\section*{Problem 1 - 101221}

All ordered pairs \((x, y)\) of real numbers are to be determined for which the system of equations

\[
x^{2}+y^{2}=1 \quad \text { (1) } \quad ; \quad x^{6}+y^{6}=\frac{7}{16}
\]

is satisfied.
}

From

\[
3 x^{2} y^{2}=3 x^{2} y^{2}\left(x^{2}+y^{2}\right)+\left(x^{6}+y^{6}-\frac{7}{16}\right)=\left(x^{2}+y^{2}\right)^{3}-\frac{7}{16}=1-\frac{7}{16}=\frac{9}{16}
\]

we immediately obtain

\[
x^{2} y^{2}=\frac{3}{16}
\]

and by substituting into (1) further

\[
x^{4}-x^{2}+\frac{3}{16}=\left(x^{2}-\frac{1}{4}\right)\left(x^{2}-\frac{3}{4}\right)=0
\]

with the solutions

\[
x \in\left\{ \pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right\}
\]

which, in conjunction with (3), yields the 8 final solutions

\[
(x, y) \in\left\{\left( \pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right),\left( \pm \frac{1}{2}, \mp \frac{\sqrt{3}}{2}\right),\left( \pm \frac{\sqrt{3}}{2}, \pm \frac{1}{2}\right),\left( \pm \frac{\sqrt{3}}{2}, \mp \frac{1}{2}\right)\right.
\]