Let $a,b,c,n \in \mathbb{N}$ with $gcd(a,b) = 1$ and $ab = c^n$. Task: Show that there exist $d,e \in \mathbb{N}$, so that $a = d^n$ and $b = e^n$.
My thoughts about that are, that because of $gcd(a,b) = 1$ the following equality holds: $$ax + by = 1 \text{ with some } x,y \in \mathbb{Z}$$
Now I tried different approaches, e.g. writing $a = \frac{c^n}{b}$ and substituting this in the equality above. This lead to $$\frac{c^n}{b}x + by = 1 \iff b^2y - b + c^nx = 0$$ But putting this into the p-q-formula, I can't show the b is of the desired form and in analog way, that a if of the desired form.
I also tried multiplying the whole equality with b and got $$abx + b^2y = b,$$ this lead me to $c^nx + b^2y = b$ and so to the same equality like above.
What could be a constructive approach to get this task solved?
Let $d=(a,c)$. In particular $d|a$, and so $(d,b)=1$. We also have $d^n|c^n=ab$, and since $d^n$ is coprime to $b$ it follows that $d^n|a$.
Now, we clearly have $(\frac{a}{d}, \frac{c}{d})=1$. Hence also $((\frac{a}{d})^n, (\frac{c}{d})^n)=(\frac{a^n}{d^n}, \frac{c^n}{d^n})=1$. But note that both $\frac{a^n}{d^n}$ and $\frac{c^n}{d^n}$ are divisible by $\frac{a}{d^n}$. Thus $\frac{a}{d^n}=1$, and so $a=d^n$. And then:
$b=\frac{c^n}{a}=(\frac{c}{d})^n$