(1) Total no. of integer ordered pair $(x,y,z)$ in $x^2 = yz\;\;,y^2=zx\;\;,z^2 = xy$
(2) Total no. of integer ordered pair $(x,y,z)$ in $x+yz = 1\;\;,y+zx = 1\;\;,z+xy = 1$
My Try:: (1) Clearly $ x = 0,y = 0,z = 0$ are the solution of given equation
and from three equation we observe that $x,y,z$ has same sign.
Now If $x\neq 0,y\neq 0$ and $z\neq 0,$ Then $x^2-y^2 =-z(x-y)\Leftrightarrow (x-y).(x+y+z) =0$
Means either $x=y$ or $x+y+z = 0$
$\bullet$ If $x = y$, The put in $z^2 = xy=x^2=y^2\Leftrightarrow z = \pm x = \pm y$
means $x = y =z$
So $(x,y,z) = (k,k,k)$ where $k\in \mathbb{Z}$
$\bullet$ If $x+y+z = 0$, Then put in third $z^2 = xy\Leftrightarrow x^2+y^2+xy = 0$
So $x^2+y^2+xy = x^2+y^2+(x+y)^2 = 0\Leftrightarrow x = 0,y = 0,x+y = 0$
So $(x,y,z) = (0,0,0)$
So Given equation has Infinite solution
My Question is I have Calculate Right or not
If not plz explain me.
Thanks
Hint to (2): subtract 2nd equation out 1st: $$(x-y)(1-z)=0.\ \ \ (1)$$
Addition:
Similarly $$(x-z)(1-y)=0,\ \ \ (2)$$ $$(z-y)(1-x)=0.\ \ \ (3)$$
If $z=1$, then $x+y=1, xy=0$ (from your $1$st and $3$rd equations) whence we obtain two solutions $$z=1, x=1, y=0,$$ $$z=1, x=0, y=1.$$ Similarly for $y=1$ we get one more solution from ($2$): $$y=1, x=1, z=0.$$
Further, let $x,y,z\ne 1$. Then $x=y=z$ and $x^2+x-1=0$ from your $1$st equation. Then we have two solutions: $$x=y=z=\frac{-1+\sqrt{5}}{2}$$ and $$x=y=z=\frac{-1-\sqrt{5}}{2}.$$