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Question Number 54786 by behi83417@gmail.com last updated on 10/Feb/19
solve for:a,b,c,d∈R a^2 =b+(√c) b^2 =c+(√d) c^2 =d+(√a) d^2 =a+(√b)
$${solve}\:{for}:{a},{b},{c},{d}\in\boldsymbol{{R}} \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}} ={b}+\sqrt{{c}} \\ $$$$\:\:\:\:\:\:{b}^{\mathrm{2}} ={c}+\sqrt{{d}} \\ $$$$\:\:\:\:\:\:{c}^{\mathrm{2}} ={d}+\sqrt{{a}} \\ $$$$\:\:\:\:\:\:{d}^{\mathrm{2}} ={a}+\sqrt{{b}} \\ $$
Commented by mr W last updated on 11/Feb/19
a=b=c=d=0 or (((((9+(√(69))))^(1/3) +((9−(√(69))))^(1/3) )/( ((18))^(1/3) )))^2 ≈1.7548 ?
$${a}={b}={c}={d}=\mathrm{0}\:{or}\:\left(\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}+\sqrt{\mathrm{69}}}+\sqrt[{\mathrm{3}}]{\mathrm{9}−\sqrt{\mathrm{69}}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{18}}}\right)^{\mathrm{2}} \approx\mathrm{1}.\mathrm{7548}\:? \\ $$
Commented by behi83417@gmail.com last updated on 11/Feb/19
right answer sir.please post your solution.thank you very much.
$${right}\:{answer}\:{sir}.{please}\:{post}\:{your}\: \\ $$$${solution}.{thank}\:{you}\:{very}\:{much}. \\ $$
Commented by mr W last updated on 11/Feb/19
due to symmetry: a=b=c=d=t^2 t^4 =t^2 +t t(t^3 −t−1)=0 ⇒t=0 ⇒t^3 −t−1=0 let t=u+v (u+v)^3 −3uv(u+v)−(u^3 +v^3 )=0 u^3 v^3 =(1/(27)) u^3 +v^3 =1 x^2 −x+(1/(27))=0 u^3 (v^3 )=(1/2)(1±(√(1−(4/(27)))))=(1/(18))(9±(√(69))) u=(((9+(√(69))))^(1/3) /( ((18))^(1/3) )) v=(((9−(√(69))))^(1/3) /( ((18))^(1/3) )) ⇒t=((((9+(√(69))))^(1/3) +((9−(√(69))))^(1/3) )/( ((18))^(1/3) )) a=b=c=d=t^2 =0 or (((((9+(√(69))))^(1/3) +((9−(√(69))))^(1/3) )/( ((18))^(1/3) )))^2
$${due}\:{to}\:{symmetry}:\:{a}={b}={c}={d}={t}^{\mathrm{2}} \\ $$$${t}^{\mathrm{4}} ={t}^{\mathrm{2}} +{t} \\ $$$${t}\left({t}^{\mathrm{3}} −{t}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{3}} −{t}−\mathrm{1}=\mathrm{0} \\ $$$${let}\:{t}={u}+{v} \\ $$$$\left({u}+{v}\right)^{\mathrm{3}} −\mathrm{3}{uv}\left({u}+{v}\right)−\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$${u}^{\mathrm{3}} {v}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{27}} \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} =\mathrm{1} \\ $$$${x}^{\mathrm{2}} −{x}+\frac{\mathrm{1}}{\mathrm{27}}=\mathrm{0} \\ $$$${u}^{\mathrm{3}} \:\left({v}^{\mathrm{3}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\pm\sqrt{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{27}}}\right)=\frac{\mathrm{1}}{\mathrm{18}}\left(\mathrm{9}\pm\sqrt{\mathrm{69}}\right) \\ $$$${u}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}+\sqrt{\mathrm{69}}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{18}}} \\ $$$${v}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}−\sqrt{\mathrm{69}}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{18}}} \\ $$$$\Rightarrow{t}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}+\sqrt{\mathrm{69}}}+\sqrt[{\mathrm{3}}]{\mathrm{9}−\sqrt{\mathrm{69}}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{18}}} \\ $$$${a}={b}={c}={d}={t}^{\mathrm{2}} =\mathrm{0}\:{or}\:\left(\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}+\sqrt{\mathrm{69}}}+\sqrt[{\mathrm{3}}]{\mathrm{9}−\sqrt{\mathrm{69}}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{18}}}\right)^{\mathrm{2}} \\ $$
Commented by behi83417@gmail.com last updated on 12/Feb/19
thanks again sir.it is perfect. is there any sulotion without using symmetry?
$${thanks}\:{again}\:{sir}.{it}\:{is}\:{perfect}. \\ $$$${is}\:{there}\:{any}\:{sulotion}\:{without}\:{using} \\ $$$${symmetry}? \\ $$
Commented by mr W last updated on 12/Feb/19
i am not sure if there are other solutions.
$${i}\:{am}\:{not}\:{sure}\:{if}\:{there}\:{are}\:{other}\:{solutions}. \\ $$

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