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Question Number 201139 by cherokeesay last updated on 30/Nov/23

Commented by Frix last updated on 30/Nov/23

$x \approx 6395.12283 \land y \approx 171.458282$ $Exact solution :$ $x = 32 (8 (5 r^{2} + 6 r + 9) \sqrt{r - 1} + 16 r^{2} + 29 r + 38)$ $y = 16 (4 (r^{2} + 4 r - 4) \sqrt{r - 1} + 16 r^{2} - 4 r - 15)$ $r = \sqrt[3]{2}$ $I got this by substituting$ $x = p^{4} \land y = p^{4} q^{4} \land p, q > 0$ $but {it}^{'} s too much work to type it here .$
	Answered by ajfour last updated on 01/Dec/23
	$(√x)=p , (√y)=q , a=(1/4) , z=((2p+q)/(p^2 +q^2 )) p(a−((2p+q)/(p^2 +q^2 )))^2 =4 q(a+((2p+q)/(p^2 +q^2 )))^2 =1 a−z=(2/( (√p))) a+z=(1/( (√q))) 2z=(((√p)−2(√q))/( (√p)(√q)))=2(((2p+q)/( p^2 +q^2 ))) 2a=(((√p)+2(√q))/( (√p)(√q))) a^2 −z^2 =(2/( (√p)(√q))) z{((16)/((a−z)^4 ))+(1/((a+z)^4 ))}=(8/((a−z)^2 ))+(1/((a+z)^2 )) 16z(a+z)^4 +z(a−z)^4 =8(a−z)^2 (a+z)^4 +(a+z)^2 (a−z)^4 or if z=acos 2θ =a(h−k) h+k=1 4(h^2 −k^2 ){(1/k^4 )+(1/(16h^4 ))} =(2/k^2 )+(1/(4h^2 )) ⇒ ((4h^2 )/k^4 )−(k^2 /(4h^4 ))+(1/(4h^2 ))−(4/k^2 )=(2/k^2 )+(1/(4h^2 )) ⇒ ((4h^2 )/k^4 )−(k^2 /(4h^4 ))=(6/k^2 ) 16h^6 −k^6 =24k^2 h^4 say tan θ=(k/h)=t t^6 +24t^2 −16=0 ⇒ t^2 =2((4)^(1/3) −(2)^(1/3) ) z=acos 2θ (√x)=p=(4/((a−z)^2 ))=(1/(a^2 sin^4 θ))=((16)/k^2 ) (√y)=q=(1/((a+z)^2 ))=(1/(4a^2 cos^4 θ))=(4/h^2 ) (k/h)=t ; h+k=1 ⇒ h=(1/(1+t))=(1/(1+2((4)^(1/3) −(2)^(1/3) ))) & k=((2((4)^(1/3) −(2)^(1/3) ))/(1+2((4)^(1/3) −(2)^(1/3) ))) (√x)=16[((1+2((4)^(1/3) −(2)^(1/3) ))/(2((4)^(1/3) −(2)^(1/3) )))]^2 ⇒ x≈ 69121.4399 (√y)=4[1+2((4)^(1/3) −(2)^(1/3) )]^2 ≈ 120.024524$
	$\sqrt{x} = p, \sqrt{y} = q, a = \frac{1}{4}, z = \frac{2 p + q}{p^{2} + q^{2}}$ $p {(a - \frac{2 p + q}{p^{2} + q^{2}})}^{2} = 4$ $q {(a + \frac{2 p + q}{p^{2} + q^{2}})}^{2} = 1$ $a - z = \frac{2}{\sqrt{p}}$ $a + z = \frac{1}{\sqrt{q}}$ $2 z = \frac{\sqrt{p} - 2 \sqrt{q}}{\sqrt{p} \sqrt{q}} = 2 (\frac{2 p + q}{p^{2} + q^{2}})$ $2 a = \frac{\sqrt{p} + 2 \sqrt{q}}{\sqrt{p} \sqrt{q}}$ $a^{2} - z^{2} = \frac{2}{\sqrt{p} \sqrt{q}}$ $z {\frac{16}{{(a - z)}^{4}} + \frac{1}{{(a + z)}^{4}}} = \frac{8}{{(a - z)}^{2}} + \frac{1}{{(a + z)}^{2}}$ $16 z {(a + z)}^{4} + z {(a - z)}^{4}$ $= 8 {(a - z)}^{2} {(a + z)}^{4} + {(a + z)}^{2} {(a - z)}^{4}$ $o r i f z = a \cos 2 θ = a (h - k)$ $h + k = 1$ $4 (h^{2} - k^{2}) {\frac{1}{k^{4}} + \frac{1}{16 h^{4}}}$ $= \frac{2}{k^{2}} + \frac{1}{4 h^{2}}$ $\Rightarrow \frac{4 h^{2}}{k^{4}} - \frac{k^{2}}{4 h^{4}} + \frac{1}{4 h^{2}} - \frac{4}{k^{2}} = \frac{2}{k^{2}} + \frac{1}{4 h^{2}}$ $\Rightarrow \frac{4 h^{2}}{k^{4}} - \frac{k^{2}}{4 h^{4}} = \frac{6}{k^{2}}$ $16 h^{6} - k^{6} = 24 k^{2} h^{4}$ $s a y \tan θ = \frac{k}{h} = t$ $t^{6} + 24 t^{2} - 16 = 0$ $\Rightarrow t^{2} = 2 (\sqrt[3]{4} - \sqrt[3]{2})$ $z = a \cos 2 θ$ $\sqrt{x} = p = \frac{4}{{(a - z)}^{2}} = \frac{1}{a^{2} \sin^{4} θ} = \frac{16}{k^{2}}$ $\sqrt{y} = q = \frac{1}{{(a + z)}^{2}} = \frac{1}{4 a^{2} \cos^{4} θ} = \frac{4}{h^{2}}$ $\frac{k}{h} = t; h + k = 1$ $\Rightarrow h = \frac{1}{1 + t} = \frac{1}{1 + 2 (\sqrt[3]{4} - \sqrt[3]{2})}$ $& k = \frac{2 (\sqrt[3]{4} - \sqrt[3]{2})}{1 + 2 (\sqrt[3]{4} - \sqrt[3]{2})}$ $\sqrt{x} = 16 {[\frac{1 + 2 (\sqrt[3]{4} - \sqrt[3]{2})}{2 (\sqrt[3]{4} - \sqrt[3]{2})}]}^{2}$ $\Rightarrow x \approx 69121.4399$ $\sqrt{y} = 4 {[1 + 2 (\sqrt[3]{4} - \sqrt[3]{2})]}^{2}$ $\approx 120.024524$
	Commented by ajfour last updated on 30/Nov/23

	$b u t i c h e c k e d m y a n s w e r s a r e$ $w r o n g!$
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