A fraction $\frac{p}{q}$ is a natural number iff $\left(\frac{p}{q}\right)^2$ is a natural number 
Let $f$ be defined as follows
$$f: \Bbb Q \to \Bbb Q$$
$$f(x) = x^2$$
Show that $f(ℕ) ∩ ℕ = f(\Bbb Q) ∩ ℕ$


My attempt at solving this:
$$f(ℕ) ∩ ℕ = \{y \in \Bbb N \quad | \quad y = x^2, \quad x \in \Bbb N\}$$
$$f(\Bbb Q) ∩ ℕ = \{y \in \Bbb N \quad | \quad y = x^2, \quad x \in \Bbb Q\}$$
So it is sufficient to show that 
$$\frac{p}{q} \in \Bbb N \iff \left(\frac{p}{q}\right)^2 \in \Bbb N$$
By definition it follows that
$$\frac{p}{q} \in \Bbb N \iff p = q \cdot l  
   \qquad p,q,l \in \Bbb N \quad q \neq 0$$
I also found out that $$\left(\frac{p}{q}\right)^2 \in \Bbb N \iff p^2 = q^2 \cdot l^2$$
(the value of $l$ stays the same, do I have to prove this?)
So $q \cdot l - p = 0 \iff q^2 \cdot l^2 - p^2 = 0$? And if so, how to proceed?
Am I on the right track?
 A: if $$\frac{p}{q}$$ is a natural number then it must be $$\frac{p}{q}=m$$ and $m$ is a natural number, then we get by squaring $$p^2=q^2m^2$$ and $$\frac{p^2}{q^2}=m^2$$ is a natural number.
if $$\frac{p^2}{q^2}=m^2$$ then we have $$p^2=q^2m^2$$ or $$(p-qm)(p+qm)=0$$ from here we get $$\frac{p}{q}=m$$ and this is a natural number.
A: I think the converse is a bit trickier than as shown in the above answer. In Travis's original post, when talking about the value of $l$ stays the same, yes it does, but we cannot assume that $l \in \mathbb{N}$ as stated in the previous line.
Here's my take at this:
It is straightforward that $\frac{p}{q} \in \mathbb{N} \implies \frac{p^2}{q^2} \in \mathbb{N}$. Let's now prove $\frac{p^2}{q^2} \in \mathbb{N} \implies \frac{p}{q} \in \mathbb{N}$: 
Suppose $\frac{p}{q} \notin \mathbb{N}$. Then, $\exists \hat{p}, \hat{q} \in \mathbb{N}: \frac{p}{q} = \frac{\hat{p}}{\hat{q}} $ and $(\hat{p},\hat{q}) = 1$. (Reduced $\frac{p}{q}$ to its simpliest form)
Hence, $\frac{p^2}{q^2} = \frac{\hat{p}^2}{\hat{q}^2}$ and $(\hat{p}^2,\hat{q}^2) = 1\implies \frac{p^2}{q^2} \notin \mathbb{N}$. This is the contrapositive of the original statement. 
