Find all triplets of strictly positive integers $(a, b, c)$ such that $a^{b} \mid b^{c}-1$ and $a^{c} \mid c^{b}-1$.
If $a=1, b, c \in \mathbb{N}^{*}$ any values work. We now assume $a>1$. If $b=1$, we have $a^{c} \mid c-1$, but $a^{c}>c-1$ if $a \geq 2$, so it must be that $c-1=0$, i.e., $c=1$. Conversely, $b=c=1$ and any $a$ provides a solution. Similarly, $c=1$ implies that $b=1$.

We now assume that $a, b, c \geq 2$. Without loss of generality, $c \geq b$.

If $a=2^{k}$ for some $k \in \mathbb{N}, b$ and $c$ are odd, so $\frac{b^{c}-1}{b-1} \equiv b^{c-1}+\ldots+b+1 \equiv 1(\bmod 2)$. Therefore, $2^{k c} \mid b-1$. Thus, $2^{c} \leq 2^{k c} \leq b-1 \leq c-1$, which is false for $c \in \mathbb{N}^{*}$. Therefore, $a$ has an odd prime divisor. Let $p$ be such an integer. We denote $\omega_{c}$ the order of $c$ modulo $p$. We apply LTE: $v_{p}\left(c^{\omega_{c}}-1\right)+v_{p}\left(b / \omega_{c}\right)=v_{p}\left(c^{b}-1\right) \geq v_{p}\left(a^{c}\right) \geq c$. Since $c \wedge p=1$, we have $v_{p}\left(c^{\omega_{c}}-1\right) \geq c$. Therefore, $c^{\omega_{c}}-1 \geq p^{c}$. Since $\omega_{c} \mid p-1$, we have $c^{p}>p^{c}$. This inequality is true only when $p>c$ (except for $p=2$ and $c=3$, but $p$ is odd). Therefore, $a>c$, so $a^{c}>c^{c}>c^{b}-1$, but $a^{c} > c^{b}-1$, contradiction. Conclusion: the suitable triplets $(a, b, c)$ are $(1, b, c)$ and $(a, 1,1)$.