Giải thích các bước giải:
a.Ta có:
$13-2n\quad\vdots\quad 3n+1$
$\to 3(13-2n)\quad\vdots\quad 3n+1$
$\to 39-6n\quad\vdots\quad 3n+1$
$\to 39+2-6n-2\quad\vdots\quad 3n+1$
$\to 41-2(3n+1)\quad\vdots\quad 3n+1$
$\to 41\quad\vdots\quad 3n+1$
$\to 3n+1\in U(41)$ vì $n\in Z$
Mà $3n+1$ chia $3$ dư $1$
$\to 3n+1\in\{-41,1\}$
$\to 3n\in\{-42, 0\}$
$\to n\in\{-14, 0\}$
b.Để $\dfrac{n^2-n+1}{n-2}$ là số nguyên
$\to n^2-n+1\quad\vdots\quad n-2$ vì $n\in Z$
$\to (n^2-2n)+(n-2)+3\quad\vdots\quad n-2$
$\to n(n-2)+(n-2)+3\quad\vdots\quad n-2$
$\to 3\quad\vdots\quad n-2$
$\to n-2\in U(3)$ vì $n\in Z$
$\to n-2\in\{1,3,-1,-3\}$
$\to n\in\{3,5,1,-1\}$
c.Ta có:
$5n^2-3n+2\quad\vdots\quad n-2$
$\to (5n^2-10n)+(7n-14)+16\quad\vdots\quad n-2$
$\to 5n(n-2)+7(n-2)+16\quad\vdots\quad n-2$
$\to 16\quad\vdots\quad n-2$
$\to n-2\in U(16)$ vì $n\in Z$
$\to n-2\in\{1,2,4,16,-1,-2,-4,-16\}$
$\to n\in\{3,4,6,18,1,0,-2,-14\}$
