Prove if $n^2$ is even, then $n$ is even.

Solution 1:

It could use a little more explanation, but yes, it works. I’d expand it to point out explicitly why $n(n+1)$ is even and that $n=(n^2+n)-n^2$ is then the difference of two even numbers and as such is even (assuming that you already have this fact available to use).

An alternative approach is to show that if $n$ is odd, then $n^2$ is odd; the desired result is the contrapositive and therefore follows at once.

Solution 2:

This proof is valid.

I suggest you add the small detail that the difference of even integers is even.

Solution 3:

If $n$ is odd say $n=2k+1$ then $$n^2=(2k+1)^2=(2k+1)\times (2k+1)=2\times(k(2k+2))+1\quad\text{is odd}$$ hence necessary $n$ is even