System of equations symmetric

Question

How do I solve the following system of equation?

$$ xyz = x+y+z $$ $$ xyt = x+y+t $$ $$ xzt = x+z+t $$ $$ yzt=y+z+t $$

I have no idea how to do.

Answer 1

$$xyz=x+y+z\tag {1}$$$$xyt=x+y+t\tag {2}$$$$xzt=x+z+t\tag {3}$$$$yzt=y+z+t\tag {4}$$

First, we subtract $(2)$ from $(1)$ to get $xy(z-t)=z-t$. From that, we see that either $z=t$ or (if $z\neq t$), $xy=1$.

Similarly, subtract $(3)$ from $(2)$ to get $xt(y-z)=y-z$ and $(4)$ from $(3)$ to get $zt(x-y)=x-y$ and $(4)$ from $(1)$ to get $yz(x-t)=x-t$.

After simplifying, we arrive at $x=y=z=t$. Substituting that into $(1)$ will give us $x^3=3x\iff x^3-3x=0\iff x(x^2-3)=0$

So we see that $\boxed{x=0}$ or $\boxed{x=\pm\sqrt{3}}$.

Answer 2

Subtracting 4th from 3rd we get $zt=1$ or $x=y$. Similarly from 2nd and 3rd we get $xt=1$ or $y=z$.

That gives four possibilities.

Case 1: $x=y=z$. Then we have $x^3=3x$ giving solutions $(0,0,0,0),(\sqrt3,\sqrt3,\sqrt3,\sqrt3),(-\sqrt3,-\sqrt3,-\sqrt3,-\sqrt3)$.

Case 2: $xt=1,x=y$. That leads to the second two solutions in Case 1.

Case 3: similarly for $zt=1,y=z$.

Case 4: $zt=xt=1$. Then going back to the original equations we have $x^2y=2x+y$ so $y=\frac{2x}{x^2-1}$, and hence $(x,y,z,t)=(x,\frac{2x}{x^2-1},x,\frac{1}{x})$. Substituting into the second equation gives $x^2=-1$. That leads to the solutions $(i,-i,i,-i),(-i,i,-i,i)$.