Giải thích các bước giải:
a.Ta có :
$3n+12\quad\vdots\quad n+2$
$\to 3n+6+6\quad\vdots\quad n+2$
$\to 3(n+2)+6\quad\vdots\quad n+2$
$\to 6\quad\vdots\quad n+2$
$\to n+2\in\{1,2,3,6,-1,-2,-3,-6\}$
$\to n\in\{-1,0,1,4,-3,-4,-5,-8\}$
b.Gọi $(2n+3,4n+8)=d$
$\to \begin{cases}2n+3\quad\vdots\quad d\\4n+8\quad\vdots\quad d\end{cases}$
$\to 4n+8-2(2n+3)\quad\vdots\quad d\to 2\quad\vdots\quad d$
Vì $2n+3\quad\vdots\quad d\to d$ lẻ
$\to d=1$
$\to 2n+3,4n+8$ là hai số nguyên tố cùng nhau.
c.Gọi $(3n+4,5n+1)=d$
$\to \begin{cases}3n+4\quad\vdots\quad d\\5n+1\quad\vdots\quad d\end{cases}$
$\to 5(3n+4)-3(5n+1)\quad\vdots\quad d$
$\to 17\quad\vdots\quad d$
$\to$Để $(3n+4,5n+1)=1$
$\to d=1$
$\to 17\quad\not\vdots\quad d$
$\to 3n+4\quad\not\vdots\quad 17$
$\to 3n+4\ne 17k$
$\to 3n\ne 17k-4$
$\to 3n\ne 17(3q+2)-4, k=3q+2$
$\to 3n\ne 51q+30$
$\to n\ne 17q+10,q\in N$
