\section*{Problem 1 - 041241}

Give all real solutions of the equation

\[
\sqrt{p+x}+\sqrt{p-x}=x
\]

where \(p\) is a positive real number (parameter)!
Assuming the real number \(x\) is a solution to the equation. Since the radicands must not be negative, and \(\sqrt{p+x}+\sqrt{p-x} \geq 0\), it follows that \(0 \leq x \leq p\). By squaring the original equation twice, we obtain \(x^{4}+4 x^{2}(1-p)=0\).

(1) \(00=x\), because \(p\) is positive. Thus, there is no solution when \(p \leq 1\)!

(2) \(11\) could have \(x=0\) and \(x=2 \sqrt{p-1}\) as solutions.

Verification: \(x=0\) is not a solution, just as in the first case.

Now test \(x=2 \sqrt{p-1}\):

\[
\begin{gathered}
s=\sqrt{p+x}+\sqrt{p-x}=\sqrt{p+2 \sqrt{p-1}}+\sqrt{p-2 \sqrt{p-1}}= \\
=\sqrt{(\sqrt{p-1}+1)^{2}}+\sqrt{(\sqrt{p-1}-1)^{2}}=|\sqrt{p-1}+1|+|\sqrt{p-1}-1|
\end{gathered}
\]

The absolute value of the first term is equal to the term itself; the absolute value of the second term is equal to the term itself for \(2 \leq p\) and the negative of it for \(1<p<2\). Therefore, for \(1<p<2\):

\[
s=\sqrt{p-1}+1+1-\sqrt{p-1}=2 \neq 2 \sqrt{p-1}
\]

For \(2 \leq p\):

\[
s=\sqrt{p-1}+1+\sqrt{p-1}-1=2 \sqrt{p-1}=x
\]

and thus it is the only solution.

The only solution to this equation is \(x=2 \sqrt{p-1}\) for the case where \(p \geq 2\).

\section*{Solution adopted from \([5]\)}