x^2 −(y−z)^2 = 3 y^2 − (z−x)^2 = 5 z^2 − (x−y)^2 = 12


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Question Number 90842 by jagoll last updated on 26/Apr/20

$x^{2} - {(y - z)}^{2} = 3$ $y^{2} - {(z - x)}^{2} = 5$ $z^{2} - {(x - y)}^{2} = 12$
Commented by john santu last updated on 26/Apr/20
$(1) (x+y−z)(x−y+z)=3 (2) (y+z−x)(y−z+x) = 5 (3) (z+x−y)(z−x+y) = 12 x^2 +y^2 +z^2 −{y^2 −2yz+z^2 +z^2 −2xz+x^2 +x^2 −2xy+y^2 } = 20 −(x^2 +y^2 +z^2 )+2xy+2yz+2xz = 20 (x^2 +y^2 +z^2 )−2(xy+xz+yz) = −20 (x+y+z)^2 −4(xy+xz+yz) = −20$
$(1) (x + y - z) (x - y + z) = 3$ $(2) (y + z - x) (y - z + x) = 5$ $(3) (z + x - y) (z - x + y) = 12$ $x^{2} + y^{2} + z^{2} - {y^{2} - 2 y z + z^{2} + z^{2} - 2 x z + x^{2}$ $+ x^{2} - 2 x y + y^{2}} = 20$ $- (x^{2} + y^{2} + z^{2}) + 2 x y + 2 y z + 2 x z = 20$ $(x^{2} + y^{2} + z^{2}) - 2 (x y + x z + y z) = - 20$ ${(x + y + z)}^{2} - 4 (x y + x z + y z) = - 20$
Commented by jagoll last updated on 26/Apr/20

$b y o b s e r v e i g o t$ $x = 2, y = 3, z = 4$ $(1) 4 - {(3 - 4)}^{2} = 3$ $(2) 9 - {(4 - 2)}^{2} = 5$ $(3) 16 - {(2 - 3)}^{2} \neq 12$ $i t h i n k i t w r o n g$
Commented by Prithwish Sen 1 last updated on 26/Apr/20

$after mr . john santu i think$ $\frac{(1)}{(2)} we get \Rightarrow \frac{x - y + z}{y + z - x} = \frac{3}{5} . . . . (4)$ $\frac{(2)}{(3)} \Rightarrow \frac{x + y - z}{x - y + z} = \frac{5}{12} . . . . . . (5)$ $\frac{(1)}{(3)} \Rightarrow \frac{x + y - z}{y - x + z} = \frac{1}{4} . . . . (6)$ $now by (4) \times (3) \Rightarrow {(x - y + z)}^{2} = \frac{36}{5}$ $\Rightarrow x - y + z = \pm \frac{6}{\sqrt{5}} . . . . (7)$ $and by (5) \times (1) \Rightarrow {(x + y - z)}^{2} = \frac{5}{4}$ $\Rightarrow x + y - z = \pm \frac{\sqrt{5}}{2} . . . . . . . (8)$ $considering only the positive value$ $by (7) + (8)$ $2 x = \frac{6}{\sqrt{5}} + \frac{\sqrt{5}}{2} \Rightarrow x = \frac{17}{2 \sqrt{5}} and so on .$
Commented by jagoll last updated on 28/Apr/20

$w a w . . g r e a t s i r . t h a n k y o u$
	Answered by ajfour last updated on 26/Apr/20
	$x^2 −y^2 −z^2 =3−2yz ..(i) y^2 −z^2 −x^2 =5−2xz ...(ii) z^2 −x^2 −y^2 =12−2xy ....(iii) Adding all (x^2 +y^2 +z^2 )−2(xy+yz+zx)=−20 ⇒ (x+y)^2 −2z(x+y)−4xy+z^2 =−20 .....(I) (i)+(ii) 2z(x+y)−2z^2 =8 ⇒ x+y=z+(4/z) from (III) ⇒ (x−y)^2 =z^2 −12 (i)−(ii) (x^2 −y^2 )−z(x−y)=−1 ⇒ (x−y)(x+y−z)=−1 (x−y)^2 {(x+y)^2 +z^2 −2z(x+y)}=1 (z^2 −12){(z+(4/z))^2 +z^2 −2z(z+(4/z))}=1 (z^2 −12)(((16)/z^2 ))=1 16z^2 −192=z^2 ⇒ z^2 =((192)/(15))=((64)/5) (x+y)^2 =((64)/5)+((16×5)/(64))+8 = ((64)/5)+(5/4)+8 = ((441)/(20)) (x−y)^2 =z^2 −12 =((64)/5)−12=(4/5) let x>y ⇒ x−y=(4/(√(20))) , x+y=((21)/(√(20))) ⇒x=((25)/(4(√5))) , y=((17)/(4(√5))) , z=((32)/(4(√5))) . these dont fit very well, so let x<y ⇒ y−x=(4/(√(20))) ⇒ x=((17)/(4(√5))) , y=((25)/(4(√5))) , z=((32)/(4(√5)))$
	$x^{2} - y^{2} - z^{2} = 3 - 2 y z . . (i)$ $y^{2} - z^{2} - x^{2} = 5 - 2 x z . . . (i i)$ $z^{2} - x^{2} - y^{2} = 12 - 2 x y . . . . (i i i)$ $A d d i n g a l l$ $(x^{2} + y^{2} + z^{2}) - 2 (x y + y z + z x) = - 20$ $\Rightarrow {(x + y)}^{2} - 2 z (x + y) - 4 x y + z^{2} = - 20$ $. . . . . (I)$ $(i) + (i i)$ $2 z (x + y) - 2 z^{2} = 8$ $\Rightarrow x + y = z + \frac{4}{z}$ $f r o m (I I I)$ $\Rightarrow {(x - y)}^{2} = z^{2} - 12$ $(i) - (i i)$ $(x^{2} - y^{2}) - z (x - y) = - 1$ $\Rightarrow (x - y) (x + y - z) = - 1$ ${(x - y)}^{2} {{(x + y)}^{2} + z^{2} - 2 z (x + y)} = 1$ $(z^{2} - 12) {{(z + \frac{4}{z})}^{2} + z^{2} - 2 z (z + \frac{4}{z})} = 1$ $(z^{2} - 12) (\frac{16}{z^{2}}) = 1$ $16 z^{2} - 192 = z^{2}$ $\Rightarrow z^{2} = \frac{192}{15} = \frac{64}{5}$ ${(x + y)}^{2} = \frac{64}{5} + \frac{16 \times 5}{64} + 8$ $= \frac{64}{5} + \frac{5}{4} + 8 = \frac{441}{20}$ ${(x - y)}^{2} = z^{2} - 12 = \frac{64}{5} - 12 = \frac{4}{5}$ $l e t x > y$ $\Rightarrow x - y = \frac{4}{\sqrt{20}}, x + y = \frac{21}{\sqrt{20}}$ $\Rightarrow x = \frac{25}{4 \sqrt{5}}, y = \frac{17}{4 \sqrt{5}}, z = \frac{32}{4 \sqrt{5}} .$ $t h e s e d o n t f i t v e r y w e l l, s o l e t$ $x < y \Rightarrow y - x = \frac{4}{\sqrt{20}}$ $\Rightarrow x = \frac{17}{4 \sqrt{5}}, y = \frac{25}{4 \sqrt{5}}, z = \frac{32}{4 \sqrt{5}}$
	Commented by ajfour last updated on 26/Apr/20

	$i c h e c k e d t h e m e v e n, h u r r a y!$
	Commented by jagoll last updated on 26/Apr/20

	$h H a h a h u r r a y s i r$
	Answered by mr W last updated on 26/Apr/20

	$(x + y - z) (x - y + z) = 3$ $(x + y + z - 2 z) (x + y + z - 2 y) = 3 . . . (i)$ $(y + z - x) (y - z + x) = 5$ $(x + y + z - 2 x) (x + y + z - 2 z) = 5 . . . (i i)$ $(z + x - y) (z - x + y) = 12$ $(x + y + z - 2 y) (x + y + z - 2 x) = 12 . . . (i i i)$ $(i) \times (i i) \times (i i i) :$ ${(x + y + z - 2 x)}^{2} {(x + y + z - 2 y)}^{2} {(x + y + z - 2 z)}^{2} = 3 \times 5 \times 12 = 180$ $(x + y + z - 2 x) (x + y + z - 2 y) (x + y + z - 2 z) = \pm 6 \sqrt{5} . . . (i v)$ $(i v) / (i) a n d s i m i l a r l y :$ $x + y + z - 2 x = \pm \frac{6 \sqrt{5}}{3} . . . I$ $x + y + z - 2 y = \pm \frac{6 \sqrt{5}}{5} . . . I I$ $x + y + z - 2 z = \pm \frac{6 \sqrt{5}}{12} . . . I I I$ $I + I I + I I I :$ $x + y + z = \pm 6 \sqrt{5} (\frac{1}{3} + \frac{1}{5} + \frac{1}{12}) . . . I V$ $I V - I a n d s i m i l a r l y :$ $2 x = \pm 6 \sqrt{5} (\frac{1}{5} + \frac{1}{12}) \Rightarrow x = \pm \frac{17 \sqrt{5}}{20}$ $2 y = \pm 6 \sqrt{5} (\frac{1}{3} + \frac{1}{12}) \Rightarrow y = \pm \frac{5 \sqrt{5}}{4}$ $2 z = \pm 6 \sqrt{5} (\frac{1}{3} + \frac{1}{5}) \Rightarrow z = \pm \frac{8 \sqrt{5}}{5}$
	Commented by mr W last updated on 26/Apr/20

	$y e s! i t m e a n s a l l + o r a l l - .$ $w e s e e i f x = a, y = b, z = c i s a s o l u t i o n, t h e n$ $x = - a, y = - b, z = - c i s a l s o a s o l u t i o n,$
	Commented by ajfour last updated on 26/Apr/20

	$a l l \pm m a y n o t b e v a l i d, S i r, i t h i n k . .$
	Commented by mr W last updated on 26/Apr/20

	$t o b e e x a c t, t h e s o l u t i o n s a r e :$ $x = \frac{17 \sqrt{5}}{20}$ $y = \frac{5 \sqrt{5}}{4}$ $z = \frac{8 \sqrt{5}}{5}$ $o r$ $x = - \frac{17 \sqrt{5}}{20}$ $y = - \frac{5 \sqrt{5}}{4}$ $z = - \frac{8 \sqrt{5}}{5}$
	Commented by ajfour last updated on 26/Apr/20

	$T h a n k s f o r c l a r i f y i n g S i r!$
	Commented by jagoll last updated on 26/Apr/20

	$g r e a t . . . .$
	Commented by I want to learn more last updated on 27/Apr/20

	$Nice .$
	Answered by MWSuSon last updated on 26/Apr/20

	$(x + y - z) (x - y + z) = 3 \dots (1)$ $(y - z + x) (y + z - x) = 5 \dots (2)$ $(z - x + y) (z + x - y) = 12 \dots (3)$ $\frac{e q n (1)}{e q n (2)} \frac{(x - y + z)}{(y + z - x)} = \frac{3}{5} \dots (4)$ $\frac{e q n (3)}{e q n (2)} \frac{(z + x - y)}{(y - z + x)} = \frac{12}{5} \dots (5)$ $\frac{e q n (1)}{e q n (3)} \frac{(x + y - z)}{(z - x + y)} = \frac{3}{12} \dots (6)$ $f r o m e q n (4) 3 y + 3 z - 3 x = 5 x - 5 y + 5 z$ $8 y - 2 z - 8 x = 0$ $[x - y = \frac{- z}{4}]$ $f r o m e q n (5) 5 z + 5 x - 5 y = 12 y - 12 z + 12 x$ $17 z - 7 x - 17 y = 0$ $[y - z = \frac{- 7 x}{17}]$ $f r o m e q n (6) 3 z - 3 x + 3 y = 12 x + 12 y - 12 z$ $[z - x = \frac{9 y}{15}]$ $i n p u t t h e v a l u e o f (x - y) i n t o e q n (3)$ $i n p u t t h e v a l u e o f (y - z) i n t o e q n (1)$ $i n p u t t h e v a l u e o f (z - x) i n t o e q n (2)$ $a n d {y o u}^{'} l l h a v e t h e a n s w e r M r W h a d .$
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