Lời giải:
Ta có:
$(n+3) \vdots (n-1)$
Lại có:
$(n-1) \vdots (n-1)$
$=>(n-1)-(n+3) \vdots (n-1)$
$=>-4 \vdots (n-1)$
$=>n-1∈U(-4)={-4;-2;-1;1;2;4}$
$n-1=-4<=>n=-3$
$n-1=-2<=>n=-1$
$n-1=-1<=>n=0$
$n-1=2<=>n=3$
$n-1=4<=>n=5$
Để $B$ nguyên $n=${$-3;-1;0;3;5$}