1. It is known that $m, n, k$ are distinct natural numbers greater than 1, the number $\log _{m} n$ is rational, and, moreover,

$$
k^{\sqrt{\log _{m} n}}=m^{\sqrt{\log _{n} k}}
$$

Find the minimum of the possible values of the sum $k+5 m+n$.
Answer: 278.

Solution. Transform the original equation, taking into account that the numbers $m, n, k$ are greater than 1 and the corresponding logarithms are positive:

$$
\begin{aligned}
& k \sqrt{\log _{m} n}=m \sqrt{\log _{n} k} \Leftrightarrow \log _{m} k \cdot \sqrt{\log _{m} n}=\sqrt{\log _{n} k} \Leftrightarrow \\
& \Leftrightarrow \log _{m} k \cdot \sqrt{\log _{m} n}=\sqrt{\frac{\log _{m} k}{\log _{m} n}} \Leftrightarrow \\
& \Leftrightarrow \sqrt{\log _{m} k} \cdot \log _{m} n=1 \Leftrightarrow \log _{m} k \cdot \log _{m}^{2} n=1
\end{aligned}
$$

By performing the substitution $k=n^{p}, m=n^{q}$, where $p \neq 0, q \neq 0$, we get

$$
\frac{p}{q} \cdot \frac{1}{q^{2}}=1 \Leftrightarrow p=q^{3}
$$

Thus, all triples $(k, m, n)$ that satisfy the condition of the problem have the form $\left(n^{q^{3}}, n^{q}, n\right)$.

It is clear that $q>0$ : otherwise, from the naturality of the number $n$ it follows that the number $n^{q}$ lies in the interval $(0 ; 1)$ and is not a natural number. The case $q=1$ contradicts the condition of the problem. If $q \in(0 ; 1)$, then $q=1 / r$, where $r>1$, and then the triple of natural numbers $\left(n^{q^{3}}, n^{q}, n\right)$ can be written in the form $\left(l, l^{r^{2}}, l^{r^{3}}\right)$ with a natural $l$. The sum $k+5 m+n$ for this triple will be greater than the same sum for the numbers in the triple $\left(l^{r^{3}}, l^{r}, l\right)$. Therefore, $q>1$.

For integer $q$, the triple of numbers with the minimum sum $k+5 m+n$ is obtained when $n=2, q=2$ : these are the numbers $k=2^{8}, m=2^{2}, n=2$, and for them the sum $k+5 m+n$ is 278. If $q$ is rational but not an integer, then for the number $n^{q^{3}}$ to be an integer, it is necessary that $n$ be at least the eighth power of an integer not less than 2, and, of course, the sum $n^{q^{3}}+5 n^{q}+n$ will be greater than 278 (since $n \geqslant 256$).