System of equations with 4 unknowns.

Question

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?

Answer 1

Subtracting $(1)$ and $(2)$, you establish

$$xy=1\lor z=t.$$

But plugging $xy=1$ in $(1)$ reduces it to

$$z=x+y+z$$ or $$x=-y,$$ which is not compatible.

Repeating with a circular permutation of the variables, the only possible solutions are trivially with $x=y=z=t$.

Answer 2

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.

Answer 3

Two trivial solutions.

Notice that by symmetry, if we let $x=y=z=t=a$, each equation is just $3a=a^3$, so we can have $(x,y,z,t)=(0,0,0,0), (\sqrt{3},\sqrt{3},\sqrt{3},\sqrt{3}), (-\sqrt{3},-\sqrt{3},-\sqrt{3},-\sqrt{3})$.

Answer 4

No. $xyz-zyt=xy(z-t)=z-t\implies (\text{$xy=1$ or $z=t$})$.