(x^2 + 1)(x − 1)^2 = 2017yz (y^2 + 1)(y − 1)^2 = 2017xz (z^2 + 1)(z − 1)^2 = 2017xy x ≥ 1 ; y ≥ 1 ; z ≥ 1 prove that x = y = z = ((1+ (√(2018)) + (√(2015+ 2(√(2018)))))/2)


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Question Number 106233 by malwaan last updated on 03/Aug/20

$(x^{2} + 1) {(x - 1)}^{2} = 2017 y z$ $(y^{2} + 1) {(y - 1)}^{2} = 2017 x z$ $(z^{2} + 1) {(z - 1)}^{2} = 2017 x y$ $x ⩾ 1; y ⩾ 1; z ⩾ 1$ $p r o v e t h a t x = y = z =$ $\frac{1 + \sqrt{2018} + \sqrt{2015 + 2 \sqrt{2018}}}{2}$
	Answered by Her_Majesty last updated on 04/Aug/20

	$o b v i o u s l y t h e r e m u s t b e a t l e a s t o n e s o l u t i o n$ $w i t h x = y = z b e c a u s e i n t h i s c a s e t h e t h r e e$ $e q u a t i o n s a r e o n e a n d t h e s a m e$ $(x^{2} + 1) {(x - 1)}^{2} = 2017 x^{2}$ $x^{4} - 2 x^{3} - 2015 x^{2} - 2 x + 1 = 0$ $(x^{2} - (1 - \sqrt{2018}) x + 1) (x^{2} - (1 + \sqrt{2018}) x + 1) = 0$ $n o w y o u c a n s o l v e i t$
	Commented by Sarah85 last updated on 04/Aug/20
	you learn fast it seems
	Commented by Her_Majesty last updated on 04/Aug/20

	$I f o r g o t m o r e t h a n {y o u}^{'} l l e v e r k n o w . . .$
	Commented by malwaan last updated on 04/Aug/20

	$t h a n k y o u$ $b u t h o w y o u f a c t o r t h e q u a r t i c$ $e q u a t i o n w h i c h h a s 4 s o l u t i o n s$ $o n l y o n e i s c o r r e c t ?$
	Commented by Her_Majesty last updated on 04/Aug/20
	$(I) x^4 +ax^3 +bx^2 +cx+d=0 x=t−(a/4) leads to t^4 +pt^2 +qt+r=0 (t^2 −αt−β)(t^2 +αt−γ)=0 t^4 −(α^2 +β+γ)t^2 +(α−β)γt+βγ=0 { ((−(α^2 +β+γ)=p)),(((α−β)γ=q)),((βγ=r)) :} solving 2 of these for β and γ we get a 3^(rd) degree polynome for α^2 (α^2 )^3 +j(α^2 )^2 +k(α^2 )+l=0 if this has a useable solution we can find the square factors of (I)$
	$(I) x^{4} + {a x}^{3} + {b x}^{2} + c x + d = 0$ $x = t - \frac{a}{4} l e a d s t o$ $t^{4} + {p t}^{2} + q t + r = 0$ $(t^{2} - α t - β) (t^{2} + α t - γ) = 0$ $t^{4} - (α^{2} + β + γ) t^{2} + (α - β) γ t + β γ = 0$ ${\begin{cases} - (α^{2} + β + γ) = p \\ (α - β) γ = q \\ β γ = r \end{cases}$ $s o l v i n g 2 o f t h e s e f o r β a n d γ w e g e t a 3^{r d} d e g r e e$ $p o l y n o m e f o r α^{2}$ ${(α^{2})}^{3} + j {(α^{2})}^{2} + k (α^{2}) + l = 0$ $i f t h i s h a s a u s e a b l e s o l u t i o n w e c a n f i n d t h e$ $s q u a r e f a c t o r s o f (I)$
	Commented by malwaan last updated on 05/Aug/20

	$I u s e d a n o t h e r m e t h o d$ $l o n g b u t e a s y$
	Answered by malwaan last updated on 05/Aug/20

	$x (x^{2} + 1) {(x - 1)}^{2} = 2017 x^{3}$ $d i v i d e b y x^{3} w e g e t$ $(\frac{x^{2} + 1}{x}) (\frac{x^{2} - 2 x + 1}{x}) = 2017$ $(x + \frac{1}{x}) (x + \frac{1}{x} - 2) = 2017$ $l e t t = x + \frac{1}{x}$ $x ⩾ 1 \Rightarrow t ⩾ 2$ $t (t - 2) = 2017$ $t^{2} - 2 t - 2017 = 0$ $\Rightarrow t = 1 + \sqrt{2018}$ $x + \frac{1}{x} = 1 + \sqrt{2018}$ $x^{2} - (1 + \sqrt{2018}) x + 1 = 0$ $\Rightarrow x = \frac{1 + \sqrt{2018} + \sqrt{2015 + 2 \sqrt{2018}}}{2}$
	Commented by Her_Majesty last updated on 05/Aug/20

	$g o o d!$
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