Solve the system of equations

$1-x_{1} x_{2} x_{3}=0$

$1+x_{2} x_{3} x_{4}=0$

$1-x_{3} x_{4} x_{5}=0$,

$1+x_{4} x_{5} x_{6}=0$

...

$1-x_{47} x_{48} x_{49}=0$,

$1+x_{48} x_{49} x_{50}=0$,

$1-x_{49} x_{50} x_{1}=0$

$1+x_{50} x_{1} x_{2}=0$.
Clearly, none of the unknowns is equal to zero. Subtracting the second equation from the first, we get $x_{2} x_{3}\left(x_{1}+x_{4}\right)$ $=0$, that is, $x_{1}=-x_{4}$. Similarly,

$x_{2}=-x_{5}, x_{3}=-x_{6}, \ldots, x_{47}=-x_{50}, x_{48}=-x_{1}, x_{49}=-x_{2}, x_{50}=-x_{3}$. From this, it is clear that $x_{1}=x_{7}, x_{2}=x_{8}, \ldots, x_{50}$ $=x_{6}$. Continuing, we get

$x_{1}=x_{3}=x_{5}=\ldots=x_{49}=-x_{2}=-x_{4}=\ldots=-x_{50}$. Substituting these formulas, for example, into the first equation, we get $x_{1}^{3}=-1$

## Answer

$(-1,1,-1,1, \ldots,-1,1)$.