\section*{Problem 3 - 031233}

Determine - depending on the real number \(p\) - all real values \(x, y\) that satisfy the following equations:

\[
x y+\frac{x}{y}=3 p\left(x^{2}+y^{2}\right) \quad ; \quad x y-\frac{x}{y}=p\left(x^{2}+y^{2}\right)
\]
}

\[
\begin{aligned}
x y-\frac{x}{y}=p\left(x^{2}+y^{2}\right) \Rightarrow x^{2}+y^{2}= & \frac{x y-\frac{x}{y}}{p} ; \quad x y+\frac{x}{y}=3 p\left(x^{2}+y^{2}\right)=3 p \frac{x y-\frac{x}{y}}{p}=3 x y-3 \frac{x}{y} \\
& \Rightarrow 4 \frac{x}{y}=2 x y \Rightarrow 2 x=x y^{2}
\end{aligned}
\]

Case 1: \(x=0\)

Substituting into the second equation: \(0=p y \rightarrow y\) must be 0. However, this is not possible since \(y\) is in the denominator. Therefore, there is only a solution for the special case where \(p=0\). In this case, \(y\) can be arbitrary \((\neq 0)\).

Case 2: \(x \neq 0\) with \(2=y^{2} \Rightarrow y= \pm \sqrt{2}\)

Case 2.1: \(y=\sqrt{2}\)

Substituting into the second equation:

\[
\begin{array}{r}
\sqrt{2} x-\frac{1}{\sqrt{2}} x=p\left(x^{2}+2\right) \\
0=p x^{2}+2 p-\sqrt{2} x+\frac{1}{\sqrt{2}} x \Rightarrow 0=x^{2}-\frac{1}{\sqrt{2} p} x+2 \\
x_{1,2}=\frac{1}{2 \sqrt{2} p} \pm \sqrt{\frac{1}{8 p^{2}}-2}
\end{array}
\]

Thus, the condition \(\frac{1}{8 p^{2}}-2 \geq 0\) must hold, which means \(-\frac{1}{4} \leq p \leq \frac{1}{4}\). To verify if a solution is found, substitute into the first equation:

\[
\begin{aligned}
&\left(\frac{1}{2 \sqrt{2} p} \pm \sqrt{\frac{1}{8 p^{2}}-2}\right) \sqrt{2}+\left(\frac{1}{2 \sqrt{2} p} \pm \sqrt{\frac{1}{8 p^{2}}-2}\right) \frac{1}{\sqrt{2}}=3 p\left(\left(\frac{1}{2 \sqrt{2} p} \pm \sqrt{\frac{1}{8 p^{2}}-2}\right)^{2}+2\right) \\
&\left(\frac{1}{2 \sqrt{2} p} \pm \sqrt{\frac{1}{8 p^{2}}-2}\right) \frac{3}{\sqrt{2}}=\frac{3}{4 p} \pm \frac{3}{\sqrt{2}} \sqrt{\frac{1}{8 p^{2}}-2}
\end{aligned}
\]

Thus, under the condition \(-\frac{1}{4} \leq p \leq \frac{1}{4}\), the solutions in this case are:

\[
x_{1}=\frac{1}{2 \sqrt{2} p}+\sqrt{\frac{1}{8 p^{2}}-2} ; \quad y_{1}=\sqrt{2} \quad \text { and } \quad x_{2}=\frac{1}{2 \sqrt{2} p}-\sqrt{\frac{1}{8 p^{2}}-2} ; \quad y_{2}=\sqrt{2}
\]

Case 2.2: \(y=-\sqrt{2}\)

Substituting into the second equation:

\[
\begin{array}{r}
-\sqrt{2} x+\frac{1}{\sqrt{2}} x=p\left(x^{2}+2\right) \\
0=p x^{2}+2 p+\sqrt{2} x-\frac{1}{\sqrt{2}} x \Rightarrow 0=x^{2}+\frac{1}{\sqrt{2} p} x+2 \\
x_{3,4}=-\frac{1}{2 \sqrt{2} p} \pm \sqrt{\frac{1}{8 p^{2}}-2}
\end{array}
\]

Thus, the condition \(\frac{1}{8 p^{2}}-2 \geq 0\) must hold, which means \(-\frac{1}{4} \leq p \leq \frac{1}{4}\). Substituting into the first equation again confirms that a solution exists.

Thus, under the condition \(-\frac{1}{4} \leq p \leq \frac{1}{4}\), the solutions in this case are:

\[
x_{3}=-\frac{1}{2 \sqrt{2} p}+\sqrt{\frac{1}{8 p^{2}}-2} ; \quad y_{3}=-\sqrt{2} \quad \text { and } \quad x_{4}=-\frac{1}{2 \sqrt{2} p}-\sqrt{\frac{1}{8 p^{2}}-2} ; \quad y_{4}=-\sqrt{2}
\]