Find the real solutions of the equations:$x+y=2$, $xy-z^2=1$ Can anyone help with this? I will show you what I did.

Find the real solutions of the equation:$$x+y=2\tag{1}$$ $$xy-z^2=1 \tag{2}$$

From equation (2), I have $$xy=1+z^2 \tag{3}$$ From equations (1) and (3), I have a new equation $$r^2-2r+(1+z^2)=0$$ The roots of this equation are the solutions. How do I find the root?
 A: Hint: From $r^2-2r+(1+z^2)=0$, you can have
$$(r-1)^2+z^2=0.$$
Hence $r=1$ and $z=0$.
A: The problem can receive a straightforward geometric interpretation/solution by considering in the plane the fixed straight line  $(L)$ with equation $x+y=2$ and the (parametricaly defined) hyperbola $(H_z)$ defined by $xy=1+z^2$. 
Curve $(H_0)$ is tangent to (D) in $A$. It is the unique $(H_z)$ that has a common point with $(L)$. All other curves $(H_z)$ ($z\neq 0$) do not intersect $(L)$. See graphics where four hyperbolas $(H_z)$ for $z=0, z=\pm 0.5, z=\pm 1, z=\pm 1.5$ (red, green, brown, blue curves resp.) are represented.
Thus the solution is $(x,y,z)=(1,1,0)$ where $(x,y)=(1,1)$ are the coordinates of $A$, value $z=0$ coming from the particular $(H_z=H_0).$

A: Your quadratic is right: you must have in fact
$$r^2-2r+(1+z^2)=0\iff r=1\pm\sqrt{1-(1+z^2)}$$ and because you are interested in real solutions the only possibility is $z=0$
Thus $$r=1\Rightarrow (x,y,z)=(1,1,0)$$
A: You have a quadratic equation; its roots are the solutions for $x$ and $y$. However the discriminant is
$$
4-4(1+z^2)=-4z^2
$$
which is non negative only if $z=0$. If you only want real solutions, you must have $z=0$ and the quadratic equation has a double solution $r=1$. Thus $x=1$ and $y=1$.
If you also allow complex solutions, the roots of the quadratic equations are
$$
\frac{2+2iz}{2}=1+iz
\qquad\text{and}\qquad
\frac{2-2iz}{2}=1-iz
$$
the solutions are
$$
(1+iz,1-iz,z)
\qquad\text{or}\qquad
(1-iz,1+iz,z)
$$
(which coincide for $z=0$).
A: Given $x+y=2$ and $xy-z^2=1$
So $xy=1+z^2\geq 1>0$.
So  $x,y>0$
Using $\bf{A.M\geq G.M},$ we get $$\frac{x+y}{2}\geq \sqrt{xy}\Rightarrow (x+y)^2\geq 4xy$$ and equality hold when $x=y$
So $4\geq 4xy\Rightarrow xy\leq 1\Rightarrow 1+z^2\leq 1\Rightarrow z^2\leq 0\Rightarrow z=0$
and $x=y=1.$ So we get $(x,y,z) = (1,1,0)$
A: See , this is a very simple approach to the problem , only using grade 6 number theory concepts .
It is given that
x + y = 2
Deductions : Both x and y are odd or both x and y are even
It cannot be that x is odd and y is even or vice versa
Also given ,
xy-z²= 1 (xy and z² are two consecutive no.s)
Deductions : Either xy is odd and z² is even ....(1)
OR xy is even and z² is odd .....(2)
Furthur Deductions: In (1),  if xy is odd , both x and y have to be odd ....(1.1)
In (1) , if z² is even , z is also even ....(1.2)
In (2) if xy is even , both x and y have to be even ....(2.1)
In (2) , if z² is odd , z is also odd ....(2.2)
BUT , because x and y add up to 2 , max value of both is 1 , therefore according to (2) if they are both even , both have to be 0 . Now ,
xy - z² = 1 , if xy = 0 , then the way that the equation is possible is if z = i , but z is a real no. , so this is not possible , and therefore we can safely discard (2.1) and (2.2) .
The result we are left with : x and y are both odd and z is even
You must have noticed it by now , but still to make it more obvious ...
(x+y)² = 4
Or ,  x²+y²+2xy = 4
Or , x²+y²+2(xy-1) = 2
Or , x²+y²+2z²= 2
Now z is even and if you add x²+y² to 2z² the result is 2
Therefore , from here it is obvious that x and y have to be 1 and z has to be 0 consequently .
(x,y,z) = (1,1,0)
P.S. : Sorry for bad indentation .
This solution is only partially true , because I brought up the concept of odd and even no.s though it is not stated x and y are whole no.s . You might consider this as a mathematical crime .
In other words , this might be correct but not fool-proof .
