Find all $x$ for which the equation $x^{2}+y^{2}+z^{2}+2 x y z=1$ (in terms of $z$) has a real solution for any $y$.

#
The discriminant $D$ of this equation is $4 x^{2} y^{2}-4 x^{2}-4 y^{2}+4=4\left(x^{2}-1\right)\left(y^{2}-1\right)$. If $x^{2}-1>0$, then $D<0$, for example, at $y=0$, if $x^{2}-1<0$, then

$D<0$, for example, at $y=2$. When $x= \pm 1$ and for any $y$, we have $D=0$, so the equation has a solution.

## Answer

$x= \pm 1$