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Question Number 14396 by tawa tawa last updated on 31/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 31/May/17

Commented by tawa tawa last updated on 31/May/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by nume1114 last updated on 31/May/17

     y^2 +x^2 =2xy+1  ⇒y^2 −2xy+x^2 −1=0  ⇒(y−x)^2 −1^2 =0  ⇒(y−x+1)(y−x−1)=0  ⇒y−x+1=0 ∨ y−x−1=0  ⇒x=y+1 ∨ x=y−1         xy−4=x  ⇒x(y−1)−4=0...(a)  (i) if x=y+1 then   (a)⇒(y+1)(y−1)−4=0          ⇒y^2 −1−4=0          ⇒y^2 =5          ⇒y=±(√5) , x=y+1=1±(√5)  (ii) if x=y−1 then   (a)⇒(y−1)^2 −4=0          ⇒y−1=±2          ⇒y=1±2=−1,3               x=y−1=−2,2  (i),(ii)  ⇒(x,y)=(−2,−1),(2,3),(1±(√5),±(√5))

$$\:\:\:\:\:{y}^{\mathrm{2}} +{x}^{\mathrm{2}} =\mathrm{2}{xy}+\mathrm{1} \\ $$$$\Rightarrow{y}^{\mathrm{2}} −\mathrm{2}{xy}+{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({y}−{x}\right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({y}−{x}+\mathrm{1}\right)\left({y}−{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{y}−{x}+\mathrm{1}=\mathrm{0}\:\vee\:{y}−{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}={y}+\mathrm{1}\:\vee\:{x}={y}−\mathrm{1} \\ $$$$ \\ $$$$\:\:\:\:\:{xy}−\mathrm{4}={x} \\ $$$$\Rightarrow{x}\left({y}−\mathrm{1}\right)−\mathrm{4}=\mathrm{0}...\left(\mathrm{a}\right) \\ $$$$\left(\mathrm{i}\right)\:\mathrm{if}\:{x}={y}+\mathrm{1}\:\mathrm{then} \\ $$$$\:\left(\mathrm{a}\right)\Rightarrow\left({y}+\mathrm{1}\right)\left({y}−\mathrm{1}\right)−\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{y}^{\mathrm{2}} −\mathrm{1}−\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{y}^{\mathrm{2}} =\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{y}=\pm\sqrt{\mathrm{5}}\:,\:{x}={y}+\mathrm{1}=\mathrm{1}\pm\sqrt{\mathrm{5}} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{if}\:{x}={y}−\mathrm{1}\:\mathrm{then} \\ $$$$\:\left(\mathrm{a}\right)\Rightarrow\left({y}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{y}−\mathrm{1}=\pm\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{y}=\mathrm{1}\pm\mathrm{2}=−\mathrm{1},\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{x}={y}−\mathrm{1}=−\mathrm{2},\mathrm{2} \\ $$$$\left(\mathrm{i}\right),\left(\mathrm{ii}\right) \\ $$$$\Rightarrow\left({x},{y}\right)=\left(−\mathrm{2},−\mathrm{1}\right),\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{1}\pm\sqrt{\mathrm{5}},\pm\sqrt{\mathrm{5}}\right) \\ $$

Commented by tawa tawa last updated on 31/May/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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