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Question Number 70847 by A8;15: last updated on 08/Oct/19

Answered by mind is power last updated on 08/Oct/19

{ ((x^2 +y^2 =2)),((2x^4 +xy^3 +x^2 y^2 −x^3 y=3)) :}∙EQ1 we have x≠0 EQ1⇔ { ((1+((y/x))^2 =(2/x^2 ))),((2+((y/x))^3 +((y/x))^2 −(y/x)=(3/x^4 ))) :}_ let u=(y/x) and z=(1/x^2 )⇒ { ((y^2 =(u^2 /z^2 ))),((x^2 =(1/z))) :} EQ1⇔ { ((1+u^2 =2z)),((2+u^3 +u^2 −u=3z^2 )) :} ⇒2+u^3 +u^2 −u=(3/4)(1+u^2 )^2 ⇒8+4u^3 +4u^2 −4u=3u^4 +6u^2 +3 ⇒3u^4 −4u^3 +2u^2 +4u−5=0 u=1 is solution ⇒(u−1)(3u^3 −u^2 +u+5)=0 3u^3 −u^2 +u+5=0 we see u=−1 is solution ⇒3u^3 −u^2 +u+5=(u+1)(3u^2 −4u+5) 3u^2 −4u+5=0 ⇒16−60=−48 u_3 =((4−4i(√3))/6) u_4 =((4+4i(√3))/6) for u=1or −1 ,z=((1+u^2 )/2)=((1+1)/2)=1 x^2 =(1/z)=1⇒x∈{−1,1} y^2 =(1/1)⇒y∈{1,−1} S′={(1,1),(1,−1),(−1,1),(−1,−1)} be contiued withe U_3 and u_4

$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}}\\{\mathrm{2}{x}^{\mathrm{4}} +{xy}^{\mathrm{3}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} −{x}^{\mathrm{3}} {y}=\mathrm{3}}\end{cases}\centerdot{EQ}\mathrm{1} \\ $$$${we}\:{have}\:{x}\neq\mathrm{0} \\ $$$${EQ}\mathrm{1}\Leftrightarrow\begin{cases}{\mathrm{1}+\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} =\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}\\{\mathrm{2}+\left(\frac{{y}}{{x}}\right)^{\mathrm{3}} +\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} −\frac{{y}}{{x}}=\frac{\mathrm{3}}{{x}^{\mathrm{4}} }}\end{cases}_{} \\ $$$${let}\:{u}=\frac{{y}}{{x}}\:{and}\:{z}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\Rightarrow\begin{cases}{{y}^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} }{{z}^{\mathrm{2}} }}\\{{x}^{\mathrm{2}} =\frac{\mathrm{1}}{{z}}}\end{cases} \\ $$$${EQ}\mathrm{1}\Leftrightarrow\begin{cases}{\mathrm{1}+{u}^{\mathrm{2}} =\mathrm{2}{z}}\\{\mathrm{2}+{u}^{\mathrm{3}} +{u}^{\mathrm{2}} −{u}=\mathrm{3}{z}^{\mathrm{2}} }\end{cases} \\ $$$$\Rightarrow\mathrm{2}+{u}^{\mathrm{3}} +{u}^{\mathrm{2}} −{u}=\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{8}+\mathrm{4}{u}^{\mathrm{3}} +\mathrm{4}{u}^{\mathrm{2}} −\mathrm{4}{u}=\mathrm{3}{u}^{\mathrm{4}} +\mathrm{6}{u}^{\mathrm{2}} +\mathrm{3} \\ $$$$\Rightarrow\mathrm{3}{u}^{\mathrm{4}} −\mathrm{4}{u}^{\mathrm{3}} +\mathrm{2}{u}^{\mathrm{2}} +\mathrm{4}{u}−\mathrm{5}=\mathrm{0} \\ $$$${u}=\mathrm{1}\:{is}\:{solution} \\ $$$$\Rightarrow\left({u}−\mathrm{1}\right)\left(\mathrm{3}{u}^{\mathrm{3}} −{u}^{\mathrm{2}} +{u}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\mathrm{3}{u}^{\mathrm{3}} −{u}^{\mathrm{2}} +{u}+\mathrm{5}=\mathrm{0} \\ $$$${we}\:{see}\:{u}=−\mathrm{1}\:{is}\:{solution} \\ $$$$\Rightarrow\mathrm{3}{u}^{\mathrm{3}} −{u}^{\mathrm{2}} +{u}+\mathrm{5}=\left({u}+\mathrm{1}\right)\left(\mathrm{3}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{5}\right) \\ $$$$\mathrm{3}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{16}−\mathrm{60}=−\mathrm{48} \\ $$$${u}_{\mathrm{3}} =\frac{\mathrm{4}−\mathrm{4}{i}\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$${u}_{\mathrm{4}} =\frac{\mathrm{4}+\mathrm{4}{i}\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$${for}\:{u}=\mathrm{1}{or}\:−\mathrm{1}\:,{z}=\frac{\mathrm{1}+{u}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{1}}{\mathrm{2}}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{1}}{{z}}=\mathrm{1}\Rightarrow{x}\in\left\{−\mathrm{1},\mathrm{1}\right\} \\ $$$${y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}}\Rightarrow{y}\in\left\{\mathrm{1},−\mathrm{1}\right\} \\ $$$${S}'=\left\{\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{1},−\mathrm{1}\right),\left(−\mathrm{1},\mathrm{1}\right),\left(−\mathrm{1},−\mathrm{1}\right)\right\} \\ $$$${be}\:{contiued}\:{withe}\:{U}_{\mathrm{3}} {and}\:{u}_{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$

Commented by MJS last updated on 08/Oct/19

you′re right but there′s an easier path

$$\mathrm{you}'\mathrm{re}\:\mathrm{right}\:\mathrm{but}\:\mathrm{there}'\mathrm{s}\:\mathrm{an}\:\mathrm{easier}\:\mathrm{path} \\ $$

Commented by mind is power last updated on 08/Oct/19

yeah ,isee your solution Sir its better

$${yeah}\:,{isee}\:{your}\:\:{solution}\:{Sir}\:\:{its}\:{better} \\ $$

Commented by MJS last updated on 08/Oct/19

thank you, but what I like in here is the variety of ideas to solve each problem

$$\mathrm{thank}\:\mathrm{you},\:\mathrm{but}\:\mathrm{what}\:\mathrm{I}\:\mathrm{like}\:\mathrm{in}\:\mathrm{here}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{variety}\:\mathrm{of}\:\mathrm{ideas}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{each}\:\mathrm{problem} \\ $$

Answered by MJS last updated on 08/Oct/19

x^2 +y^2 =2 2x^4 −x^3 y+x^2 y^2 +xy^3 =3 let y=px (p^2 +1)x^2 =2 ⇒ x^2 =(2/(p^2 +1)) (p+2)(p^2 −p+1)x^4 =3 ⇒ ⇒ p^4 −(4/3)p^3 +(2/3)p^2 +(4/3)p−(5/3)=0 (p−1)(p+1)(p^2 −(4/3)p+(5/3))=0 ⇒ p=±1∨p=(2/3)±((√(11))/3)i ⇒ x=±1; y=±1 and we can also calculate the complex solutions

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$$\mathrm{2}{x}^{\mathrm{4}} −{x}^{\mathrm{3}} {y}+{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{xy}^{\mathrm{3}} =\mathrm{3} \\ $$$$\mathrm{let}\:{y}={px} \\ $$$$\left({p}^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{2}} =\mathrm{2}\:\Rightarrow\:{x}^{\mathrm{2}} =\frac{\mathrm{2}}{{p}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\left({p}+\mathrm{2}\right)\left({p}^{\mathrm{2}} −{p}+\mathrm{1}\right){x}^{\mathrm{4}} =\mathrm{3}\:\Rightarrow \\ $$$$\Rightarrow\:{p}^{\mathrm{4}} −\frac{\mathrm{4}}{\mathrm{3}}{p}^{\mathrm{3}} +\frac{\mathrm{2}}{\mathrm{3}}{p}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{3}}{p}−\frac{\mathrm{5}}{\mathrm{3}}=\mathrm{0} \\ $$$$\left({p}−\mathrm{1}\right)\left({p}+\mathrm{1}\right)\left({p}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}{p}+\frac{\mathrm{5}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{p}=\pm\mathrm{1}\vee{p}=\frac{\mathrm{2}}{\mathrm{3}}\pm\frac{\sqrt{\mathrm{11}}}{\mathrm{3}}\mathrm{i} \\ $$$$\Rightarrow\:{x}=\pm\mathrm{1};\:{y}=\pm\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{can}\:\mathrm{also}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{solutions} \\ $$

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