1. Find the $x$ in $\mathbb{Z}$ such that $x-1$ divides $x-3$
2. Find the $x$ in $\mathbb{Z}$ such that $x+2$ divides $x^{2}+3$.
1. $x-1$ divides $x-3$ if and only if $x-1$ divides $x-3-(x-1)$, hence if and only if $x-1$ divides -2. Thus, $x-1$ belongs to the set $\{-2,-1,1,2\}$, which means that $x \in\{-1,0,2,3\}$.
2. $x+2$ divides $x^{2}+3$ if and only if $x+2$ divides $\left(x^{2}+3\right)-x(x+2)=-2 x+3$, if and only if $x+2$ divides $-2 x+3+2(x+2)=7$. Therefore, $x+2$ belongs to the set $\{-7,-1,1,7\}$, which means that $x \in\{-9,-3,-1,5\}$.