The proof given in This question is incorrect (the proof will be posted at the end for convenience). **However**, the question seem to address the fact that the statement to be proven is not stated correctly, and not the fact the the proof is incorrect.

Stated different, what I am asking is...

What step in the following (see below) proof of "If a number $n^2$ is even, then $n$ is even" is incorrect?

(Some step must be incorrect since If I change $n^2$ to $x$ and $n$ to $\sqrt{x}$, we would have a proof saying "if you give me any even number $x$, its square root is even")

The proof (by contrapositive) given is as follows:

If a number $n^2$ is even, then $n$ is even. The contrapositive is that is that if $n$ is not even (odd), then $n^2$ must also be not be even (be odd).

We represent n as $n=2p+1$. $n^2=4p^2 + 4p + 1 = 2(p^2+2) + 1$. We see that $n^2$ is odd. Therefore, the original statement must be true.

Is the problem that, in general, the negation of "$n$ is even", but rather "$n$ is odd OR $n$ is not an integer$?