problem 66 // project euler

Diophantine equation

Consider quadratic Diophantine equations of the form:

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 ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.

丢番图方程
考虑如下形式的二次丢番图方程：

举例而言，当D=13时，x的最小值出现在649² – 13×180² = 1。

可以断定，当D是平方数时，这个方程不存在正整数解。

对于D= {2, 3, 5, 6, 7}分别求出x取最小值的解，我们得到：

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

因此，对于所有D ≤ 7，当D=5时x的最小值最大。

对于D ≤ 1000，求使得x的最小值最大的D值。