System of equations with 4 unknowns. I'm trying to solve this system of equations but I'm reaching a dead end.
$$\begin{array}{lcl}
xyz & = & x+y+z \ \ \ \ \ \ \ (1)\\
xyt & = & x+y+t \ \ \ \ \ \ \ (2)\\ 
xzt & = & x+z+t \ \ \ \ \ \ \ (3)\\ 
yzt & = & y+z+t \ \ \ \ \ \ \ (4)\\
\end{array}$$
So, (1)-(2) gives $$xyz-zyt=xy(z-t)=z-t\Rightarrow xy=1.$$
(3)-(4) gives $$xzt-yzt=zt(x-y)=x-y\Rightarrow zt=1.$$
This system reduces to $$\begin{array}{lcl}
xy & = & 1 \ \ \ \ \ \ \ (5)\\
zt & = & 1 \ \ \ \ \ \ \ (6)\\ 
\end{array}$$
One trivial solution is $x=y=z=t=\pm1,$ but this can't be because then I'd have divided by zero earlier in my simplifications.
How do I solve this one?

We will find all solutions using only (1), (2), and (3).
Making the difference between the first two equations you get $xy=1$ or $x=y$. Let us suppose that $xy=1$. Then $x,y$ have the same sign and are both $\neq 0$. This implies that
$$
z=xyz=x+y+z=x+\frac{1}{x}+z \implies x+\frac{1}{x}=0,
$$
which is impossibile since $|x+\frac{1}{x}|\ge 2$ for all non-zero reals $x$ (by AM-GM). Therefore $x=y$ ($\neq 1$ and $\neq -1$, otherwise $xy$ would still be $1$). 
The first and second equations become (*)
$$
z(x^2-1)=2x \text{ and }t(x^2-1)=2x.
$$
In particular, $z=t$. This means the only two first equations imply $x=y \notin \{\pm 1\}$ and $z=t$.
Lastly, using (3), we get $xz^2=x+2z$, hecen we reduce to 
$$
x(z^2-1)=2z\text{ and }z(x^2-1)=2x.
$$
If $x=0$ then also $z=0$ and viceversa. Otherwise substituing the first into the second we get
$$
z=\frac{2x}{x^2-1}=\frac{4z}{\left(z^2-1\right) \left(\left(\frac{2z}{z^2-1}\right)^2-1\right)}=\frac{4z(z^2-1)}{4z^2-(z^2-1)^2}.
$$
Since $z\neq 0$ then 
$$
4z^2-(z^2-1)^2=4(z^2-1) \Leftrightarrow (z^2-3)(z^2+1)=0.
$$
Hence $|z|=\sqrt{3}$. Thus we have only the trivial solutions.
