freeCodeCamp/curriculum/challenges/italian/10-coding-interview-prep/project-euler/problem-66-diophantine-equa...

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---
id: 5900f3ae1000cf542c50fec1
title: 'Problem 66: Diophantine equation'
challengeType: 5
forumTopicId: 302178
dashedName: problem-66-diophantine-equation
---
# --description--
Consider quadratic Diophantine equations of the form:
<div style='text-align: center;'>x<sup>2</sup> Dy<sup>2</sup> = 1</div>
For example, when D=13, the minimal solution in x is 649<sup>2</sup> 13×180<sup>2</sup> = 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:
<div style='margin-left: 2em;'>
3<sup>2</sup> 2×2<sup>2</sup> = 1<br>
2<sup>2</sup> 3×1<sup>2</sup> = 1<br>
<strong><span style='color: red;'>9</span></strong><sup>2</sup> 5×4<sup>2</sup> = 1<br>
5<sup>2</sup> 6×2<sup>2</sup> = 1<br>
8<sup>2</sup> 7×3<sup>2</sup> = 1<br>
</div>
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.
```js
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`.
```js
assert.strictEqual(diophantineEquation(1000), 661);
```
# --seed--
## --seed-contents--
```js
function diophantineEquation(n) {
return true;
}
diophantineEquation(7);
```
# --solutions--
```js
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;
}
```