Giải thích các bước giải:
a.Ta có :
$3n^2+5\quad\vdots\quad n-1$
$\to (3n^2-3)+8\quad\vdots\quad n-1$
$\to 3(n^2-1)+8\quad\vdots\quad n-1$
$\to 8\quad\vdots\quad n-1$ vì $n^2-1=n^2-1^2\quad\vdots\quad n-1$
$\to n-1\in\{1,2,4,8,-1,-2,-4,-8\}$
$\to n\in\{2,3,5,9,0,-1,-3,-7\}$
b.Ta có :
$2n^2+11\quad\vdots\quad 3n+1$
$\to 9(2n^2+11)\quad\vdots\quad 3n+1$
$\to 2.(3n)^2+99\quad\vdots\quad 3n+1$
$\to 2((3n)^2-1)+101\quad\vdots\quad 3n+1$
$\to 101\quad\vdots\quad 3n+1$
$\to 3n+1\in\{-101,1\}$ vì $3n+1$ chia 3 dư 1
$\to n\in\{-34,0\}$
