Đáp án: $n\in\{\dfrac{2+1}{3},\dfrac{ 2^3+1}{3},\dfrac{ 2^5+1}{3}, ...,\dfrac{ 2^{2k+1}+1}{3}, ...,\dfrac{ 2^{2019}+1}{3}\}$
Giải thích các bước giải:
Để $A\in Z$
$\to \dfrac{2^{2020}}{3n-1}\in Z$
$\to 2^{2020}\quad\vdots\quad 3n-1$
$\to 3n-1\in U(2^{2020})$
Mà $3n-1$ chia $3$ dư $2$
$\to 3n-1\in\{2, 2^3, 2^5, ..., 2^{2k+1}, ..., 2^{2019}\}$
$\to 3n\in\{2+1, 2^3+1, 2^5+1, ..., 2^{2k+1}+1, ..., 2^{2019}+1\}$
$\to n\in\{\dfrac{2+1}{3},\dfrac{ 2^3+1}{3},\dfrac{ 2^5+1}{3}, ...,\dfrac{ 2^{2k+1}+1}{3}, ...,\dfrac{ 2^{2019}+1}{3}\}$
