Find all pairs of natural numbers $(a, b)$ such that $b^{a} \mid a^{b}-1$.
We quickly eliminate the cases $a=1$ or $b=1$. Let $p$ be the smallest prime divisor of $b$, and $m$ the order of $a$ modulo $p$. $m$ divides $b$ and $p-1$, so it is 1. According to LTE, if $p$ is odd, $a v_{p}(b) \leq v_{p}(a-1)+v_{p}(b)$, hence $(a-1) v_{p}(b) \leq v_{p}(a-1)<a-1$. Since $v_{p}(b) \geq 1$, this is a contradiction. Therefore, $p=2$.

We write $p=2c$, and see that $a$ is odd. By LTE, we find $a=3$ and $v_{2}(b)=1$. Thus, $8c^{3} \mid 3^{2c}-1$ and $c$ is odd. Let $q$ be its smallest prime divisor. We consider its order modulo 3, and quickly arrive at the conclusion: $a=3$ and $b=2$ provides the only solution (for $a, b \neq 1$).