How many Triple?

If we take ([T=(a, b, c)), we got $a+b+c=x-y+y-z+z-x=0$ so $b=-(a+c)$

Let $r\in\mathbb{C}$ , $x'=x+r, y'=y+r, z'=z+r$ , then $a=x'-y', b=y'-z', c=z'-x'$ So we can suppose $x'=x-x=0$ , and then $x'=0, y'=-a, z'=c$ And the equallity become $-2ac=-a(-a-1)=c(c-1)$ $-2ac=a(a+1)=c(c-1)$ If $c=0$ then $a(a+1)=0$

So the possible solutions are $T=(0, 0, 0)$ $T=(-1, 1, 0)$ Else $c\neq0$ and $-2ac=c(c-1)$

So $-2a=c-1$ ; $c=1-2a$

Taking the equallity $a(a+1)=c(c-1)$ we have $a^2+a=4a^2-2a$ $3a^2=3a$ So, the possible solutions are $T=(0, -1, 1)$ $T=(1, 0, -1)$

Then, the answer is $\boxed{n=4}$