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Question-200972




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

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