Find all possible values of $c^2$ in a system of equations. Numbers $x,y,z,c\in \Bbb R$ satisfy the following system of equations:
$$x(y+z)=20$$
$$y(z+x)=13$$
$$z(x+y)=c^2$$
Find all possible values of $c^2$.
To try to solve this, I expanded the equations:
$$xy+xz=20$$
$$yz+xy=13$$
$$xz+yz=c^2$$
Then I subtracted the first equation from the second one to get:
$$xz-yz=7$$
I added and subtracted this equation with the 3rd and got the following equations:
$$2yz=7-c^2$$
$$2xz=7+c^2$$
I then added these equations, factored out $2z$ and divided by $2$ to get:
$$z(x+y)=7$$
So, I found one possible value of $c^2 = 7$. How do I find the other values if they exist or how do I prove that there are no other values if they don't? Thanks!

Let $c^2 = s$.  Eliminating $x$ and $y$, you get an equation in $s$ and $z$:
$$ s^2 + 2 z^2 s - 66 z^2 - 49 $$
Thus 
$$ z^2 = \dfrac{s^2 - 49}{66 - 2 s}$$
Since $z^2 \ge 0$, we need either $s \le -7$ or $7 \le s < 33$.  This corresponds to $\sqrt{7} \le c < \sqrt{33}$.   We then have
$$ \eqalign{y &= \dfrac{z (33 - c^2)}{c^2 + 7}\cr
x &= \dfrac{z(33 - c^2)}{c^2-7} \cr}$$
In particular, $c = \sqrt{7}$ is not possible.  So the result is that
$\sqrt{7} < c < \sqrt{33}$.
