Question

Find the number of ordered triplets $(xy,yz,zx)$ where x,y,z satisfy
$x(x-1)+2yz=y(y-1)+2zx=z(z-1)+2xy.$

Answer 1

Let $x(x-1)+2yz$$=$ $\lambda$ ...........(1)
$y(y-1)+2zx$$=$ $\lambda$ .............(2)
$z(z-1)+2xy$ $=$ $\lambda$.............(3)
(2) -(1) $\Rightarrow y^2-x^2-y+x+2z(x-y)=0$
$(x-y)(x+y-1-2z)$ $=$ 0
Similarly $(y-z)(y+z-1-2x)$$=$ 0
$(z-x)(z+x-1-2y)$ $=$ 0
Case $-Ix\ne y\ne z$
$x+y-1-2z$ $=$0
$y+z-1-2x$$=$0
$z+x-1-2y$$=$0
Adding $\Rightarrow$ -3$=$0 contradiction
No solution
$Case-II$ Any two are equal
$x=y\ne z\Rightarrow (y-z)(z-1-x)=0,$
$z-1-x$ $=$0 $\Rightarrow$ z $=$ $x+1$
$\Rightarrow$ $(x-y,y-z,z-x)$$=$ $(0,-1,1)$
Similarly $x\ne y=z\Rightarrow (1,0,-1)$
$x\ne y=z\Rightarrow (-1,1,0)$
Three solutions
$Case-IIIx=y=z\Rightarrow (x-y,y-z,z-x)=(0,0,0)$
One solution
Total 4 solutions