Find all pairs of strictly positive integers such that $a^{b}=b^{\left(a^{2}\right)}$.
Lemme 4.14. If $a^{b} \geqslant b^{a}$, then $b \geqslant a$ or $b=2$ or $b=1$. In the second case, we also have $a \leqslant 4$.

## Proof.

$$
a^{b} \geqslant b^{a} \Leftrightarrow b \ln (a) \geqslant a \ln (b) \Leftrightarrow \frac{b}{\ln (b)} \geqslant \frac{a}{\ln (a)}
$$

The function $f: x \mapsto \frac{x}{\ln x}$ is strictly increasing for $x \geqslant e$, and $f(2)=f(4)$, which concludes the proof.

If $a=1$, then $b=1$ and vice versa. From now on, we assume that $a \geqslant 2$ and $b \geqslant 2$.

We have $a\left(a^{2}\right)^{b} \geqslant a^{b}=b^{\left(a^{2}\right)}$, so, by the lemma, $b \geqslant a^{2}$ or $\left(b=2\right.$ and $\left.a^{2}=3\right)$ or $(b=2$ and $\left.a^{2}=4\right) . a^{2}=3$ is absurd, and $(2,2)$ is not a solution.

If $b \geqslant a^{2}$, and $p$ is a prime number, we have $b \cdot v_{p}(a)=a^{2} \cdot v_{p}(b)$, so $v_{p}(a) \leqslant v_{p}(b)$.

Therefore, $a \mid b$. We set $b=a c$, then we have $a^{a c}=(a c)^{\left(a^{2}\right)}$, so $a^{c-a}=c^{a} \cdot c^{a}$ is an integer, so $c-a \geqslant 0$. We have $a^{c-a} \geqslant(c-a)^{a}$, so, as before, either $(c-a) \leqslant 2$, or $c-a \geqslant a$, so $a \mid c$. If $c-a=0$, we find the solution $(1,1)$, and $c-a \in\{1,2\}$ leads to an absurdity.

We note then $\mathrm{c}=\mathrm{da}$. We then have $\mathrm{a}^{\mathrm{a}(\mathrm{d}-1)}=(\mathrm{ad})^{\mathrm{a}}$ which is $\mathrm{a}^{\mathrm{d}-2}=d$. This implies $\mathrm{d} \leqslant 4$.

$\mathrm{d}=2$ gives $1=0$, which is absurd.

$\mathrm{d}=3$ gives $\mathrm{a}=3$, then $\mathrm{b}=27$, which is indeed a solution.

$\mathrm{d}=4$ gives $\mathrm{a}=2$, then $\mathrm{b}=16$, which is indeed a solution.