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Question Number 176718 by Frix last updated on 25/Sep/22

x^3 +y^3 =z^2 x^3 +z^3 =y^2 y^3 +z^3 =x^2 obviously x=y=z=0∨(1/2) trying to totally solve it let y=px∧z=qx (p^3 +1)x^3 =q^2 x^2 (q^3 +1)x^3 =p^2 x^2 (p^3 +q^3 )x^3 =x^2 ⇒ x=0 ⇒ y=0∧z=0 ★ x=(q^2 /(p^3 +1)) x=(p^2 /(q^3 +1)) x=(1/(p^3 +q^3 )) ⇒ (p^2 /(q^3 +1))=(q^2 /(p^3 +1)) (p^2 /(q^3 +1))=(1/(p^3 +q^3 )) ========== p^5 +p^2 −q^5 −q^2 =0 p^5 +p^2 q^3 −q^3 −1=0 subtracting both p^2 (q^3 −1)+q^5 −q^3 +q^2 −1=0 (q−1)(p^2 (q^2 +q+1)+q^4 +q^3 +q+1)=0 ⇒ q=1 ⇒ p^5 +p^2 −2=0 (p−1)(p^4 +p^3 +2p+2)=0 ⇒ p=1 ⇒ x=y=z=(1/2) ★ p^4 +p^3 +2p+2=0 no useable exact solutions p≈−.975564±.528237i ⇒ x≈.33635∓.515329i ∧ y≈−.0559113±.680406i ∧ z=x ★ p≈.475564±1.18273i ⇒ x≈−.586346±.562464i ∧ y≈−.944089∓.426001i ∧ z=x ★ p^2 (q^2 +q+1)+q^4 +q^3 +q+1=0 p^2 =−(((q+1)^2 (q^2 −q+1))/(q^2 +q+1)) we can be sure that (q≠−1⇒p=0) ⇒ p^2 <0 ⇒ p=±(√((q^2 −q+1)/(q^2 +q+1)))(q+1)i ...I′ll continue later

$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} ={z}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +{z}^{\mathrm{3}} ={y}^{\mathrm{2}} \\ $$$${y}^{\mathrm{3}} +{z}^{\mathrm{3}} ={x}^{\mathrm{2}} \\ $$$$\mathrm{obviously}\:{x}={y}={z}=\mathrm{0}\vee\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{trying}\:\mathrm{to}\:\mathrm{totally}\:\mathrm{solve}\:\mathrm{it} \\ $$$$\mathrm{let}\:{y}={px}\wedge{z}={qx} \\ $$$$\left({p}^{\mathrm{3}} +\mathrm{1}\right){x}^{\mathrm{3}} ={q}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\left({q}^{\mathrm{3}} +\mathrm{1}\right){x}^{\mathrm{3}} ={p}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right){x}^{\mathrm{3}} ={x}^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}=\mathrm{0}\:\Rightarrow\:{y}=\mathrm{0}\wedge{z}=\mathrm{0}\:\bigstar \\ $$$${x}=\frac{{q}^{\mathrm{2}} }{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$${x}=\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}} \\ $$$${x}=\frac{\mathrm{1}}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} } \\ $$$$\Rightarrow \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}}=\frac{{q}^{\mathrm{2}} }{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}}=\frac{\mathrm{1}}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} } \\ $$$$========== \\ $$$${p}^{\mathrm{5}} +{p}^{\mathrm{2}} −{q}^{\mathrm{5}} −{q}^{\mathrm{2}} =\mathrm{0} \\ $$$${p}^{\mathrm{5}} +{p}^{\mathrm{2}} {q}^{\mathrm{3}} −{q}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{subtracting}\:\mathrm{both} \\ $$$${p}^{\mathrm{2}} \left({q}^{\mathrm{3}} −\mathrm{1}\right)+{q}^{\mathrm{5}} −{q}^{\mathrm{3}} +{q}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({q}−\mathrm{1}\right)\left({p}^{\mathrm{2}} \left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right)+{q}^{\mathrm{4}} +{q}^{\mathrm{3}} +{q}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:{q}=\mathrm{1}\:\Rightarrow\:{p}^{\mathrm{5}} +{p}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\left({p}−\mathrm{1}\right)\left({p}^{\mathrm{4}} +{p}^{\mathrm{3}} +\mathrm{2}{p}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:{p}=\mathrm{1}\:\Rightarrow\:{x}={y}={z}=\frac{\mathrm{1}}{\mathrm{2}}\:\bigstar \\ $$$$\:\:\:\:\:{p}^{\mathrm{4}} +{p}^{\mathrm{3}} +\mathrm{2}{p}+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{p}\approx−.\mathrm{975564}\pm.\mathrm{528237i} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}\approx.\mathrm{33635}\mp.\mathrm{515329i} \\ $$$$\:\:\:\:\:\wedge\:{y}\approx−.\mathrm{0559113}\pm.\mathrm{680406i} \\ $$$$\:\:\:\:\:\wedge\:{z}={x}\:\bigstar \\ $$$$\:\:\:\:\:{p}\approx.\mathrm{475564}\pm\mathrm{1}.\mathrm{18273i} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}\approx−.\mathrm{586346}\pm.\mathrm{562464i} \\ $$$$\:\:\:\:\:\wedge\:{y}\approx−.\mathrm{944089}\mp.\mathrm{426001i} \\ $$$$\:\:\:\:\:\wedge\:{z}={x}\:\bigstar \\ $$$${p}^{\mathrm{2}} \left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right)+{q}^{\mathrm{4}} +{q}^{\mathrm{3}} +{q}+\mathrm{1}=\mathrm{0} \\ $$$${p}^{\mathrm{2}} =−\frac{\left({q}+\mathrm{1}\right)^{\mathrm{2}} \left({q}^{\mathrm{2}} −{q}+\mathrm{1}\right)}{{q}^{\mathrm{2}} +{q}+\mathrm{1}} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{be}\:\mathrm{sure}\:\mathrm{that}\:\left({q}\neq−\mathrm{1}\Rightarrow{p}=\mathrm{0}\right) \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\:{p}=\pm\sqrt{\frac{{q}^{\mathrm{2}} −{q}+\mathrm{1}}{{q}^{\mathrm{2}} +{q}+\mathrm{1}}}\left({q}+\mathrm{1}\right)\mathrm{i} \\ $$$$...\mathrm{I}'\mathrm{ll}\:\mathrm{continue}\:\mathrm{later} \\ $$

Commented by Tawa11 last updated on 25/Sep/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by behi834171 last updated on 26/Sep/22

y^3 −z^3 =z^2 −y^2 ⇒(y−z)(y^2 +z^2 +yz)=−(y−z)(y+z) ⇒x=y=z=0 or (1/2) y^2 +z^2 +yz=−(y+z) z^2 +x^2 +xz=−(x+z) x^2 +y^2 +xy=−(x+y) ⇒y^2 −x^2 +z(y−x)=−(y−x)⇒ ⇒(y−x)(y+x+z+1)=0 ⇒y+x+z=−1 ...

$${y}^{\mathrm{3}} −{z}^{\mathrm{3}} ={z}^{\mathrm{2}} −{y}^{\mathrm{2}} \Rightarrow\left({y}−{z}\right)\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{yz}\right)=−\left({y}−{z}\right)\left({y}+{z}\right) \\ $$$$\Rightarrow{x}={y}={z}=\mathrm{0}\:\:{or}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{yz}=−\left({y}+{z}\right) \\ $$$${z}^{\mathrm{2}} +{x}^{\mathrm{2}} +{xz}=−\left({x}+{z}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=−\left({x}+{y}\right) \\ $$$$\Rightarrow{y}^{\mathrm{2}} −{x}^{\mathrm{2}} +{z}\left({y}−{x}\right)=−\left({y}−{x}\right)\Rightarrow \\ $$$$\Rightarrow\left({y}−{x}\right)\left({y}+{x}+{z}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{y}+{x}+{z}=−\mathrm{1} \\ $$$$... \\ $$$$ \\ $$

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