\section*{Problem 1 - 161231}

Give all pairs \((x, y)\) of real numbers for which the following holds:

\[
x^{2}+y=2 \quad(1) \quad \text { and } \quad y^{2}+x=2
\]
}

By subtracting the two equations from each other, we obtain the new equation

\[
x^{2}-y^{2}=x-y
\]

We make the following case distinction here:

1. Case: \(x=y\)

Substituting into (1) then yields

\[
x^{2}+x-2=(x-1)(x+2)=0
\]

which leads to the solutions

\[
(x, y) \in\{(1,1),(-2,-2)\}
\]

2. Case: \(x \neq y\).

By dividing \((*)\) by \(x-y\), we obtain

\[
x+y=1 \quad \text { or } \quad y=1-x
\]

Substituting this into (1) yields

\[
x^{2}-x-1=0
\]

which leads to the additional two solutions

\[
(x, y) \in\{(1 \pm \sqrt{5}) / 2,(1 \mp \sqrt{5}) / 2)\}
\]