4. find all pairs $(a, b)$ of natural numbers so that the following equation is satisfied:

$$
a^{6 a}=b^{b}
$$
## 1st solution:

There is $6 a>b$, because otherwise $b^{b} \geq(6 a)^{6 a}>a^{6 a}$. Division by $a^{b}$ results in the new equation

$$
a^{6 a-b}=\left(\frac{b}{a}\right)^{b}
$$

in which the left-hand side is whole according to what has just been said, so the right-hand side is also whole. Thus $a$ is a divisor of $b=k a$ and the inequality $6 a>b$ translates to $1 \leq k<6$. If you insert $b=k a$ into the original equation, take the $a$th root and transform, then $a^{6-k}=k^{k}$ follows. Testing the cases $k=1, \ldots, 5$ results in the following possibilities for the pair $(a, b)$:

$$
(1,1), \quad\left(3,3^{2}\right), \quad\left(4^{2}, 4^{3}\right), \quad\left(5^{5}, 5^{6}\right)
$$