Let $a,b\in\Bbb{N}$ such that $\gcd(a,b)=1$ and $n,r\in\Bbb{N}^*$. If $ab=n^r$ then there exists $k,l\in\Bbb{N}$ such that $a=k^r$ and $b=l^r$.

Question

If $ab=n^r$ then there exists $k, l \in \mathbb{N}$ such that $a=k^r$ and $b=l^r$.

To prove this statement, I try to start by writing : $ab=n^r=p_{1}^r...p_{n}^r$ where the $p_i$ are pairwise distincts prime numbers (in fact pairwise coprime).

So, $a$ and $b$ in their prime factor decomposition meet some of the $p_i^r$. Then if we consider all the $p_i^{r}$ which divide for example $a$ then the product of the $p_i^{r}$ divide $a$. Moreover this product is coprime with $b$ so we can deduce using Gauss that : $a=x \prod p_i^{r}$.

Same with $b= y \prod p_j^{r}$.

Then as $a$ and $b$ are coprime necessarily $x$ and $y$ are coprime too. So $x=1$ or $y=1$ or both are equal to $1$. We can also have $x=p_i$ and $y=p_j$.

In every case we have found the positive integer $k$ and $l$.

I think my attempt to proof this statement is a bit unclear with the second paragraph. Is it possible to improve it or try a different approach ?

Thanks in advance !