{
  "id": "1702.03861",
  "version": "v1",
  "published": "2017-02-09T09:41:35.000Z",
  "updated": "2017-02-09T09:41:35.000Z",
  "title": "Is there an algorithm that decides the solvability of a Diophantine equation with a finite number of solutions?",
  "authors": [
    "Apoloniusz Tyszka"
  ],
  "comment": "7 pages. arXiv admin note: text overlap with arXiv:1109.3826",
  "categories": [
    "math.NT"
  ],
  "abstract": "For a positive integer n, let {\\theta}(n) denote the smallest positive integer b such that for each system S \\subseteq {x_i \\cdot x_j=x_k, x_i+1=x_k: i,j,k \\in {1,...,n}} which has a solution in positive integers x_1,...,x_n and which has only finitely many solutions in positive integers x_1,...,x_n, there exists a solution of S in ([1,b] \\cap N)^n. We conjecture that there exists an integer {\\delta} \\geq 9 such that the inequality {\\theta}(n) \\leq (2^{2^{n-5}}-1)^{2^{n-5}}+1 holds for every integer n \\geq {\\delta}. We prove: (1) for every integer n>9, the inequality {\\theta}(n)<(2^{2^{n-5}}-1)^{2^{n-5}}+1 implies that 2^{2^{n-5}}+1 is composite, (2) the conjecture implies that there exists an algorithm which takes as input a Diophantine equation D(x_1,...,x_p)=0 and returns the message \"Yes\" or \"No\" which correctly determines the solvability of the equation D(x_1,...,x_p)=0 in positive integers, if the solution set is finite, (3) if a function f:N\\{0} \\to N\\{0} has a finite-fold Diophantine representation, then there exists a positive integer m such that f(n)<{\\theta}(n) for every integer n>m.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2017-02-09T09:41:35.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "11U05"
    ],
    "keywords": [
      "diophantine equation",
      "finite number",
      "solvability",
      "smallest positive integer"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 7,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}