\section*{Problem 5 - 061045}

Let \(a\) be any given real number. Determine all real \(x\) that satisfy the equation:

\[
\sqrt{a+x}-\sqrt{\frac{a^{2}}{a+x}}=\sqrt{2 a+x}
\]
}

Multiplying the equation by \(\sqrt{a+x}\) transforms it into (the equivalent equation due to \(a+x \neq 0\)) \(a+x-|a|=\sqrt{(2 a+x) \cdot(a+x)}\).

1. Case: \(a>0\). Then \(|a|=a\) and by squaring we obtain \(x^{2}=x^{2}+3 a x+2 a^{2}\) or \(x=-\frac{2}{3} \cdot a\). This gives \(a+x=\frac{1}{3} \cdot a\) or

\[
\sqrt{a+x}-\sqrt{\frac{a^{2}}{a+x}}=\sqrt{\frac{1}{3} a}-\sqrt{\frac{a^{2}}{\frac{1}{3} a}}=\frac{1}{\sqrt{3}} \cdot \sqrt{a}-\sqrt{3 a}=\sqrt{a} \cdot\left(\frac{\sqrt{3}}{3}-\sqrt{3}\right)<0 \leq \sqrt{2 a-x}
\]

which is a contradiction to the original equation, so there are no solutions in this case.

2. Case: \(a \leq 0\). Then \(|a|=-a\), so the equation becomes \(2 a+x=\sqrt{(2 a+x) \cdot(a+x)}\).

Case 2.1: \(2 a+x=0\). (Since \(a+x \neq 0\), in this case \(a \neq 0\), because otherwise from \(2 a+x=0\) by subtracting \(a=0\) we would immediately get \(a+x=0\).) Then \(x=-2 a\) and \(\sqrt{a+x}-\sqrt{\frac{a^{2}}{a+x}}=\) \(\sqrt{-a}-\sqrt{-a}=0=\sqrt{2 a+x}\), so these are solutions to the original equation.

Case 2.2: \(2 a+x \neq 0\), so \(\sqrt{2 a+x} \neq 0\), allowing us to divide by it. This gives us the equation \(\sqrt{2 a+x}=\sqrt{a+x}\) or, after squaring, \(2 a+x=a+x\), thus \(a=0\). However, in this case, the original equation simplifies to \(\sqrt{x}-\sqrt{\frac{0}{x}}=\sqrt{x}\), which is satisfied for all positive real numbers \(x\).

Summary: For positive \(a\), the equation has no solutions; for negative \(a\), \(x=-2 a\) is the only solution; and for \(a=0\), any positive real number \(x\) satisfies the equation.