If $a$, $b$, $z$ and $y$ are integers such that $\gcd(a,b)=1$ and $x^a=y^b$, show that $x=n^b$ and $y=n^a$ for some integer $n>1$

Question

I first noted that the set of primes dividing $x$ will be the same as the set of primes dividing $y$. Then i assumed $x=p^l$ and $y=p^m$ (where $p$ is any prime dividing $x$ and $y$ and $p^l$ is the largest power of $p$ dividing $x$ and $p^m$ is the largest power of $p$ diving $y$) Hence $x^a=y^b$ can be written as $p^{la}=p^{mb}$, thus $la=mb$. This would mean that $a$ divides $m$ and $b\mid l$ since $(a,b)=1$. I though a lot but couldn't go any further. Am i going in the right direction? please help me further.

Answer 1

$la=mb,\ \gcd(a,b)=1\Longrightarrow l=bt, m=at$ for some integer $t$. Now, $p^l=p^{bt}=(p^t)^b\mid x$ and $p^m = (p^t)^a\mid y$. And it's true for all primes $p$ dividing $x$ (and $y$). So, $x=(p_1^{t_1})^b\cdots (p_k^{t_k})^b = n^b$ and $y=(p_1^{t_1})^a\cdots (p_k^{t_k})^a = n^a$.