Solve the equation in positive integers for $ x,y $
$$ \frac{47}{\sqrt x} \ + \ \frac{43}{\sqrt y} = \frac{1}{\sqrt {2021}}$$ I tried factoring , squaring , putting integer restrictions but didn't seem to go anywhere. Please provide the way through.
2026-04-24 00:37:45.1776991065
Solve the equation in positive integers.
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Clearing denominators yields $$47\sqrt{2021}\sqrt{y}+43\sqrt{2021}\sqrt{x}=\sqrt{x}\sqrt{y},$$ and hence also $$\sqrt{x}\sqrt{y}-43\sqrt{2021}\sqrt{x}-47\sqrt{2021}\sqrt{y}=0.\tag{1}$$ Next note that $$(\sqrt{x}-47\sqrt{2021})(\sqrt{y}-43\sqrt{2021})=\sqrt{x}\sqrt{y}-43\sqrt{2021}\sqrt{x}-47\sqrt{2021}\sqrt{y}+43\cdot47\cdot2021,$$ so equation $(1)$ is equivalent to $$(\sqrt{x}-47\sqrt{2021})(\sqrt{y}-43\sqrt{2021})=43\cdot47\cdot2021=43^2\cdot47^2.$$ Can you take it from here?