Question Number 200972 by Mingma last updated on 27/Nov/23
Answered by deleteduser1 last updated on 27/Nov/23
$${a}^{\mathrm{2}} ={b}^{\mathrm{2}} \left({b}+\mathrm{2}\right);{a}={b}\sqrt{{b}+\mathrm{2}} \\ $$$${b}=\left(\frac{{x}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}\Rightarrow\left({a},{b}\right)=\left(\frac{{x}^{\mathrm{3}} −\mathrm{2}{xy}^{\mathrm{2}} }{{y}^{\mathrm{3}} },\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }−\mathrm{2}\right)\:{for}\:{x},{y}\in\mathbb{Z} \\ $$
Commented by Mingma last updated on 27/Nov/23
very elegant!