Question Number 172824 by Mikenice last updated on 01/Jul/22
Commented by MJS_new last updated on 02/Jul/22
$$\mathrm{how}\:\mathrm{many}\:\%\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}? \\ $$$$\mathrm{I}\:\mathrm{can}. \\ $$
Commented by cortano1 last updated on 02/Jul/22
$${one}\:{solution}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$
Answered by mr W last updated on 02/Jul/22
$${let}\:{x}=\sqrt{{X}}>\mathrm{0},\:{y}=\sqrt{{Y}}>\mathrm{0} \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){y}={x}\:\:\:\:…\left({i}\right) \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){x}=\mathrm{3}{y}\:\:\:…\left({ii}\right) \\ $$$${x}={y}=\mathrm{0}\:{is}\:{solution}\:\Rightarrow{X}={Y}=\mathrm{0} \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{3} \\ $$$$\Rightarrow{x}^{\mathrm{4}} −{y}^{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{2}}\:\:\:…\left({I}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\frac{\mathrm{2}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} }{\mathrm{3}{y}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{4}} −\mathrm{5}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} \right)=\mathrm{0}\:\:\:…\left({II}\right) \\ $$$${case}\:\mathrm{1}:\:{x}^{\mathrm{2}} =\mathrm{2}{y}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{y}^{\mathrm{4}} −{y}^{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{y}^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\Rightarrow{Y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\sqrt{\mathrm{2}}\:\Rightarrow{X}=\sqrt{\mathrm{2}} \\ $$$${case}\:\mathrm{2}:\:{x}^{\mathrm{2}} =\mathrm{3}{y}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}{y}^{\mathrm{4}} −{y}^{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{y}^{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{16}}\:\Rightarrow{y}^{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow{Y}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow{X}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$${summary}: \\ $$$$\left({X},{Y}\right)=\left(\mathrm{0},\mathrm{0}\right),\left(\sqrt{\mathrm{2}},\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right),\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}},\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right) \\ $$
Commented by Tawa11 last updated on 02/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by MJS_new last updated on 02/Jul/22
$$\begin{cases}{\mathrm{2}{y}^{\mathrm{1}/\mathrm{2}} {x}−{x}^{\mathrm{1}/\mathrm{2}} −\mathrm{2}{y}^{\mathrm{3}/\mathrm{3}} =\mathrm{0}}\\{{x}^{\mathrm{3}/\mathrm{2}} +{yx}^{\mathrm{1}/\mathrm{2}} −\mathrm{3}{y}^{\mathrm{1}/\mathrm{2}} =\mathrm{0}}\end{cases} \\ $$$${y}={px} \\ $$$$\begin{cases}{\sqrt{{x}}\left(\mathrm{2}\sqrt{{p}}\left({p}−\mathrm{1}\right){x}+\mathrm{1}\right)=\mathrm{0}}\\{\sqrt{{x}}\left(\left({p}+\mathrm{1}\right){x}−\mathrm{3}\sqrt{{p}}\right)=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} =\mathrm{0}\:\Rightarrow\:{y}_{\mathrm{1}} =\mathrm{0} \\ $$$$\begin{cases}{{x}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{p}}\left({p}−\mathrm{1}\right)}}\\{{x}=\frac{\mathrm{3}\sqrt{{p}}}{{p}+\mathrm{1}}}\end{cases} \\ $$$$\Rightarrow \\ $$$${p}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{6}}{p}+\frac{\mathrm{1}}{\mathrm{6}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${p}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{x}_{\mathrm{2}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow\:{y}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$${p}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{x}_{\mathrm{3}} =\sqrt{\mathrm{2}}\:\Rightarrow\:{y}_{\mathrm{3}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 03/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$