\section*{Problem 4 - 131224}

Determine all pairs \((x, y)\) of real numbers that are solutions to the system of equations:

\[
\begin{aligned}
& x^{3}+y^{2}+x+1=0 \\
& y^{3}+x^{2}+y+1=0
\end{aligned}
\]
}

If we multiply equation (2) by \(x\) and subtract (1) from it, we obtain

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

Here, we are allowed to divide by \(y^{2}+1\) because \(y^{2}+1>0\), which leads to the simple relationship

\[
x y=1
\]

Now, if we multiply (1) by \(x^{2}\) and then substitute \(x y=1\) into it, we obtain the new equation

\[
x^{5}+x^{3}+x^{2}+1=0
\]

in the variable \(x\) alone.

An obvious solution to this is \(x=-1\), and after dividing by the linear factor \(x+1\), we get

\[
x^{4}-x^{3}+2 x^{2}-x+1=0
\]

We now divide this by \(x^{2} \neq 0\) and obtain, with the substitution \(z:=x+\frac{1}{x}\), the simple equation

\[
z^{2}-z=0
\]

in \(z\) with the two solutions \(z \in\{0,1\}\). For both values of \(z\),

\[
x+\frac{1}{x}=z \quad \text { or } \quad x^{2}-z x+1=0
\]

is unsolvable in real numbers, as one can easily verify.

In summary, the only solution for \(x\) remains \(x=-1\), and since \(x y=1\), it follows that \(y=-1\). Indeed, \((x, y)=(-1,-1)\) satisfies both equations (1) and (2) and is thus the only real solution here.