Question Number 217732 by ArshadS last updated on 19/Mar/25
$$\begin{cases}{{x}+{y}={xy}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{25}}\end{cases};\:\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} =? \\ $$
Commented by Ghisom last updated on 19/Mar/25
$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\mathrm{571}\pm\mathrm{4}\sqrt{\mathrm{26}} \\ $$
Answered by Wuji last updated on 19/Mar/25
$${S}={x}+{y}\:,\:{P}={xy} \\ $$$$\left({xy}\right)^{\mathrm{2}} ={P}^{\mathrm{2}} ={S}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({xy}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\left(\mathrm{25}\right)^{\mathrm{2}} −\mathrm{2}{S}^{\mathrm{2}} \:\:=\mathrm{625}−\mathrm{2}{S}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={S}^{\mathrm{2}} −\mathrm{2}{P}\:\:\Rightarrow{S}^{\mathrm{2}} −\mathrm{2}{S}=\mathrm{25} \\ $$$${S}^{\mathrm{2}} =\mathrm{2}{S}+\mathrm{25} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\mathrm{625}−\mathrm{2}\left(\mathrm{2}{S}+\mathrm{25}\right)=\mathrm{625}−\mathrm{4}{S}−\mathrm{50} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\mathrm{575}−\mathrm{4}{S} \\ $$$${from}\:{S}^{\mathrm{2}} −\mathrm{2}{S}−\mathrm{25}=\mathrm{0} \\ $$$${S}=\frac{\mathrm{2}\pm\sqrt{\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}×\left(−\mathrm{25}\right)}}{\mathrm{2}}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{100}}}{\mathrm{2}}\:=\frac{\mathrm{2}\pm\sqrt{\mathrm{104}}}{\mathrm{2}} \\ $$$${S}=\frac{\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{26}}}{\mathrm{2}}\:=\mathrm{1}\pm\sqrt{\mathrm{26}} \\ $$$${S}\:=\mathrm{1}+\sqrt{\mathrm{26}}\:\:\:{S}>\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\mathrm{575}−\mathrm{4}\left(\mathrm{1}+\sqrt{\mathrm{26}}\right) \\ $$$$\left.{x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\mathrm{575}−\mathrm{4}−\mathrm{4}\sqrt{\mathrm{26}}\right)=\mathrm{571}−\mathrm{4}\sqrt{\mathrm{26}} \\ $$$$\therefore{x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\mathrm{571}−\mathrm{4}\sqrt{\mathrm{26}} \\ $$
Commented by ArshadS last updated on 22/Mar/25
$${Thanks}\:{a}\:{lot}! \\ $$
Answered by Rasheed.Sindhi last updated on 19/Mar/25
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{25} \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{25} \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+{y}\right)=\mathrm{25} \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+{y}\right)−\mathrm{25}=\mathrm{0} \\ $$$${x}+{y}={xy}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{100}}}{\mathrm{2}}=\mathrm{1}\pm\sqrt{\mathrm{26}}\: \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{25}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +\mathrm{2}\left({xy}\right)^{\mathrm{2}} =\mathrm{625} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\mathrm{625}−\mathrm{2}\left(\mathrm{1}\pm\sqrt{\mathrm{26}}\:\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{625}−\mathrm{2}\left(\mathrm{1}+\mathrm{26}\pm\mathrm{2}\sqrt{\mathrm{26}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{625}−\mathrm{54}\mp\mathrm{4}\sqrt{\mathrm{26}}\: \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{571}\mp\mathrm{4}\sqrt{\mathrm{26}}\: \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{571}\pm\mathrm{4}\sqrt{\mathrm{26}}\: \\ $$
Commented by ArshadS last updated on 22/Mar/25
$${Thanks}\:{sir}! \\ $$