130 lines
2.9 KiB
Markdown
130 lines
2.9 KiB
Markdown
---
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id: 5900f3ae1000cf542c50fec1
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title: 'Problem 66: Diophantine equation'
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challengeType: 5
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forumTopicId: 302178
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dashedName: problem-66-diophantine-equation
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---
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# --description--
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Consider quadratic Diophantine equations of the form:
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<div style='text-align: center;'>x<sup>2</sup> – Dy<sup>2</sup> = 1</div>
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For example, when D=13, the minimal solution in x is 649<sup>2</sup> – 13×180<sup>2</sup> = 1.
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It can be assumed that there are no solutions in positive integers when D is square.
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By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
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<div style='margin-left: 2em;'>
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3<sup>2</sup> – 2×2<sup>2</sup> = 1<br>
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2<sup>2</sup> – 3×1<sup>2</sup> = 1<br>
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<strong><span style='color: red;'>9</span></strong><sup>2</sup> – 5×4<sup>2</sup> = 1<br>
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5<sup>2</sup> – 6×2<sup>2</sup> = 1<br>
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8<sup>2</sup> – 7×3<sup>2</sup> = 1<br>
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</div>
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Hence, by considering minimal solutions in `x` for D ≤ 7, the largest `x` is obtained when D=5.
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Find the value of D ≤ `n` in minimal solutions of `x` for which the largest value of `x` is obtained.
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# --hints--
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`diophantineEquation(7)` should return a number.
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```js
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assert(typeof diophantineEquation(7) === 'number');
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```
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`diophantineEquation(7)` should return `5`.
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```
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assert.strictEqual(diophantineEquation(7), 5);
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```
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`diophantineEquation(100)` should return `61`.
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```
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assert.strictEqual(diophantineEquation(100), 61);
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```
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`diophantineEquation(409)` should return `409`.
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```
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assert.strictEqual(diophantineEquation(409), 409);
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```
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`diophantineEquation(500)` should return `421`.
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```
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assert.strictEqual(diophantineEquation(500), 421);
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```
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`diophantineEquation(1000)` should return `661`.
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```js
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assert.strictEqual(diophantineEquation(1000), 661);
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```
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# --seed--
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## --seed-contents--
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```js
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function diophantineEquation(n) {
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return true;
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}
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diophantineEquation(7);
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```
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# --solutions--
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```js
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function diophantineEquation(n) {
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// Based on https://www.mathblog.dk/project-euler-66-diophantine-equation/
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function isSolution(D, numerator, denominator) {
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return numerator * numerator - BigInt(D) * denominator * denominator === 1n;
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}
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let result = 0;
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let biggestX = 0;
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for (let D = 2; D <= n; D++) {
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let boundary = Math.floor(Math.sqrt(D));
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if (boundary ** 2 === D) {
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continue;
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}
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let m = 0n;
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let d = 1n;
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let a = BigInt(boundary);
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let [numerator, prevNumerator] = [a, 1n];
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let [denominator, prevDenominator] = [1n, 0n];
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while (!isSolution(D, numerator, denominator)) {
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m = d * a - m;
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d = (BigInt(D) - m * m) / d;
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a = (BigInt(boundary) + m) / d;
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[numerator, prevNumerator] = [a * numerator + prevNumerator, numerator];
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[denominator, prevDenominator] = [
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a * denominator + prevDenominator,
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denominator
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];
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}
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if (numerator > biggestX) {
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biggestX = numerator;
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result = D;
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}
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}
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return result;
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}
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```
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