\section*{Problem 4 - 081244}

Solve the system of equations:

\[
\begin{aligned}
& \left|\log _{2}(x+y)\right|+\left|\log _{2}(x-y)\right|=3 \\
& x y=3
\end{aligned}
\]
}

Due to the second equation, both variables have the same sign. If both were negative, so would \(x+y\). However, \(\log _{2}(x+y)\) would then not be defined. Therefore, both \(x\) and \(y\) are positive. For \(\log _{2}(x-y)\) to be defined, \(x>y\) must hold, so due to the second equation, \(x+y>x>\sqrt{3}\). Thus, \(\log _{2}(x+y)\) is always positive, and the absolute value bars can be omitted.

There remain two cases:

1. Case: \(x-y \geq 1\).

Then \(\log _{2}(x-y) \geq 0\) as well, and the first equation becomes

\[
3=\log _{2}(x+y)+\log _{2}(x-y)=\log _{2}\left(x^{2}-y^{2}\right)
\]

so \(x^{2}-y^{2}=8\) or \(x^{4}-(x y)^{2}=x^{4}-9=8 x^{2}\), which, with the substitution \(t=x^{2}\), transforms into the quadratic equation \(t^{2}-8 t-9=0\), which has the solutions \(4 \pm \sqrt{16+9}=4 \pm 5\). Since \(t=x^{2} \geq 0\), the negative solution -1 is discarded, leaving \(t=x^{2}=9\), so \(x=3\) and \(y=1\). The check confirms that this assignment of variables also solves the original system of equations.

2. Case: \(x-y<1\).

Then \(\log _{2}(x-y) < 0\) and the first equation becomes

\[
3=\left|\log _{2}(x+y)\right|+\left|\log _{2}(x-y)\right|=\log _{2}(x+y)-\log _{2}(x-y)=\log _{2}\left(\frac{x+y}{x-y}\right)
\]

so \(\frac{x+y}{x-y}=8\), or \(x+y=8(x-y)\), which simplifies to \(9y=7x\), or \(y=\frac{7}{9}x\). Substituting \(y=\frac{7}{9}x\) into the second equation \(xy=3\) yields \(x \cdot \frac{7}{9}x=3\), or \(x^{2}=\frac{27}{7}\), so \(x=3 \cdot \sqrt{\frac{3}{7}}=3 \cdot \frac{\sqrt{21}}{7}\) and \(y=\frac{7}{9} x=\frac{\sqrt{21}}{3}\).

For verification: \(x \cdot y=\frac{3}{7} \cdot \sqrt{21} \cdot \frac{1}{3} \cdot \sqrt{21}=\frac{1}{7} \cdot 21=3\), so the second equation is satisfied. Furthermore,

\[
x+y=\frac{3}{7} \cdot \sqrt{21}+\frac{1}{3} \cdot \sqrt{21}=\frac{9+7}{21} \cdot \sqrt{21}=\frac{16}{\sqrt{21}}
\]

so \(\log _{2}(x+y)=\log _{2}(16)-\log _{2}(\sqrt{21})=4-\frac{1}{2} \log _{2}(21)\). Since \(21>4\), \(\frac{1}{2} \log _{2}(21)>\frac{1}{2} \log _{2}(4)=\frac{1}{2} \cdot 2=1\), so \(\left|\log _{2}(x-y)\right|=-\log _{2}(x-y)=\frac{1}{2} \log _{2}(21)-1\). Together with the previously calculated value, \(\left|\log _{2}(x+y)\right|+\left|\log _{2}(x-y)\right|=4-\frac{1}{2} \log _{2}(21)-\frac{1}{2} \log _{2}(21)-1=4-1=3\), so the second equation is also satisfied by this pair. Thus, there are exactly two solution pairs, namely \((x, y)=(3,1)\) or \((x, y)=\left(\frac{3}{7} \cdot \sqrt{21}, \frac{1}{3} \cdot \sqrt{21}\right)\)