Find all couples $(a,b) \in \mathbb{N}^2$ such that $a^{2b}=b^{3a}$ 
Find all couples $(a,b) \in \mathbb{N}^2$ such that $a^{2b}=b^{3a}$

I thought I could use prime factorizations, to show that $b|a$ if $3a \geq 2b$ and that $a|b$ if $3a < 2b$ but not sure how to put it together. 

Let $p$ be a prime. If $p$ occurs in $a$ to the $k$th power and in $b$ to the $m$th power, then it occurs in $a^{2b}$ to the $2bk$th power and in $b^{3a}$ to the $3am$th power. Thus for all $p$, we must have $2bk=3am$.
In particular, if $2b<3a$ then $k\ge m$, i.e., $b\mid a$, and if $2b>3a$ then $k\le m$, i.e., $a\mid b$. We can exclude $2b=3a>0$ as that would imply $k=m$ for all $p$.
Consider the case $2b<3a$, $a=cb$, so $(cb)^{2b}=b^{3cb}$.
Taking $b$th roots and rearranging, we have $c^2=b^{3c-2}$.
If $c=1$, this implies $b=1$, then $a=1$, indeed a solution.
If $c\ge 2$, we need $b\ge2$, but then $b^{3c-2}\ge 2^{3c-2}>c^2$ (by induction. for example).
Now consider the case $2b>3a$, $b=ca$, so $a^{2ca}=(ca)^{3a}$.
Taking $a$th roots and rearranging, $a^{2c-3} = c^3$.
From $2ca>3a$, we see $c\ge 2$.
$c=2$ leads to $a=8$, $b=16$, a valid solution.
$c=3$ leads to $a=3$, $b=9$, anothr solution.
$c=4$ leads to $a^5=64$, which is impossible.
$c\ge5$ leads to $a^{2c-3}\ge 2^{2c-3}>c^3$ (by induction. for example).
In summary, we have $(a,b)=(1,1)$, $(8,16)$, or $(3,9)$ (or if we allow $0\in\Bbb N$, also $(0,0)$).
