Giải thích các bước giải:
Ta có:
$A=1+2+2^2+2^3+...+2^{2019}$
$\to 2A=2+2^2+2^3+2^4+...+2^{2020}$
$\to 2A-A=2^{2020}-1$
$\to A=2^{2020}-1$
Vì $2\quad\vdots\quad 2\to 2^{2020}\quad\vdots\quad 2$
$\to 2^{2020}-1\quad\not\vdots\quad 2$
$\to A\quad\not\vdots\quad 2$
Ta có:
$2^{2020}-1=(2^2)^{1010}-1=4^{1010}-1\quad\vdots\quad 4-1=3$
$\to 2^{2020}-1\quad\vdots\quad 3$
$\to A\quad\vdots\quad 3$
Lại có:
$2^{2020}=2\cdot 2^{2019}=2\cdot 2^{3\cdot 673}=2\cdot (2^3)^{673}=2\cdot 8^{673}$
Vì $8$ chia $7$ dư $1$
$\to 8^{673}$ chia $7$ dư $1$
$\to 2\cdot 8^{673}$ chia $7$ dư $2$
$\to 2\cdot 8^{673}-1$ chia $7$ dư $1$
$\to2^{2020}-1$ chia $7$ dư $1$
$\to A$ chia $7$ dư $1$
$\to A\quad\not\vdots\quad 7$
$\to A\quad\not\vdots\quad 70$ vì $70=7\cdot 10$
Ta có:
$A=2^{2020}-1$
$\to A+1=2^{2020}$
$\to A+1=(2^{1010})^2$ là số chính phương
