74th Romanian Mathematical Olympiad

If $b\ge 4a$, then $b>a$ and $b^b>a^{4a}$. Therefore, in this case the equality is impossible.

If $b<4a$, we have $b^b=a^{4a}=a^{4a-b}{a}^b$, so $b^b$ is divisible by $a^b$. Therefore, $b$ is divisible by $a$. It follows that $b=na$, with $n=1$ (I), $n=2$ (II), or $n=3$ (III).

We get $(a^4{)}^a=((na{)}^n{)}^a$, then $a^4=n^n{a}^n$, or $a^{4-n}=n^n$. In case (I), we obtain $a=b=1$, in case (II) we obtain $a=2,b=4$, while $a=27,b=81$ in case (III).