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.