{
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  {
   "cell_type": "markdown",
   "id": "ca2d9c5a",
   "metadata": {},
   "source": [
    "# [Diophantine Equation](https://projecteuler.net/problem=66)\n",
    "\n",
    "SageMath can solve these [Diophantine equations](https://en.wikipedia.org/wiki/Diophantine_equation) for us. For equations of the form $x^2 - Dy^2 = 1$ (which are called [Pell equations](https://en.wikipedia.org/wiki/Pell%27s_equation)), it will give parameterizations in $t$ for both $x$ and $y$. Here's $D=2$ as an example:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "d6c3ebd4",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[(sqrt(2)*(2*sqrt(2) + 3)^t - sqrt(2)*(-2*sqrt(2) + 3)^t + 3/2*(2*sqrt(2) + 3)^t + 3/2*(-2*sqrt(2) + 3)^t,\n",
       "  -3/4*sqrt(2)*(2*sqrt(2) + 3)^t + 3/4*sqrt(2)*(-2*sqrt(2) + 3)^t - (2*sqrt(2) + 3)^t - (-2*sqrt(2) + 3)^t),\n",
       " (-sqrt(2)*(2*sqrt(2) + 3)^t + sqrt(2)*(-2*sqrt(2) + 3)^t - 3/2*(2*sqrt(2) + 3)^t - 3/2*(-2*sqrt(2) + 3)^t,\n",
       "  3/4*sqrt(2)*(2*sqrt(2) + 3)^t - 3/4*sqrt(2)*(-2*sqrt(2) + 3)^t + (2*sqrt(2) + 3)^t + (-2*sqrt(2) + 3)^t)]"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "var('x,y')\n",
    "solve_diophantine(x^2 - 2*y^2 == 1)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "519c091f",
   "metadata": {},
   "source": [
    "$t=0$ seems to correspond to the minimal solution we are after."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "4b9b6240",
   "metadata": {},
   "outputs": [],
   "source": [
    "def minimal_solution(d):\n",
    "    sols = solve_diophantine(x^2 - d*y^2 == 1)\n",
    "    u, v = sols[0]\n",
    "    return (abs(u(t=0)), abs(v(t=0)))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "42ba15f7",
   "metadata": {},
   "source": [
    "Now we can just iterate (although it is pretty slow)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "300b1afb",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "661"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "max((d for d in range(1, 1001) if not is_square(d)), key=lambda k: minimal_solution(k)[0])"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b16636f7",
   "metadata": {},
   "source": [
    "Let's dig a little deeper so we can optimize.\n",
    "\n",
    "## Solving Pell equations\n",
    "Lagrange proved that if $(x_0, y_0)$ is a solution to\n",
    "$$x^2 - dy^2 = 1$$\n",
    "then $\\frac{x_0}{y_0}$ is a [convergent of the continued fraction](https://en.wikipedia.org/wiki/Simple_continued_fraction) of $\\sqrt{d}$. Specifically, if $p$ is the period of the continued fraction, then the first solution will be the $(p-1)$th convergent if $p$ is even, and the $(2p-1)$th convergent if $p$ is odd.\n",
    "\n",
    "This is great for us, since there are algorithms to compute these convergents. We'll use SageMath here; see [problem 64](https://projecteuler.net/problem=64) for how to compute the partial denominators of the continued fraction of a square root, and see [problem 65](https://projecteuler.net/problem=65) for an algorithm that uses partial denominators to compute convergents of continued fractions (SageMath's constructions make this implementation a little slow, but it makes the code easier to read - and it's still considerably faster than using `solve_diophantine`)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "5d95125c",
   "metadata": {},
   "outputs": [],
   "source": [
    "def pell_fundamental_solution(d):\n",
    "    K.<s> = QuadraticField(d)\n",
    "    f = continued_fraction(s)\n",
    "    p = f.period_length()\n",
    "    if p % 2 == 0:\n",
    "        n = p - 1\n",
    "    else:\n",
    "        n = 2 * p - 1\n",
    "        \n",
    "    x, y = f.convergent(n).as_integer_ratio()\n",
    "    return x, y"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b0bec674",
   "metadata": {},
   "source": [
    "Now we'll just iterate with this function instead."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "03d7ba9d",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "661"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "max((d for d in range(1, 1001) if not is_square(d)), key=lambda x: pell_fundamental_solution(x)[0])"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3f573e65",
   "metadata": {},
   "source": [
    "And in case you want to know the minimal $x$ for $x^2 - 661y^2 = 1$:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "b4b119a0",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(16421658242965910275055840472270471049, 638728478116949861246791167518480580)"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "pell_fundamental_solution(661)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cfb42d20",
   "metadata": {},
   "source": [
    "## Relevant sequences\n",
    "* Minimal values of $x$ for solutions to the Pell equation: [A002350](https://oeis.org/A002350)\n",
    "\n",
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
   ]
  }
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