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Question Number 161071 by naka3546 last updated on 11/Dec/21

F o r a, b, c > 0 .

F i n d (x, y, z) t h a t s a t i s f y t h i s e q u a t i o n s y s t e m

a x + b y = {(x - y)}^{2}

b y + c z = {(y - z)}^{2}

c z + a x = {(z - x)}^{2}

Answered by MJS_new last updated on 11/Dec/21

I think the only solution is x = y = z = 0

Commented by naka3546 last updated on 12/Dec/21

H o w t o g e t i t, s i r ?

S h o w y o u r w o r k i n g s, p l e a s e .

T h a n k y o u \dots

Answered by MJS_new last updated on 12/Dec/21

transformed equations

x^{2} - 2 x y + y^{2} - a x - b y = 0

y^{2} - 2 y z + z^{2} - b y - c z = 0

x^{2} - 2 x z + z^{2} - a x - c z = 0

each of these can be seen as a conic section .

the 1^{st} is in the x - y - plane

the 2^{nd} is in the y - z - plane

the 3^{rd} is in the x - z - plane

their only common point is (0 ∣ 0 ∣ 0)

Commented by MJS_new last updated on 12/Dec/21

you are right . pairs of the cones have common

points in the axes

Commented by mr W last updated on 12/Dec/21

i t h i n k o t h e r s o l u t i o n s a r e p o s s i b l e,

e . g .

(x, y, z) = (a, 0, 0)

(x, y, z) = (0, b, 0)

(x, y, z) = (0, 0, c)

m a y b e {t h a t}^{'} s a l l .

Commented by naka3546 last updated on 13/Dec/21

T h a n k y o u s o m u c h, s i r .