Given that x,y∈R ∀ x^2 −y^2 =32, (x+y)^4 +(x−y)^4 =4352, Find the value of x^2 +y^2 .


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Question Number 115555 by ZiYangLee last updated on 26/Sep/20

$Given that x, y \in R \forall x^{2} - y^{2} = 32,$ ${(x + y)}^{4} + {(x - y)}^{4} = 4352,$ $Find the value of x^{2} + y^{2} .$
Commented by Rasheed.Sindhi last updated on 27/Sep/20

$A n O ther W ay$ $(B y c h a n g i n g i n t o$ $r e c i p r o c a l f o r m)$ ${(x + y)}^{4} + {(x - y)}^{4} = 4352$ $\frac{{(x + y)}^{4}}{{(x + y)}^{2} {(x - y)}^{2}} + \frac{{(x - y)}^{4}}{{(x + y)}^{2} {(x - y)}^{2}}$ $= \frac{4352}{{(x + y)}^{2} {(x - y)}^{2}}$ $\frac{{(x + y)}^{2}}{{(x - y)}^{2}} + \frac{{(x - y)}^{2}}{{(x + y)}^{2}} = \frac{4352}{{(x^{2} - y^{2})}^{2}}$ $\frac{{(x + y)}^{2}}{{(x - y)}^{2}} + \frac{{(x - y)}^{2}}{{(x + y)}^{2}} = \frac{4352}{{(32)}^{2}} = \frac{17}{4}$ $\frac{{(x + y)}^{2}}{{(x - y)}^{2}} = z$ $z + \frac{1}{z} = \frac{17}{4}$ $4 z^{2} - 17 z + 4 = 0$ $(z - 4) (4 z - 1) = 0$ $z = 4 \lor z = \frac{1}{4}$ $\frac{{(x + y)}^{2}}{{(x - y)}^{2}} = 4, \frac{1}{4}$ $\frac{{(x + y)}^{2}}{{(32 / (x + y))}^{2}} = 4, \frac{1}{4}$ ${(x + y)}^{4} = 4 {(32)}^{2}$ ${(x + y)}^{2} = \pm 64$ ${(x - y)}^{2} = {(32 / (x + y))}^{2} = 32^{2} / (\pm 64) = \pm 16$ $2 (x^{2} + y^{2}) = {(x + y)}^{2} + {(x - y)}^{2}$ $x^{2} + y^{2} = \frac{1}{2} (\pm 64 \pm 16) = \pm 40$
	Answered by mr W last updated on 26/Sep/20

	$u = x + y$ $v = x - y$ $x^{2} - y^{2} = (x + y) (x - y) = 32$ $\Rightarrow u v = 32$ $u^{4} + v^{4} = 4352$ $u^{4} + v^{4} + 2 u^{2} v^{2} = 4352 + 2 {(u v)}^{2}$ ${(u^{2} + v^{2})}^{2} = 4352 + 2 \times 32^{2} = 6400 = 80^{2}$ $\Rightarrow u^{2} + v^{2} = 80$ $u^{2} + v^{2} + 2 u v = 80 + 2 (u v)$ ${(u + v)}^{2} = 80 + 2 \times 32 = 144 = 12^{2}$ $\Rightarrow u + v = \pm 12$ $u^{2} + v^{2} - 2 u v = 80 - 2 (u v)$ ${(u - v)}^{2} = 80 - 2 \times 32 = 16 = 4^{2}$ $\Rightarrow u - v = \pm 4$ $x = \frac{u + v}{2} = \pm 6$ $y = \frac{u - v}{2} = \pm 2$ $\Rightarrow x^{2} + y^{2} = 36 + 4 = 40$
	Commented by ZiYangLee last updated on 26/Sep/20

	$Thanks very much . .$
	Answered by ruwedkabeh last updated on 26/Sep/20

	${(x + y)}^{4} + {(x - y)}^{4} = 4352,$ $x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 {x y}^{3} + y^{4} + x^{4} - 4 x^{3} y + 6 x^{2} y^{2} - 4 {x y}^{3} + y^{4} = 4352$ $2 x^{4} + 12 x^{2} y^{2} + 2 y^{4} = 4352$ $x^{4} + 6 x^{2} y^{2} + y^{4} = 2176$ ${(x^{2} - y^{2})}^{2} + 8 x^{2} y^{2} = 2176$ ${(32)}^{2} + 8 x^{2} y^{2} = 2176$ $1024 + 8 x^{2} y^{2} = 2176$ $8 x^{2} y^{2} = 1152$ $x^{2} y^{2} = 144$ ${(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2} = 1024 + 576 = 1600$ $x^{2} + y^{2} = 40$
	Commented by ZiYangLee last updated on 26/Sep/20

	$Thank you!$
	Answered by Olaf last updated on 26/Sep/20

	${(x + y)}^{4} + {(x - y)}^{4} =$ ${[{(x + y)}^{2} + {(x - y)}^{2}]}^{2} - 2 {(x + y)}^{2} {(x - y)}^{2} =$ ${(2 x^{2} + 2 y^{2})}^{2} - 2 {(x^{2} - y^{2})}^{2} =$ $4 {(x^{2} + y^{2})}^{2} - 2 \times 32^{2} = 4352$ ${(x^{2} + y^{2})}^{2} = \frac{4352 + 2048}{4} = \frac{6400}{4} = 1600$ $x^{2} + y^{2} = 40$
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