Question Number 211416 by AlagaIbile last updated on 08/Sep/24
Commented by AlagaIbile last updated on 08/Sep/24
$$\:{Kindly}\:{post}\:{the}\:{solution} \\ $$
Answered by Frix last updated on 09/Sep/24
![x+y=1 ⇒ y=1−x x^2 +y^2 =s ⇒ x=(1/2)±((√(2s−1))/2) Let x=(1/2)+((√(2s−1))/2) ⇒ we only get (1/2)Σ (x^2 +y^2 ) x^(30) +y^(30) =z [the value of z is irrelevant] x^(30) +(1−x)^(30) =z ⇒ (s^(15) /(16384))+((105s^(14) )/(8192))+... s^(15) +210s^(14) +... ⇒ (1/2)Σ s =(1/2)Σ (x^2 +y^2 ) =−210 ⇒ Answer is −420](https://www.tinkutara.com/question/Q211426.png)
$${x}+{y}=\mathrm{1}\:\Rightarrow\:{y}=\mathrm{1}−{x} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={s}\:\Rightarrow\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{2}{s}−\mathrm{1}}}{\mathrm{2}} \\ $$$$\mathrm{Let}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}{s}−\mathrm{1}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{we}\:\mathrm{only}\:\mathrm{get}\:\frac{\mathrm{1}}{\mathrm{2}}\Sigma\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$${x}^{\mathrm{30}} +{y}^{\mathrm{30}} ={z}\:\:\:\:\:\left[\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{z}\:\mathrm{is}\:\mathrm{irrelevant}\right] \\ $$$${x}^{\mathrm{30}} +\left(\mathrm{1}−{x}\right)^{\mathrm{30}} ={z} \\ $$$$\Rightarrow \\ $$$$\frac{{s}^{\mathrm{15}} }{\mathrm{16384}}+\frac{\mathrm{105}{s}^{\mathrm{14}} }{\mathrm{8192}}+… \\ $$$${s}^{\mathrm{15}} +\mathrm{210}{s}^{\mathrm{14}} +… \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\Sigma\:{s}\:=\frac{\mathrm{1}}{\mathrm{2}}\Sigma\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:=−\mathrm{210} \\ $$$$\Rightarrow \\ $$$$\mathrm{Answer}\:\mathrm{is}\:−\mathrm{420} \\ $$
Commented by AlagaIbile last updated on 09/Sep/24
$$\:{Great}\:.\:{The}\:{required}\:{answer}\:{is}\:−\mathrm{210}\:{though} \\ $$
Commented by Frix last updated on 09/Sep/24
$$\mathrm{I}\:\mathrm{tbink}\:\mathrm{that}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{must}\:\mathrm{be}\:−\mathrm{420}. \\ $$$$\mathrm{Easy}\:\mathrm{example}: \\ $$$${x}+{y}=\mathrm{1} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} ={z} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={s}\:\Rightarrow\:{x}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}{s}−\mathrm{1}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +\left(\mathrm{1}−{x}\right)^{\mathrm{4}} ={z} \\ $$$${s}^{\mathrm{2}} +\mathrm{2}{s}−\mathrm{1}=\mathrm{2}{z} \\ $$$$\overset{???} {\Rightarrow}\:\mathrm{Answer}\:\mathrm{is}\:−\mathrm{2} \\ $$$$\mathrm{But} \\ $$$${x}^{\mathrm{4}} +\left(\mathrm{1}−{x}\right)^{\mathrm{4}} ={z} \\ $$$$\mathrm{Transforms}\:\mathrm{to} \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} +\frac{\mathrm{3}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{8}{z}−\mathrm{1}}{\mathrm{16}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{−\frac{\mathrm{3}}{\mathrm{4}}\pm\frac{\sqrt{{z}+\mathrm{1}}}{\:\sqrt{\mathrm{2}}}}\:\left(\mathrm{4}\:\mathrm{solutions}\right) \\ $$$${y}=\mathrm{1}−{x} \\ $$$$\Sigma\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:=−\mathrm{4} \\ $$$$\mathrm{Try}\:\mathrm{with}\:{z}=\mathrm{7}\:\Rightarrow \\ $$$${x}=\left\{\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}},\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}},\:\frac{\mathrm{1}−\sqrt{\mathrm{11}}\mathrm{i}}{\mathrm{2}},\:\frac{\mathrm{1}+\sqrt{\mathrm{11}}\mathrm{i}}{\mathrm{2}}\right\} \\ $$$${y}=\left\{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}},\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}},\:\frac{\mathrm{1}+\sqrt{\mathrm{11}}\mathrm{i}}{\mathrm{2}},\:\frac{\mathrm{1}−\sqrt{\mathrm{11}}\mathrm{i}}{\mathrm{2}}\right\} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left\{\mathrm{3},\:\mathrm{3},\:−\mathrm{5},\:−\mathrm{5}\right\} \\ $$$$\Sigma\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:=−\mathrm{4} \\ $$
Commented by Frix last updated on 09/Sep/24
$$\mathrm{I}\:\mathrm{missed}\:“\mathrm{all}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ''.\:\mathrm{Each} \\ $$$$\mathrm{value}\:\mathrm{occurs}\:\mathrm{2}\:\mathrm{times}\:\mathrm{since}\:\mathrm{we}\:\mathrm{have}\:\mathrm{pairs} \\ $$$$\left({x},\:{y}\right)=\left({u}−\sqrt{{v}},\:{u}+\sqrt{{v}}\right)\:\mathrm{and}\:\left({u}+\sqrt{{v}},\:{u}−\sqrt{{v}}\right) \\ $$$$\mathrm{which}\:\mathrm{give}\:\mathrm{the}\:\mathrm{same}\:\mathrm{value}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} . \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{Answer}\:\mathrm{is}\:−\mathrm{210} \\ $$