69th Belarusian Mathematical Olympiad

(ii) $a=b=c=1$.

It is easy to verify that $a=n^n$, $b=n^{n-1}$, $c=n^n$ is a solution of the given equation for all $n\in \mathbb{N}$, thus (iii) is solved.

Note that if $a$, $b$ or $c$ equals $1$, then $a=b=c=1$, which is a solution. Now we prove that if $(a,b,c)$ is a solution of $$a^b\cdot b^c=c^a$$ with $a$, $b$ and $c$ greater than $1$, then $b<a$ (this gives the answer $a=b=c=1$ of (ii)).

Suppose, in contrary, that $b\ge a$. Then $c^a=a^b\cdot b^c>a^b\ge a^a$, whence $c>a$. Further, $c^a=a^b\cdot b^c\ge a^a\cdot a^c>a^a\cdot a^a=(a^2{)}^a$, so $c>a^2$. Suppose that $c\ge a^m$, where $m\ge 2$. Then $c^a=a^b\cdot b^c\ge a^a\cdot a^{m-1}$, whence $c\ge a^{1+a^{m-1}}\ge a^{m+1}$ (for $a,m\ge 2$ holds $a^{m-1}\ge m$). Proceeding this way, we obtain that $c\ge a^n$ for any positive integer $n$, which is absurd.

It remains to prove (i). Suppose that some prime $p$ divides $a$, but doesn't divide $b$. Let ${\nu }_p(a)=\alpha$, $a=p^{\alpha }{a}_0$ and ${\nu }_p(c)=\gamma$, $c=p^{\gamma }{c}_0$. Then ${\nu }_p(a^b\cdot b^c)={\nu }_p(a^b)=\alpha \cdot b$ and ${\nu }_p(c^a)=\gamma \cdot a$. Hence $\alpha \cdot b=p^{\alpha }{a}_0\cdot \gamma$, so $\alpha$ is divisible by $p^{\alpha }$, which is impossible.