{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Diophantine equation (Euler Problem 66)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "Consider quadratic Diophantine equations of the form:\n", "\n", "$x^2 – Dy^2 = 1$\n", "\n", "For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1.\n", "\n", "It can be assumed that there are no solutions in positive integers when D is square.\n", "\n", "By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:\n", "\n", "$3^2 – 2×2^2 = 1$\n", "\n", "$2^2 – 3×1^2 = 1$\n", "\n", "$9^2 – 5×4^2 = 1$\n", "\n", "$5^2 – 6×2^2 = 1$\n", "\n", "$8^2 – 7×3^2 = 1$\n", "\n", "Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.\n", "\n", "Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.\n" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": true }, "outputs": [], "source": [ "import math\n", "\n", "def get_floor_sqrt(n):\n", " return math.floor(math.sqrt(n))\n", "\n", "assert(get_floor_sqrt(5) == 2)\n", "assert(get_floor_sqrt(23) == 4)\n", "\n", "def next_expansion_1(current_a, current_nominator, current_denominator, original_number):\n", " cn = current_nominator\n", " cd = current_denominator\n", " cn, cd = cd, (original_number - cd * cd) // cn\n", " next_a = math.floor((math.sqrt(original_number) + cn) // cd)\n", " cn = cn - next_a * cd\n", " cn *= -1\n", " \n", " return next_a, cn, cd\n", "\n", "def get_continued_fraction_sequence(n):\n", " \n", " # If number is a square number we return it.\n", " floor_sqrt = get_floor_sqrt(n)\n", " if n == floor_sqrt * floor_sqrt:\n", " return ((floor_sqrt), [])\n", " \n", " # Otherwise, we calculate the next expansion till we\n", " # encounter a step a second time. \n", " a = floor_sqrt\n", " cn = a\n", " cd = 1\n", " sequence = []\n", " previous_steps = []\n", " \n", " while not (a, cn, cd) in previous_steps :\n", " #print(\"a: {} cn: {} cd: {}\".format(a, cn, cd))\n", " previous_steps.append((a, cn, cd))\n", " a, cn, cd = next_expansion_1(a, cd, cn, n)\n", " sequence.append(a)\n", " sequence.pop()\n", " return ((floor_sqrt), sequence)\n", "\n", "def get_continued_fraction_sequence(n):\n", " \n", " # If number is a square number we return it.\n", " floor_sqrt = get_floor_sqrt(n)\n", " if n == floor_sqrt * floor_sqrt:\n", " return ((floor_sqrt), [])\n", " \n", " # Otherwise, we calculate the next expansion till we\n", " # encounter a step a second time. \n", " a = floor_sqrt\n", " cn = a\n", " cd = 1\n", " sequence = []\n", " previous_steps = []\n", " \n", " while not (a, cn, cd) in previous_steps :\n", " #print(\"a: {} cn: {} cd: {}\".format(a, cn, cd))\n", " previous_steps.append((a, cn, cd))\n", " a, cn, cd = next_expansion_1(a, cd, cn, n)\n", " sequence.append(a)\n", " sequence.pop()\n", " return ((floor_sqrt), sequence)\n", "\n", "assert(get_continued_fraction_sequence(1) == ((1), []))\n", "assert(get_continued_fraction_sequence(4) == ((2), []))\n", "assert(get_continued_fraction_sequence(25) == ((5), []))\n", "assert(get_continued_fraction_sequence(2) == ((1), [2]))\n", "assert(get_continued_fraction_sequence(3) == ((1), [1,2]))\n", "assert(get_continued_fraction_sequence(5) == ((2), [4]))\n", "assert(get_continued_fraction_sequence(13) == ((3), [1,1,1,1,6]))" ] }, { "cell_type": "code", "execution_count": 43, "metadata": {}, "outputs": [], "source": [ "def gcd(a, b):\n", " if b > a:\n", " a, b = b, a\n", " while a % b != 0:\n", " a, b = b, a % b\n", " return b\n", "\n", "def add_fractions(n1, d1, n2, d2):\n", " d = d1 * d2\n", " n1 = n1 * (d // d1)\n", " n2 = n2 * (d // d2)\n", " n = n1 + n2\n", " p = gcd(n, d)\n", " return (n // p, d // p)\n", "\n", "def next_expansion_2(previous_numerator, previous_denumerator, value):\n", " if previous_numerator == 0:\n", " return (value, 1)\n", " return add_fractions(previous_denumerator, previous_numerator, value, 1)\n", " \n", "def get_fractions(n, x):\n", " #print(\"get_fractions(n={}, x={})\".format(n, x))\n", " # Get sequence of a_x\n", " first_value, sequence = get_continued_fraction_sequence(n)\n", " sequence = [first_value] + math.ceil(x / len(sequence)) * sequence\n", " sequence = sequence[:x + 1]\n", " sequence = sequence[::-1]\n", " \n", " n, d = 0, 1\n", " for s in sequence:\n", " n, d = next_expansion_2(n, d, s)\n", " return (n, d)\n", "\n", "assert(get_fractions(7, 0) == (2, 1))\n", "assert(get_fractions(7, 1) == (3, 1))\n", "assert(get_fractions(7, 2) == (5, 2))\n", "assert(get_fractions(7, 3) == (8, 3))" ] }, { "cell_type": "code", "execution_count": 48, "metadata": {}, "outputs": [], "source": [ "def get_minimal_solution(d):\n", " for i in range(0, 100):\n", " x, y = get_fractions(d, i)\n", " if x * x - d * y * y == 1:\n", " return((x, y))\n", "\n", "assert(get_minimal_solution(2) == (3, 2))\n", "assert(get_minimal_solution(3) == (2, 1))\n", "assert(get_minimal_solution(85) == (285769, 30996))\n", "assert(get_minimal_solution(109) == (158070671986249, 15140424455100))" ] }, { "cell_type": "code", "execution_count": 52, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "True\n" ] } ], "source": [ "from math import sqrt\n", "\n", "def is_square(n):\n", " return sqrt(n).is_integer()\n", "\n", "print(is_square(4))" ] }, { "cell_type": "code", "execution_count": 55, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "d: 661 x: 16421658242965910275055840472270471049\n" ] } ], "source": [ "x_max = 0\n", "d_max = 0\n", "\n", "for d in range(2, 1001):\n", " if is_square(d):\n", " continue\n", " \n", " x, y = get_minimal_solution(d)\n", " if x > x_max:\n", " x_max = x\n", " d_max = d\n", "\n", "print(\"d: {} x: {}\".format(d_max, x_max))\n", "s = d_max\n", "assert(s == 661)" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [] } ], "metadata": { "completion_date": "Tue, 22 Jan 2019, 05:10", "kernelspec": { "display_name": "Python 3", "language": "python3.6", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.6.5" }, "tags": [ "fractions", "square roots", "sequence", "periodic" ] }, "nbformat": 4, "nbformat_minor": 2 }