3. Determine all pairs of positive integers $a$ and $b$ such that $a^{a-b}=b^{a}$.

(Jaromír Šimša)
SOLUTION. After multiplying the given equality by the number $a^{b} / b^{a}$, we get

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

The power on the left side is an integer precisely when its base $k=a / b$ is an integer (in our case, a positive integer). ${ }^{*}$ Substituting $a=k b$ transforms the rewritten equation into the form $k^{k b}=(k b)^{b}$, from which, after dividing the exponents by the number $b$, we get $k^{k}=k b$ or $b=k^{k-1}$, hence $a=k b=k^{k}$. Therefore, all sought pairs are of the form $(a, b)=\left(k^{k}, k^{k-1}\right)$, where $k \geqq 1$ is any integer. Verification is not necessary due to the equivalent transformations.

NOTE. Let's show how the key insight that the number $a$ is a multiple of the number $b$ can also be obtained without rewriting the given equation in the form of the fraction $a / b$. We use the fact that the numbers $a$ and $b$ are the bases of equal powers $a^{a-b}$ and $b^{a}$, for which the exponents satisfy $a-b0$ we get $b_{p} / a_{p}=1-b / a<1$. This proves the desired inequality $b_{p} \leqq a_{p}$.