$ \cfrac{n+4} { n+1}=1+\cfrac3{n+1}\\\text{Để A là số nguyên tố thì }\cfrac{3}{n+1} \text{ nguyên}\\\to3\vdots x+1\\\to x+1\in Ư_{(3)}=(-3;-1;1;3)\\\begin{array}{|c|c|}\hline x+1&-3&-1&1&3\\\hline x&-4&-2&0&2\\\hline\end{array}\\\text{Vậy }x=(-4;-2;0;2)$
