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freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-66-diophantine-equation.md

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---
id: 5900f3ae1000cf542c50fec1
title: 问题66丢番图方程
challengeType: 5
videoUrl: ''
---
# --description--
考虑形式的二次丢番图方程x2 - Dy2 = 1例如当D = 13时x中的最小解是6492 - 13×1802 = 1.可以假设当D是正整数时没有解广场。通过在D中找到D = {2,3,5,6,7}的最小解我们得到以下结果32 - 2×22 = 1 22 - 3×12 = 192 - 5×42 = 1 52 - 6× 22 = 1 82 - 7×32 = 1因此通过考虑D中对于D≤7的最小解当D = 5时获得最大的x。在x的最小解中找到D≤1000的值其中获得x的最大值。
# --hints--
`euler66()`应返回661。
```js
assert.strictEqual(euler66(), 661);
```
# --solutions--