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Question Number 120185 by mr W last updated on 29/Oct/20
solve for x, y∈Z  (√x)+(√y)=(√(2020))
$${solve}\:{for}\:{x},\:{y}\in{Z} \\ $$$$\sqrt{{x}}+\sqrt{{y}}=\sqrt{\mathrm{2020}} \\ $$
Answered by floor(10²Eta[1]) last updated on 29/Oct/20
(√y)=(√(2020))−(√x)  (y=2020+x−2(√(2020x))) ∈Z  ⇒(√(2020x)) ∈Z  (√(2020x))=2(√(505x))  ⇒(√(505x)) ∈Z⇒x=505a^2 , a∈Z  Similarly: y=505b^2 , b∈Z  (√(505a^2 ))+(√(505b^2 ))=2(√(505))  ⇒a+b=2  a=0, b=2  a=1, b=1  a=2, b=0  (x, y)∈{(0, 2020), (505, 505), (2020, 0)}
$$\sqrt{\mathrm{y}}=\sqrt{\mathrm{2020}}−\sqrt{\mathrm{x}} \\ $$$$\left(\mathrm{y}=\mathrm{2020}+\mathrm{x}−\mathrm{2}\sqrt{\mathrm{2020x}}\right)\:\in\mathbb{Z} \\ $$$$\Rightarrow\sqrt{\mathrm{2020x}}\:\in\mathbb{Z} \\ $$$$\sqrt{\mathrm{2020x}}=\mathrm{2}\sqrt{\mathrm{505x}} \\ $$$$\Rightarrow\sqrt{\mathrm{505x}}\:\in\mathbb{Z}\Rightarrow\mathrm{x}=\mathrm{505a}^{\mathrm{2}} ,\:\mathrm{a}\in\mathbb{Z} \\ $$$$\mathrm{Similarly}:\:\mathrm{y}=\mathrm{505b}^{\mathrm{2}} ,\:\mathrm{b}\in\mathbb{Z} \\ $$$$\sqrt{\mathrm{505a}^{\mathrm{2}} }+\sqrt{\mathrm{505b}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{505}} \\ $$$$\Rightarrow\mathrm{a}+\mathrm{b}=\mathrm{2} \\ $$$$\mathrm{a}=\mathrm{0},\:\mathrm{b}=\mathrm{2} \\ $$$$\mathrm{a}=\mathrm{1},\:\mathrm{b}=\mathrm{1} \\ $$$$\mathrm{a}=\mathrm{2},\:\mathrm{b}=\mathrm{0} \\ $$$$\left(\mathrm{x},\:\mathrm{y}\right)\in\left\{\left(\mathrm{0},\:\mathrm{2020}\right),\:\left(\mathrm{505},\:\mathrm{505}\right),\:\left(\mathrm{2020},\:\mathrm{0}\right)\right\} \\ $$
Commented by mr W last updated on 30/Oct/20
thanks!
$${thanks}! \\ $$

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