Đáp án: $m\in\{2021,1011,675, 341\}$
Giải thích các bước giải:
Ta có:
$x^2-mx+2021=m$
$\to x^2+2021=m+mx$
$\to x^2-1+2022=m(x+1)$
$\to (x-1)(x+1)+2022=m(x+1)$
$\to m=x-1+\dfrac{2022}{x+1}$
Vì $m,x\in Z$
$\to m(x+1)\quad\vdots\quad x+1$
$\to (x-1)(x+1)+2022\quad\vdots\quad x+1$
$\to 2022\quad\vdots\quad x+1$
$\to x+1\in\{1,2,3,337,6,674,1011,2022\}$ vì $x\in N$
$\to x\in\{0,1,2,336,5,673,1010,2021\}$
$\to m\in\{2021,1011, 675,341, 341,675,1011, 2021\}$
Thử lại:
$\to m\in\{2021,1011,675, 341\}$
