I define a relation on $\Bbb N$ as follows:
$x \sim y \Longleftrightarrow \ \exists \ j,k \in \Bbb Z$ s.t. $x \mid y^j \ \wedge \ y \mid x^k$
I have shown that $\sim$ is an equivalence relation by proving symmetry, reflexivity, and transitivity, so now I am trying to determine the equivalence classes $[1], [2], [9], [10], [20]$
My Work
\begin{align*} \left[1\right] &= \{1\}\\ \left[2\right] &= \{2^n \mid n \in \Bbb N\} \\ \left[9\right] &= \{3^n \mid n \in \Bbb N\} \\ \left[10\right]&= \{2^n\cdot 5^m \mid n,m \in \Bbb N\} \\ \left[20\right]&= \left[10\right] \end{align*}
In general, $$\left[x\right] = \{p_1^{n_1}\cdots p_m^{n_m} \mid n_i \in \Bbb N \ \forall i \}$$ where $p_1 \cdots p_m$ is the prime factorization of $x$, which each prime raised to the first power, which is why $[10] = [20]$.
My Question
Is this a valid way to answer this question? This is just what makes sense to me, but I'm not sure if more explicit reasoning is needed.
You should state explicitly and prove the following general lemma. Your post shows that you are perfectly aware of it.
Lemma: Let $p_1,p_2,\dots, p_k$ be distinct primes, and let $m=p_1p_2\cdots p_k$. Let $n$ be a positive integer. Then $[n]=[m]$ iff there exist positive integers $a_1,a_2,\dots,a_k$ such that $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$.
After that, all of the specific cases are immediate consequences of the lemma.