10. let $m$ be any natural number. As a function of $m$, determine the smallest natural number $k$ for which the following applies: If $\{m, m+1, \ldots, k\}=A \cup B$ is any decomposition into two sets $A$ and $B$, then $A$ or $B$ contains three elements $a, b, c$ (which do not necessarily have to be different) with $a^{b}=c$.

## Solution
Answer: $k=m^{m^{m+2}}$

We first show that $\left\{m, m+1, \ldots, m^{m^{m+2}}-1\right\}$ can be split into two sets so that there are no three elements $a, b, c$, as required in the problem. Set

$$
\begin{aligned}
& A=A_{1} \cup A_{2}=\left\{m, \ldots, m^{m}-1\right\} \cup\left\{m^{m^{m+1}}, \ldots, m^{m^{m+2}}-1\right\} \\
& B=\left\{m^{m}, \ldots, m^{m^{m+1}}-1\right\}
\end{aligned}
$$

We will now look at various cases:

1. $a, b \in A_{1}$. Then $a^{b} \geq m^{m}$ and $a^{b}<\left(m^{m}\right)^{m^{m}}=m^{m^{m+1}}$, so $a^{b}$ is in $B$ and therefore not in $A$.
2. $a \in A_{2}, b \in A_{1}$. The following applies $a^{b} \geq\left(m^{m^{m+1}}\right)^{m}=m^{m^{m+2}}$, and $a^{b}$ is not in $A$.
3. $a \in A, b \in A_{2}$. In this case, $a^{b} \geq m^{\left(m^{m} m^{m+1}\right)} \geq m^{m^{m+2}}$, so $a^{b}$ is not in $A$.
4. $a, b \in B$. Then $a^{b} \geq\left(m^{m}\right)^{m^{m}}=m^{m^{m+1}}$ and therefore $a^{b}$ is not in $B$.

Now we show that for $k \geq m^{m^{m+2}}$ there are always three elements $a, b, c$ in $A$ or $B$ such that $a^{b}=c$. Suppose this is false and there exists a decomposition $\{m, m+$ $\left.1, \ldots, m^{m^{m+2}}\right\}=A \cup B$ such that neither $A$ nor $B$ contains three elements with the required properties. We can also assume that $m \in A$. Then $m^{m}$ must lie in $B$ and therefore $\left(m^{m}\right)^{m^{m}}=m^{m^{m+1}}$ in $A$. Then $\left(m^{m^{m+1}}\right)^{m}=m^{m^{m+2}}$ lies in $B$ again.

1. if $m^{m+1}$ is in $A$, then $a^{b}=c$ with $a=m, b=m^{m+1}$ and $c=m^{m^{m+1}}$, in contradiction to our assumption.
2. if $m^{m+1}$ lies in $B$, then $a^{b}=c$ applies with $a=m^{m}, b=m^{m+1}$ and $c=m^{m^{m+2}}$, this is again a contradiction to the assumption.