An algebra problem by A Former Brilliant Member

From equation $1$ we obtain $x=2-y$ . Substituting this into equation $2$ results in $2y-y^2-z^2=1$ which can be written as $z^2+y^2-2y+1=0$ . Hence, $z^2+(y-1)^2=0$ . Each of the summands on the left hand side of the last equality is non-negative and, consequently, it must be equal to zero. It follows that

$z=0$ and $y=1$ . Hence $x=1$ .

Thus, the system of equations possesses a single real solution.

We rewrite the equations as $\begin{aligned} x+y &= 2 \\ xy &= z^2 + 1 \end{aligned}$ From Vieta's relations , it follows that $x$ and $y$ are solutions to the quadratic equation $p^2 - 2p+(z^2 + 1)=0$ We take the discriminant of this equation and find that $\Delta = (-2)^2 - 4(1)(z^2 + 1) = 4 - 4z^2 - 4 = -4z^2$ If $z \neq 0$ , then $\Delta = -4z^2 <0$ , and thus the solutions $x$ and $y$ of the equation are not real numbers, a contradiction. Thus $z=0$ , and the quadratic equation becomes $p^2 - 2p + (0^2 + 1)=p^2 -2p+1=(p-1)^2 = 0$ This gives $x=1,y=1$ as the only possible solution to the equation (and ultimately, to the system itself). Therefore, there is only $\boxed{1}$ real numbered ordered triple solutions.