Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 73047 by mathmax by abdo last updated on 05/Nov/19

solve inside Z^3 x^2 +y^2 +z^(2 ) =2xyz

$${solve}\:{inside}\:{Z}^{\mathrm{3}} \:\:\:\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}\:} =\mathrm{2}{xyz} \\ $$

Answered by mind is power last updated on 07/Nov/19

2∣2xyz⇒2∣x^2 +y^2 +z^2 ⇒x,y,z can not bee all odd lets asum x=2x1⇔ 4a^2 +y^2 +z^2 =4ayz⇒y^2 +z^2 =4(x_1 yz−x^2 1) ⇒4∣y^2 +z^2 ⇒y=2y_1 &z=2z_1 equation becom 4(x_1 ^2 +y_1 ^2 +z_1 ^2 )=8x_1 y_1 z_1 ⇔x_1 ^2 +y_1 ^2 +z_1 ^2 =2x_1 y_1 z_1 ⇒2∣x_1 ,y_1 ,z_1 so ∀n∈IN 2^n ∣x,y,z⇒x=y=z=0

$$\mathrm{2}\mid\mathrm{2xyz}\Rightarrow\mathrm{2}\mid\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x},\mathrm{y},\mathrm{z}\:\mathrm{can}\:\mathrm{not}\:\mathrm{bee}\:\mathrm{all}\:\mathrm{odd} \\ $$$$\mathrm{lets}\:\mathrm{asum}\:\mathrm{x}=\mathrm{2x1}\Leftrightarrow \\ $$$$\mathrm{4a}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{4ayz}\Rightarrow\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{4}\left(\mathrm{x}_{\mathrm{1}} \mathrm{yz}−\mathrm{x}^{\mathrm{2}} \mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{4}\mid\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \Rightarrow\mathrm{y}=\mathrm{2y}_{\mathrm{1}} \&\mathrm{z}=\mathrm{2z}_{\mathrm{1}} \\ $$$$\mathrm{equation}\:\mathrm{becom} \\ $$$$\mathrm{4}\left(\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{y}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} \right)=\mathrm{8x}_{\mathrm{1}} \mathrm{y}_{\mathrm{1}} \mathrm{z}_{\mathrm{1}} \\ $$$$\Leftrightarrow\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{y}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{2x}_{\mathrm{1}} \mathrm{y}_{\mathrm{1}} \mathrm{z}_{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}\mid\mathrm{x}_{\mathrm{1}} ,\mathrm{y}_{\mathrm{1}} ,\mathrm{z}_{\mathrm{1}} \\ $$$$\mathrm{so}\:\forall\mathrm{n}\in\mathrm{IN}\:\:\mathrm{2}^{\mathrm{n}} \mid\mathrm{x},\mathrm{y},\mathrm{z}\Rightarrow\mathrm{x}=\mathrm{y}=\mathrm{z}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com