$(x + \sqrt{x^2 + 2020})(y + \sqrt{y^2 + 2020}) = 2020 \qquad (*)$
$(*)\Leftrightarrow (x - \sqrt{x^2 + 2020})(x + \sqrt{x^2 + 2020})(y + \sqrt{y^2 + 2020}) = 2020(x - \sqrt{x^2 + 2020})$
$\Leftrightarrow [x^2 - (x^2 + 2020)](y + \sqrt{y^2 + 2020}) = 2020(x - \sqrt{x^2 + 2020})$
$\Leftrightarrow -2020(y + \sqrt{y^2 + 2020}) = 2020(x - \sqrt{x^2 + 2020})$
$\Leftrightarrow y + \sqrt{y^2 + 2020} = \sqrt{x^2 + 2020} - x\qquad (1)$
$(*)\Leftrightarrow (x + \sqrt{x^2 + 2020})(y + \sqrt{y^2 + 2020})(y - \sqrt{y^2 + 2020}) = 2020(y- \sqrt{y^2 + 2020})$
$\Leftrightarrow [y^2 - (y^2 + 2020)](x + \sqrt{x^2 + 2020}) = 2020(y - \sqrt{y^2 + 2020})$
$\Leftrightarrow -2020(x + \sqrt{x^2 + 2020}) = 2020(y - \sqrt{y^2 + 2020})$
$\Leftrightarrow x+ \sqrt{x^2 + 2020} = \sqrt{y^2 + 2020} - y\qquad (2)$
Lấy $(1)+(2)$ ta được:
$x + y + \sqrt{x^2 + 2020} + \sqrt{y^2 + 2020} = \sqrt{x^2 + 2020} + \sqrt{y^2 + 2020} - x - y$
$\Leftrightarrow 2(x + y) = 0$
$\Leftrightarrow x = - y$
Do đó:
$x^{2020} + y^{2020}$
$= (-y)^{2020} + y^{2020}$
$= 2y^{2020}$
Trường hợp:
$x^{2021} + y^{2021}$
$= (-y)^{2021} + y^{2021}$
$= -y^{2021}+ y^{2021} = 0$
