\section*{Problem 4 - 271034}

Determine all positive rational numbers \(x\) for which \(x^{2 x}=\frac{1}{2}\) holds!
}

Every positive rational number \(x\) has its representation

\[
x=\frac{p}{q}
\]

with natural numbers \(p, q>0\), which are coprime. For this, the required equation is successively equivalent to

\[
\begin{aligned}
\left(\frac{p}{q}\right)^{\left(\frac{2 p}{q}\right)} & =\frac{1}{2} \\
\left(\frac{p}{q}\right)^{2 p} & =\left(\frac{1}{2}\right)^{q} \\
2^{q} \cdot p^{2 p} & =q^{2 p}
\end{aligned}
\]

Due to the coprimality of \(p, q\), it follows from (2) that in the prime factorization of \(q\) only the prime factor 2 can occur, and in that of \(p\) no prime factor at all, i.e., it must be

\[
q=2^{m}
\]

with a natural number \(m\) and \(p=1\) (4). For these \(p, q\), (2) is equivalent to

\[
2^{2^{m}}=2^{2 m} \quad \text { with } \quad 2^{m}=2 m
\]

Among the natural numbers \(m=0,1,2\), (5) holds exactly for \(m=1\) and \(m=2\). If one increases \(m\) for \(m \geq 2\) by 1 each time, the right side of (5) increases by 2, while the left side doubles, i.e., increases by \(2^{m}\) and thus by at least 4. For all \(m \geq 3\), (5) is therefore not satisfied. Thus, \(m=1\) or \(m=2\).

Consequently, according to (1), (3), (4), the positive rational numbers \(x\) for which the required equation holds are exactly the numbers

\[
x=\frac{1}{2} \quad \text { and } \quad x=\frac{1}{4}
\]

Adapted from \([5]\)