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id	title	challengeType	forumTopicId	dashedName
5900f3ae1000cf542c50fec1	Problem 66: Diophantine equation	5	302178	problem-66-diophantine-equation

--description--

Consider quadratic Diophantine equations of the form:

x² – Dy² = 1

For example, when D=13, the minimal solution in x is 649² – 13×180² = 1.

It can be assumed that there are no solutions in positive integers when D is square.

By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:

3² – 2×2² = 1
2² – 3×1² = 1
9² – 5×4² = 1
5² – 6×2² = 1
8² – 7×3² = 1

Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.

Find the value of D ≤ n in minimal solutions of x for which the largest value of x is obtained.

--hints--

diophantineEquation(7) should return a number.

assert(typeof diophantineEquation(7) === 'number');

diophantineEquation(7) should return 5.

assert.strictEqual(diophantineEquation(7), 5);

diophantineEquation(100) should return 61.

assert.strictEqual(diophantineEquation(100), 61);

diophantineEquation(409) should return 409.

assert.strictEqual(diophantineEquation(409), 409);

diophantineEquation(500) should return 421.

assert.strictEqual(diophantineEquation(500), 421);

diophantineEquation(1000) should return 661.

assert.strictEqual(diophantineEquation(1000), 661);

--seed--

--seed-contents--

function diophantineEquation(n) {

  return true;
}

diophantineEquation(7);

--solutions--

function diophantineEquation(n) {
  // Based on https://www.mathblog.dk/project-euler-66-diophantine-equation/
  function isSolution(D, numerator, denominator) {
    return numerator * numerator - BigInt(D) * denominator * denominator === 1n;
  }

  let result = 0;
  let biggestX = 0;

  for (let D = 2; D <= n; D++) {
    let boundary = Math.floor(Math.sqrt(D));
    if (boundary ** 2 === D) {
      continue;
    }

    let m = 0n;
    let d = 1n;
    let a = BigInt(boundary);

    let [numerator, prevNumerator] = [a, 1n];

    let [denominator, prevDenominator] = [1n, 0n];

    while (!isSolution(D, numerator, denominator)) {
      m = d * a - m;
      d = (BigInt(D) - m * m) / d;
      a = (BigInt(boundary) + m) / d;

      [numerator, prevNumerator] = [a * numerator + prevNumerator, numerator];
      [denominator, prevDenominator] = [
        a * denominator + prevDenominator,
        denominator
      ];
    }

    if (numerator > biggestX) {
      biggestX = numerator;
      result = D;
    }
  }
  return result;
}

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