials"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 4,
      "language": "en",
      "license": "arXiv",
      "status": "editable",
      "adsabs": "2016arXiv160405019G"
    }
  }
}