11. Solve the system of equations:

$$
\left\{\begin{array}{l}
X^{2}=2 Y-1 \\
X^{4}+Y^{4}=2
\end{array}\right.
$$
70.11. Substituting $x^{2}=2 y-1$ into the second equation, we get $y^{4}+4 y^{2}-4 y-1=0$. Dividing by $y-1$, we obtain (*) $y^{3}+y^{2}+5 y+1=0$. But since $2 y-1=x^{2} \geqslant 0$, $y$ is positive. However, the equation (*) clearly has no positive solutions. Therefore, $y=1$ is the only option. Accordingly, $x=1$ or $x=-1$.