{ ((x^2 +y^2 =2yz+2−z^2 )),((z=4022−x−y)) :} x,y,z∈Z^+ . then z=


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Question Number 142556 by iloveisrael last updated on 02/Jun/21
${ ((x^2 +y^2 =2yz+2−z^2 )),((z=4022−x−y)) :} x,y,z∈Z^+ . then z=$
${\begin{cases} x^{2} + y^{2} = 2 y z + 2 - z^{2} \\ z = 4022 - x - y \end{cases}$ $x, y, z \in Z^{+} . t h e n z =$
	Answered by iloveisrael last updated on 02/Jun/21
	$⇒y^2 +z^2 −2yz = 2−x^2 ⇒(y−z)^2 = 2−x^2 ...(i) ⇒y=4022−x−z...(ii) ⇒(4022−x−2z)^2 =2−x^2 test for x=1 ⇒(4021−2z)^2 =1 ⇒ { ((4021−2z =1⇒2z=4020⇒z=2010)),((4021−2z=−1⇒4022=2z⇒z=2011)) :} if x=1∧z=2010 then y=2011 if x=1 ∧z=2011 then y=2010 solution (x,y,z)=(1,2011,2010) or (1,2010,2011)$
	$\Rightarrow y^{2} + z^{2} - 2 y z = 2 - x^{2}$ $\Rightarrow {(y - z)}^{2} = 2 - x^{2} . . . (i)$ $\Rightarrow y = 4022 - x - z . . . (i i)$ $\Rightarrow {(4022 - x - 2 z)}^{2} = 2 - x^{2}$ $t e s t f o r x = 1$ $\Rightarrow {(4021 - 2 z)}^{2} = 1$ $\Rightarrow {\begin{cases} 4021 - 2 z = 1 \Rightarrow 2 z = 4020 \Rightarrow z = 2010 \\ 4021 - 2 z = - 1 \Rightarrow 4022 = 2 z \Rightarrow z = 2011 \end{cases}$ $i f x = 1 \land z = 2010 t h e n y = 2011$ $i f x = 1 \land z = 2011 t h e n y = 2010$ $s o l u t i o n (x, y, z) = (1, 2011, 2010)$ $o r (1, 2010, 2011)$
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