\section*{Problem 3 - 091213}

All natural numbers \(a\) are to be determined for which the equation \(a^{a^{a}}=\left(a^{a}\right)^{a}\) is satisfied.

Note: \(a^{a^{a}}\) means \(a^{\left(a^{a}\right)}\).
}

For all positive integers

\[
\left(a^{a}\right)^{a}=a^{a \cdot a}=a^{\left(a^{2}\right)}
\]

the given equation can also be written in the form \(a^{a^{a}}=a^{\left(a^{2}\right)}\).

(1) For \(a \neq 1\), it follows from this that the exponents must be equal, so that \(a^{a}=a^{2}\) can be concluded. Because \(a \neq 1\), this further implies the condition \(a=2\). Therefore, for \(a \neq 1\), the only natural number solution can be 2.

Indeed, \(2^{2^{2}}=2^{4}=16\) and \(\left(2^{2}\right)^{2}=4^{2}=16\).

(2) If the case \(a=1\) that was previously excluded is directly checked by substituting it into the given equation, it is shown that the natural number 1 also solves the given equation; because \(1^{1^{1}}=1=\left(1^{1}\right)^{1}\).

There are no further solutions.

Adapted from \([5]\)