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Question Number 14535 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17
$hello dears. this is an ancient system of equiations that have not solved by any one yet. but i think by studing several ways that used for solveing Q.14157,now we can attack this. note:it is a first time in world that quistions such that Q.14157,solved. Congraglotion to evry one in this forum! by puting: (p=x^2 ,q=y^2 ) we can write that system as below: { ((x^2 +y=a^2 )),((x+y^2 =b^2 )) :} (A) i have an idea for solving system(A). but at first i have one numeric case: { ((p+(√q)=5)),(((√p)+q=3)) :} p−4+(√q)=1⇒((√p)−2)((√p)+2)+(√q)−1=0 ((√p)−2)+((√q)−1)((√q)+1)=0⇒ ((√p)−2)=−((√q)−1)((√q)+1)⇒ −((√q)−1)((√q)+1)((√p)+2)+(√q)−1=0 ((√q)−1)(−((√q)+1)((√p)+2)+1)=0 ⇒(√q)−1=0⇒(q=1⇒p=4) but we can not use this numeric way for parameters: a^2 ,b^2 . geometric ways can useful for this. by writing system as (A),and adding equations ,we have: x^2 +x+y^2 +y=a^2 +b^2 ⇒ ⇒(x+(1/2))^2 +(y+(1/2))^2 =a^2 +b^2 +(1/2). i.e :answers of system are located on the premeter of a circle whit center point: M(((−1)/2),((−1)/2)) and radius of:R=(√(a^2 +b^2 +(1/2).)) i have a drawing in commenet below. and i want members of this forum continue my way to geting answers, or another ideas that you have. excuse me for long commenet.$
$h e l l o d e a r s .$ $t h i s i s a n a n c i e n t s y s t e m o f e q u i a t i o n s$ $t h a t h a v e n o t s o l v e d b y a n y o n e y e t .$ $b u t i t h i n k b y s t u d i n g s e v e r a l w a y s$ $t h a t u s e d f o r s o l v e i n g Q . 14157, n o w$ $w e c a n a t t a c k t h i s .$ $n o t e : i t i s a f i r s t t i m e i n w o r l d t h a t$ $q u i s t i o n s s u c h t h a t Q . 14157, s o l v e d .$ $C o n g r a g l o t i o n t o e v r y o n e i n t h i s f o r u m!$ $b y p u t i n g : (p = x^{2}, q = y^{2}) w e c a n w r i t e$ $t h a t s y s t e m a s b e l o w :$ ${\begin{cases} x^{2} + y = a^{2} \\ x + y^{2} = b^{2} \end{cases} (A)$ $i h a v e a n i d e a f o r s o l v i n g s y s t e m (A) .$ $b u t a t f i r s t i h a v e o n e n u m e r i c c a s e :$ ${\begin{cases} p + \sqrt{q} = 5 \\ \sqrt{p} + q = 3 \end{cases}$ $p - 4 + \sqrt{q} = 1 \Rightarrow (\sqrt{p} - 2) (\sqrt{p} + 2) + \sqrt{q} - 1 = 0$ $(\sqrt{p} - 2) + (\sqrt{q} - 1) (\sqrt{q} + 1) = 0 \Rightarrow$ $(\sqrt{p} - 2) = - (\sqrt{q} - 1) (\sqrt{q} + 1) \Rightarrow$ $- (\sqrt{q} - 1) (\sqrt{q} + 1) (\sqrt{p} + 2) + \sqrt{q} - 1 = 0$ $(\sqrt{q} - 1) (- (\sqrt{q} + 1) (\sqrt{p} + 2) + 1) = 0$ $\Rightarrow \sqrt{q} - 1 = 0 \Rightarrow (q = 1 \Rightarrow p = 4)$ $b u t w e c a n n o t u s e t h i s n u m e r i c w a y$ $f o r p a r a m e t e r s : a^{2}, b^{2} .$ $g e o m e t r i c w a y s c a n u s e f u l f o r t h i s .$ $b y w r i t i n g s y s t e m a s (A), a n d a d d i n g$ $e q u a t i o n s, w e h a v e :$ $x^{2} + x + y^{2} + y = a^{2} + b^{2} \Rightarrow$ $\Rightarrow {(x + \frac{1}{2})}^{2} + {(y + \frac{1}{2})}^{2} = a^{2} + b^{2} + \frac{1}{2} .$ $i . e : a n s w e r s o f s y s t e m a r e l o c a t e d o n$ $t h e p r e m e t e r o f a c i r c l e w h i t c e n t e r$ $p o i n t : M (\frac{- 1}{2}, \frac{- 1}{2}) a n d r a d i u s o f : R = \sqrt{a^{2} + b^{2} + \frac{1}{2} .}$ $i h a v e a d r a w i n g i n c o m m e n e t b e l o w .$ $a n d i w a n t m e m b e r s o f t h i s f o r u m$ $c o n t i n u e m y w a y t o g e t i n g a n s w e r s,$ $o r a n o t h e r i d e a s t h a t y o u h a v e .$ $e x c u s e m e f o r l o n g c o m m e n e t .$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 02/Jun/17

$M (\frac{- 1}{2}, \frac{- 1}{2}), G (0, y), H (x, 0), M D = M B = R$ $y o u a r e r i g h t m r W 1 . c o r r e c t e d . t h a n k s .$ $w a i t i n g f o r y o u r s o l u t i o n . . . . . . . . . . . . . .$
Commented by ajfour last updated on 03/Jun/17

$k i n d l y l e t m e k n o w, h o w t o f i n d$ $x s a t i s f y i n g :$ $x^{4} + x = a$
Commented by mrW1 last updated on 02/Jun/17

$y o u m e a n M D = M B = R, n o t o D = o B = R$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 03/Jun/17

$i t i s n o t e a s y . i f a i s a n u m b e r, n u m e r i c$ $m e t h o d s, s u c h t h a t : t r i a l a n d e r r o r, n e w t o n - r a p h s o n$ $d r a w i n g a g r a p h, m y b e h e l p f u l . b y$ $f a c t r i z i n g t o : x (x + 1) (x^{2} - x + 1) = a, a n d$ $t h i s f a c t t h a t : a s h o u l d d i v i d a b l e b y$ $x, x + 1, x^{2} - x + 1, y o u c a n f i n d a n s w e r s .$ $x (x + 1) (x^{2} - x + 1) = (x^{2} + x) (x^{2} - x + 1) =$ $(x^{2} + x) (x^{2} + x - 2 x + 1) =$ $\Rightarrow {(x^{2} + x)}^{2} - (2 x - 1) (x^{2} + x) - a = 0$ $x^{2} + x = \frac{2 x - 1 \pm \sqrt{{(2 x - 1)}^{2} + 4 a}}{2}$ $2 x^{2} + 2 x = 2 x - 1 \pm \sqrt{{(2 x - 1)}^{2} + 4 a}$ $\Rightarrow {(2 x^{2} + 1)}^{2} = {(2 x - 1)}^{2} + {(2 \sqrt{a})}^{2}$ $n o w y o u h a v e a r i g h t a n g l e t r i a n g l e$ $w i t h s i d s o f : 2 x^{2} + 1, 2 x - 1, 2 \sqrt{a} .$ $m y b e h a v e a g e o m e t r i c w a y . . . . . .$
	Answered by ajfour last updated on 03/Jun/17

	$\sqrt{p} + {(a^{2} - p)}^{2} = b^{2}, a n d$ $\sqrt{q} + {(b^{2} - q)}^{2} = a^{2} .$
	Commented by ajfour last updated on 03/Jun/17

	Commented by ajfour last updated on 03/Jun/17

	$w h e t h e r \sqrt{p} = x a n d \sqrt{q} = y$ $w e s u b s t i t u t e o r n o t d i f f i c u l t y$ $i s t h e s a m e, i t h i n k . .$
	Commented by ajfour last updated on 03/Jun/17

	$b e y o n d t h i s i b a r e l y c a n d o$ $a n y t h i n g . . .$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 03/Jun/17

	$b y t h i s s u b s t i t u t i o n, d i f f c u l t y o f e q .$ $a d d e d m u c h m o r e .$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 03/Jun/17

	$v e r y n i c e p i c . m r A j f o u r! i t h i n k t h e r e$ $i s a f o o t a g e o f^{'} g o l d e n {r a t i o}^{'} t h a t c a n b e$ $e x p l o r e d . b u t i {d o n}^{'} t k n o w w h e r e a n d$ $h o w ?$
	Commented by ajfour last updated on 03/Jun/17

	$t h a n k s, c a n y o u n o t l e t m e k n o w$ $a m o r e d e f i n i t e a n s w e r t o t h i s$ $q u e s t i o n; w h a t g o l d e n r a t i o,$ $i h a v e h e a r d n o t .$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 03/Jun/17

	$i n a s i m p l e w o r d,^{'} g o l d e n {r a t i o}^{'} i s t h e$ $p o s i t i v e r o o t o f e q u a t i o n : φ (φ - 1) = 1$ $You can't use 'macro parameter character #' in math mode$ $g o l d e n r a t i o i s o b t a i n e d w h e n a$ $l i n e A B i s d i v i d e d b y 2 p a r t$ $w h i t l e n g t h s o f : a & b, (a > b) s u c h t h a t$ $r a t i o o f g r e a t e r p a r t t o w h o l e A B$ $i s e q u a l t o r a t i o o f s m a l l e r p a r t t o$ $g r e a t e r p a r t . i . e : \frac{b}{a + b} = \frac{a}{b}$ $\Rightarrow b^{2} - a b - a^{2} = 0 \Rightarrow b = \frac{a + \sqrt{5 a^{2}}}{2} = \frac{\sqrt{5} + 1}{2} a$ $o r : \frac{b}{a} = \frac{\sqrt{5} + 1}{2} = 1.618 .$ $g o l d e n r a t i o i s u s e d i n$ $v e r y p a r t o f s c i e n c e a n d a r t .$ $i n a r c h i t e c t u r e, d e s i g n a n d d r a w i n g$ $o f h u m a n b u d y, e t c . . . . .$ $a r e c t a n g l e t h a t h a v e g o l d e n r a t i o$ $b e t w e e n i t s s i d e s, n a m e d a s :$ $g o l d e n r e c t a n g l e . . . . . . .$
	Answered by mr W last updated on 05/Aug/19
	$let x=(√p), y=(√q) let u=a^2 , v=b^2 x^2 +y=u x+y^2 =v x+(u−x^2 )^2 =v ⇒x^4 −2ux^2 +x+u^2 −v=0 trying to use method from ajfour sir: a=−2u b=1 c=u^2 −v ⇒x^2 −px+(1/2)(−2u+p^2 +(1/p))=0 ⇒x=(1/2)(p±(√(4u−p^2 −(2/p)))) where p^2 =P with P^3 −4uP^2 +4vP−1=0 let P=Q+((4u)/3) Q^3 +3(((4u)/3))Q^2 +3(((4u)/3))^2 Q+(((4u)/3))^3 −4u[Q^2 +2(((4u)/3))Q+(((4u)/3))^2 ]+4v(Q+((4u)/3))−1=0 Q^3 +4(v−((4u^2 )/3))Q+((16uv)/3)−1=0 Δ=[(4/3)(v−((4u^2 )/3))]^3 +[(1/2)(((16uv)/3)−1)]^2 Δ=((64)/(729))(3v−4u^2 )^3 +(1/(36))(16uv−3)^2 >0, say Q=(((√Δ)−(1/6)(16uv−3)))^(1/3) −(((√Δ)+(1/6)(16uv−3)))^(1/3) P=p^2 =((4u)/3)+(((√Δ)−(1/6)(16uv−3)))^(1/3) −(((√Δ)+(1/6)(16uv−3)))^(1/3) ⇒p=±(√(((4u)/3)+(((√Δ)−(1/6)(16uv−3)))^(1/3) −(((√Δ)+(1/6)(16uv−3)))^(1/3) )) ⇒x=(1/2){±(√(((4u)/3)+(((√Δ)−(1/6)(16uv−3)))^(1/3) −(((√Δ)+(1/6)(16uv−3)))^(1/3) ))±(√(((8u)/3)−(((√Δ)−(1/6)(16uv−3)))^(1/3) +(((√Δ)+(1/6)(16uv−3)))^(1/3) ∓(2/(√(((4u)/3)+(((√Δ)−(1/6)(16uv−3)))^(1/3) −(((√Δ)+(1/6)(16uv−3)))^(1/3) )))))} ⇒p=(1/4){±(√(((4a^2 )/3)+(((√Δ)−(1/6)(16a^2 b^2 −3)))^(1/3) −(((√Δ)+(1/6)(16a^2 b^2 −3)))^(1/3) ))±(√(((8a^2 )/3)−(((√Δ)−(1/6)(16a^2 b^2 −3)))^(1/3) +(((√Δ)+(1/6)(16a^2 b^2 −3)))^(1/3) ∓(2/(√(((4a^2 )/3)+(((√Δ)−(1/6)(16a^2 b^2 −3)))^(1/3) −(((√Δ)+(1/6)(16a^2 b^2 −3)))^(1/3) )))))}^2 with Δ=((64)/(729))(3b^2 −4a^4 )^3 +(1/(36))(16a^2 b^2 −3)^2$
	$l e t x = \sqrt{p}, y = \sqrt{q}$ $l e t u = a^{2}, v = b^{2}$ $x^{2} + y = u$ $x + y^{2} = v$ $x + {(u - x^{2})}^{2} = v$ $\Rightarrow x^{4} - 2 {u x}^{2} + x + u^{2} - v = 0$ $t r y i n g t o u s e m e t h o d f r o m a j f o u r s i r :$ $a = - 2 u$ $b = 1$ $c = u^{2} - v$ $\Rightarrow x^{2} - p x + \frac{1}{2} (- 2 u + p^{2} + \frac{1}{p}) = 0$ $\Rightarrow x = \frac{1}{2} (p \pm \sqrt{4 u - p^{2} - \frac{2}{p}})$ $w h e r e p^{2} = P w i t h$ $P^{3} - 4 {u P}^{2} + 4 v P - 1 = 0$ $l e t P = Q + \frac{4 u}{3}$ $Q^{3} + 3 (\frac{4 u}{3}) Q^{2} + 3 {(\frac{4 u}{3})}^{2} Q + {(\frac{4 u}{3})}^{3} - 4 u [Q^{2} + 2 (\frac{4 u}{3}) Q + {(\frac{4 u}{3})}^{2}] + 4 v (Q + \frac{4 u}{3}) - 1 = 0$ $Q^{3} + 4 (v - \frac{4 u^{2}}{3}) Q + \frac{16 u v}{3} - 1 = 0$ $Δ = {[\frac{4}{3} (v - \frac{4 u^{2}}{3})]}^{3} + {[\frac{1}{2} (\frac{16 u v}{3} - 1)]}^{2}$ $Δ = \frac{64}{729} {(3 v - 4 u^{2})}^{3} + \frac{1}{36} {(16 u v - 3)}^{2} > 0, s a y$ $Q = \sqrt[3]{\sqrt{Δ} - \frac{1}{6} (16 u v - 3)} - \sqrt[3]{\sqrt{Δ} + \frac{1}{6} (16 u v - 3)}$ $P = p^{2} = \frac{4 u}{3} + \sqrt[3]{\sqrt{Δ} - \frac{1}{6} (16 u v - 3)} - \sqrt[3]{\sqrt{Δ} + \frac{1}{6} (16 u v - 3)}$ $\Rightarrow p = \pm \sqrt{\frac{4 u}{3} + \sqrt[3]{\sqrt{Δ} - \frac{1}{6} (16 u v - 3)} - \sqrt[3]{\sqrt{Δ} + \frac{1}{6} (16 u v - 3)}}$ $\Rightarrow x = \frac{1}{2} {\pm \sqrt{\frac{4 u}{3} + \sqrt[3]{\sqrt{Δ} - \frac{1}{6} (16 u v - 3)} - \sqrt[3]{\sqrt{Δ} + \frac{1}{6} (16 u v - 3)}} \pm \sqrt{\frac{8 u}{3} - \sqrt[3]{\sqrt{Δ} - \frac{1}{6} (16 u v - 3)} + \sqrt[3]{\sqrt{Δ} + \frac{1}{6} (16 u v - 3)} \mp \frac{2}{\sqrt{\frac{4 u}{3} + \sqrt[3]{\sqrt{Δ} - \frac{1}{6} (16 u v - 3)} - \sqrt[3]{\sqrt{Δ} + \frac{1}{6} (16 u v - 3)}}}}}$ $\Rightarrow p = \frac{1}{4} {\pm \sqrt{\frac{4 a^{2}}{3} + \sqrt[3]{\sqrt{Δ} - \frac{1}{6} (16 a^{2} b^{2} - 3)} - \sqrt[3]{\sqrt{Δ} + \frac{1}{6} (16 a^{2} b^{2} - 3)}} \pm \sqrt{\frac{8 a^{2}}{3} - \sqrt[3]{\sqrt{Δ} - \frac{1}{6} (16 a^{2} b^{2} - 3)} + \sqrt[3]{\sqrt{Δ} + \frac{1}{6} (16 a^{2} b^{2} - 3)} \mp \frac{2}{\sqrt{\frac{4 a^{2}}{3} + \sqrt[3]{\sqrt{Δ} - \frac{1}{6} (16 a^{2} b^{2} - 3)} - \sqrt[3]{\sqrt{Δ} + \frac{1}{6} (16 a^{2} b^{2} - 3)}}}}}^{2}$ $w i t h Δ = \frac{64}{729} {(3 b^{2} - 4 a^{4})}^{3} + \frac{1}{36} {(16 a^{2} b^{2} - 3)}^{2}$
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