1. Given that $m$ and $n$ are integers, the equation
$$
x^{2}+(n-2) \sqrt{n-1} x+m+18=0
$$

has two distinct real roots, and the equation
$$
x^{2}-(n-6) \sqrt{n-1} x+m-37=0
$$

has two equal real roots. Find the minimum value of $n$, and explain the reasoning.
1. $\left\{\begin{array}{l}n \geqslant 1, \\ (n-2)^{2}(n-1)-4(m+18)>0, \\ (n-6)^{2}(n-1)-4(m-37)=0 .\end{array}\right.$
(2) - (3), and rearrange to get
$$
(n-4)(n-1)>27.5 \text {, }
$$

then $n \geqslant 8$.
When $n=8$, $m=44$, which satisfies the given conditions. Therefore, the minimum value of $x$ is 8.