# Find all $x,y,z$ satisfying $xy=z-x-y$ and cyclic permutations

Find all ordered pairs $(x,y,z)$ real numbers, which satisfy the following system of equations: $$xy=z-x-y\\xz=y-x-z\\yz=x-y-z$$

-
What brought you to tagging this with [logic]? –  Lord_Farin Oct 25 '13 at 14:53
I think there is no solution when x,y,z are distinct –  lokesh sangabattula Aug 30 at 14:47

Hint: $$xy=z-x-y \quad \iff \quad (x+1)(y+1)=z+1.$$

Hence $(x+1)(y+1)=z+1$, $(x+1)(z+1)=y+1$, $(z+1)(y+1)=x+1$.

-
what to do next? –  lokesh sangabattula Oct 25 '13 at 15:01
@lokeshsangabattula, then multiply the equations and solve. –  njguliyev Oct 25 '13 at 15:06

Another hint: Subtracting the third equation from the second reveals a constant value for z as long as x ≠ y (Thank you Calvin).

$z(x-y)=-2(x-y)$

-
Of course, we can only conclude that it's constant if $x\neq y$. –  Calvin Lin Oct 25 '13 at 15:05
Mark, +1 for the hint. But you don't need to delete your answer and write a new one. Just click "edit" under your post. –  njguliyev Oct 25 '13 at 15:08
I know. I screwed up somehow, and had 2 answers. So I killed the other one. –  Mark Bailey Oct 25 '13 at 15:09

The groebner basis in lex order is (try sympy)

$$x - z^4 - z^3 + 3z^2 + 2z=0$$ $$y + z^4 + 2z^3 - z^2 - 3z=0$$ $$z^5 + 3z^4 - 6z^2 - 4z=0$$

Note that now the first equation only contains $x,z$, the second only $y,z$ and the third only $z$.

If you factor out one $z$ from the third equation you have a degree four polynomial that is solvable in formulas (or you carry on guessing and factoring for that polynomial).

-