Is this an issue with the law of the excluded middle, or an issue with the proof? Part of a proof requiring you to prove that if $x^2$ is odd then $x$ is odd (given that $x \in \mathbb{N}$). It is my understanding that the contrapositive is used for this as follows.
$x=2n, n \in \mathbb{N}$
$\Rightarrow x^2 = 4n^2$
$\Rightarrow x^2$ is even
Then using the contrapositive:
$\Rightarrow \lnot Even(x^2) \rightarrow \lnot Even(x)$
Now the Law of the excluded middle:
$\Rightarrow Odd(x^2) \rightarrow Odd(x)$
So reasonably straight forward. However, my issue with this is whilst $x^2$ is even, it is even more strictly defined as a multiple of $4$. So in the contrapositive sense it doesn't feel right that it can be any even number. So what happens if that said number is 6? Not strictly divisible by 4 but still even. This is a bit difficult because this proof's conclusion is actually correct and any squared integer is either divible by 4 or odd. But that was found through exhaustion in a different way. Using the law of the excluded middle after saying that it wasn't just any even number seems spurious.
Could someone please clarify if I am right with reservation about this? If not, please explain (without just saying it is contrapositive, therefore). I feel like there should be a continuing connection between the definition of what type of even and what can be deduced from that.

If you're in Paris, then you're in Europe. Even more particularly, you're in France. However, "if you're not in Europe, then you're not in Paris," is still a valid application of the contrapositive, even though we've "forgotten" the information about being in France.
"You're in Paris." $\longleftrightarrow  x$ is even.
"You're in Europe." $\longleftrightarrow  x^2$ is even.
"You're in France." $\longleftrightarrow  x^2$ is a multiple of $4$.
