6. 阅读：能够成为直角三角形三条边长的三个正整数$a$，$b$，$c$称为勾股数. 世界上第一次给出勾股数通解公式的是我国古代数学著作《九章算术》，其勾股数组公式为$\begin{cases}a=\frac{1}{2}(m^{2}-n^{2}),\\b = mn,\\c=\frac{1}{2}(m^{2}+n^{2}),\end{cases}$其中$m ＞ n ＞ 0$，$m$，$n$是互质的奇数.
应用：当$n = 1$时，求有一边长为5的直角三角形的另外两条边长.

当$n = 1$,$a = \frac{1}{2}(m^{2} - 1)$ ①,$b = m$ ②,$c = \frac{1}{2}(m^{2} + 1)$ ③.$\because$ 直角三角形中有一边长为 5,$\therefore$ 当$a = 5$时,$\frac{1}{2}(m^{2} - 1) = 5$,解得$m = ±\sqrt{11}$(不合题意,舍去).当$b = 5$时,即$m = 5$,代入①③,解得$a = 12$,$c = 13$.当$c = 5$时,$\frac{1}{2}(m^{2} + 1) = 5$,解得$m = ±3$.$\because m ＞ 0$,$\therefore m = 3$,代入①②,解得$a = 4$,$b = 3$.综上所述,直角三角形的另外两条边长分别为 12,13 或 3,4.

7. 如图17-9，在$\triangle ABC$中，$\angle A$，$\angle B$，$\angle C$所对的边长分别为$a = n^{2}-1$，$b = 2n$，$c = n^{2}+1$，且$n ＞ 1$.
（1）判断该三角形的形状，并说明理由；
（2）若$\angle B = 60^{\circ}$，求$\triangle ABC$的三边长.

(1)$△ABC$是直角三角形.理由如下:$\because a = n^{2} - 1$,$b = 2n$,$c = n^{2} + 1$,$\therefore a^{2} + b^{2} = (n^{2} - 1)^{2} + (2n)^{2} = n^{4} - 2n^{2} + 1 + 4n^{2} = n^{4} + 2n^{2} + 1 = (n^{2} + 1)^{2}$,$c^{2} = (n^{2} + 1)^{2}$,$\therefore a^{2} + b^{2} = c^{2}$,$\therefore △ABC$是直角三角形,且$∠C = 90^{\circ}$. (2)$\because ∠B = 60^{\circ}$,$∠C = 90^{\circ}$,$\therefore c = 2a$,即$n^{2} + 1 = 2(n^{2} - 1)$,$\therefore n^{2} = 3$,解得$n = \sqrt{3}$或$n = -\sqrt{3}$(不合题意,舍去).当$n = \sqrt{3}$时,$a = n^{2} - 1 = (\sqrt{3})^{2} - 1 = 3 - 1 = 2$,$b = 2n = 2\sqrt{3}$,$c = 2a = 4$.