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Problem 66

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Diophantine equation

Consider quadratic Diophantine equations of the form:

$$x^2-Dy^2=1$$

For example, when $D=13$, the minimal solution in $x$ is $649^2 – 13\times 180^2 = 1$.

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:

$$\begin{aligned}
3^2 - 2\times 2^2 &= 1 \\
2^2 - 3\times 1^2 &= 1 \\
{\color{red}9}^2 - 5\times 4^2 &= 1 \\
5^2 - 6\times 2^2 &= 1 \\
8^2 - 7\times 3^2 &= 1 \\
\end{aligned}$$

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

Find the value of $D \le 1000$ in minimal solutions of $x$ for which the largest value of $x$ is obtained.

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丢番图方程

考虑如下形式的二次丢番图方程：

$$x^2-Dy^2=1$$

举例而言，当$D=13$时，$x$的最小值出现在$649^2 – 13\times 180^2 = 1$。

可以断定，当$D$是平方数时，这个方程不存在正整数解。

对于$D= \{2, 3, 5, 6, 7\}$，$x$取最小值的解分别是：

$$\begin{aligned}
3^2 - 2\times 2^2 &= 1 \\
2^2 - 3\times 1^2 &= 1 \\
{\color{red}9}^2 - 5\times 4^2 &= 1 \\
5^2 - 6\times 2^2 &= 1 \\
8^2 - 7\times 3^2 &= 1 \\
\end{aligned}$$

因此，对于所有$D \le 7$，当$D=5$时$x$的最小值最大。

对于所有$D \le 1000$，求使得$x$的最小值最大时$D$的取值。

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