Find $a,b$ for which $xyz+z=a,\quad xyz^2+z=b,\quad x^2+y^2+z^2=4$ has unique solution Find all values of $a,b$ for which the system of equations $xyz+z=a,\quad xyz^2+z=b,\quad x^2+y^2+z^2=4$ has only one real solution.

$xyz+z=a$
$xyz^2+z=b$
So,$\frac{xy+1}{xyz+1}=\frac{a}{b}$
I can think no method by so as this system of equations has only one real solution. What should I do? Please help me. 

If $(x, y, z)$ is a solution, then $(y, x, z)$ is a solution. Hence if there is a unique solution, we need $x=y$. This reduces the problem to $2$ variables.
$$\begin{align*}
x^2z+z&=a\\
x^2z^2+z&=b\\
2x^2+z^2&=4\\
\end{align*}$$
If $(x, x, z)$ is a solution, note that $(-x, -x, z)$ is also a solution. Hence, $x=0$ if the solution is unique.
$$\begin{align*}
z&=a\\
z&=b\\
z^2&=4\\
\end{align*}$$
Hence $a=b$. For there to be a solution, we need $a=b=2$ or $a=b=-2$.
We now check the $2$ solutions of $(a, b)$ by solving the equations.
$$\begin{align*}
xyz+z&=a\\
xyz^2+z&=b\\
x^2+y^2+z^2&=4\\
\end{align*}$$
Since $a=b$, we subtract the first equation from the second equation to get $xy(z^2-z)=0$. There are a few cases:
If $z=0$, then $xyz+z=0\neq a$.
Otherwise, if $x=0$ (or $y=0$ symmetrically), we have $z=a$ and $z^2=4$, which, substituting into the third equation, means that $y^2=0$ (or $x^2=0$), which means that $(x, y, z)=(0, 0, a)$ is one solution.
The final case $z=1$ should have no solutions. Let's check the $2$ cases $a=b=2$ and $a=b=-2$.
If $a=b=2$, the system of equations reduces to:
$$\begin{align*}
xy&=1\\
x^2+y^2&=3\\
\end{align*}$$
Solving, we find another solution, $(x, y)=\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$, so $a=b=2$ does not admit a unique solution.
If $a=b=-2$, the system of equations reduces to:
$$\begin{align*}
xy&=-3\\
x^2+y^2&=3\\
\end{align*}$$
Note that $(x+y)^2=x^2+y^2+2xy=-3$. As such $x+y$ is not real and there is no extra solution here.
Hence the only $(a, b)$ for the system to have a unique solution is $(-2, -2)$.
