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Question Number 141749 by Huy last updated on 23/May/21

 { ((x^2 −y+(√(y^2 +5))=xy−(√(x−1)))),((y^2 +(√(xy+2))=2(x+y))) :}  Find x,y

$$\begin{cases}{\mathrm{x}^{\mathrm{2}} −\mathrm{y}+\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{5}}=\mathrm{xy}−\sqrt{\mathrm{x}−\mathrm{1}}}\\{\mathrm{y}^{\mathrm{2}} +\sqrt{\mathrm{xy}+\mathrm{2}}=\mathrm{2}\left(\mathrm{x}+\mathrm{y}\right)}\end{cases} \\ $$$$\mathrm{Find}\:\mathrm{x},\mathrm{y} \\ $$

Answered by MJS_new last updated on 23/May/21

assuming a “nice” solution I tried  (√(y^2 +5))∈N ⇒ y=2  (√(x−1))∈N∧(√(2x+2))∈N ⇒ x=1∨x=17∨x=577 ...  ⇒ x=1∧y=2

$$\mathrm{assuming}\:\mathrm{a}\:``\mathrm{nice}''\:\mathrm{solution}\:\mathrm{I}\:\mathrm{tried} \\ $$$$\sqrt{{y}^{\mathrm{2}} +\mathrm{5}}\in\mathbb{N}\:\Rightarrow\:{y}=\mathrm{2} \\ $$$$\sqrt{{x}−\mathrm{1}}\in\mathbb{N}\wedge\sqrt{\mathrm{2}{x}+\mathrm{2}}\in\mathbb{N}\:\Rightarrow\:{x}=\mathrm{1}\vee{x}=\mathrm{17}\vee{x}=\mathrm{577}\:... \\ $$$$\Rightarrow\:{x}=\mathrm{1}\wedge{y}=\mathrm{2} \\ $$

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