Use a contrapositive proof to show that $mn$ is a perfect square if $m$ and $n$ are

Question

Definition: $a\in \Bbb Z$ is a perfect square if there is a $b\in\Bbb Z$ and $a = b^2$

To prove: if $m$ and $n$ are perfect squares, then $mn$ is a perfect square.

I know that this can most easily be proved with a direct proof. I want to prove this using contrapositive. Is the below valid, and if not, what would be? Also, tips on correctness of how I wrote it would be appreciated.

Proof: by contrapositive

Assume $mn$ is not a perfect square. Let $b$, $c$ $\in \Bbb Z$ where $m \ne b^2$, $n \ne c^2$. Then $mn \ne b^2c^2 \ne (bc)^2$. QED.

Answer 1

Using here the contrapositive of $p \wedge q \Rightarrow r$ which is $\sim r \Rightarrow \sim(p \wedge q) \equiv (\sim p) \vee (\sim q) $. We have

If $mn$ is not a perfect square then either $m$ is not a perfect square or $n$ is not a perfect square.

Answer 2

Your proof is not entirely correct. As it is, you proved that $mn\neq (bc)^2$, for very specific $b$ and $c$ This is not at all close to a full proof.

To prove that $mn$ not a perfect square $\implies $ either $m$ or $n$ is not a perfect square, here is something you could do:
Using the fundamental theorem of airithmetic $$mn=p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}\text{ not a perfect square}\iff \text{ at least one of the $r_i$ is odd}.$$ We may say without loss of generality that $r_1$ is odd. This means that $m$ and $n$ cannot both have an even number of $p_1$'s in their prime factorization, i.e. one of them has an odd number of $p_1$'s in its prime factorization. This is (as we said) equivalent to one of them not being a perfect square. This proves the statement.