"Since $2k^2$ is an integer, $n^2$ is even." Why? I don't understand the correlation between $2k^2$ being an integer and $n^2$ being even.

Prove: For all integers $n$, if $n$ is even, then $n^2$ is even.
To fill in the details, we will basically just explain what it means for $n$ to be even, and then see what that means for $n^2$. Here is a
complete proof.
Proof: Let $n$ be an arbitrary integer. Suppose $n$ is even. Then $n = 2k$ for some integer k. Now $n^2 = (2k)^2 = 4k^2 = 2(2k^2)$. Since $2k^2$ is an
integer, $n^2$ is even.

This is supposed to be an introductory proof, but while the first part is easy to follow, I don't get where the "since $2k^2$ is an integer, $n^2$ is even" conclusion comes from. Why does $2k^2$ have to be an integer in order to prove that if n is even, so is $n^2$?
 A: Asserting that an integer is even is the same thing as asserting that it is twice another integer. So, since $2k^2$ is an integer, $2(2k^2)$ is an even integer. But $n^2=2(2k^2)$.
A: Because n is even we can write it of the from $n=2k$ where k is an integer (because if n is even then $\frac{n}{2}$ must be integer. As you have written in the proof $n^2=2(2k^2)$, and because of the fact $2k^2$ is an integer, 2 multiplied by an integer must be even.
Hopefully that helps you :)
A: To show that $n^2$ is even, by definition we need to show that it is twice an integer—it is not enough for it to simply be twice any number. For example, $1^2=1=2\times\dfrac{1}{2}$, but $1^2$ is not even.
A: If $n^2$ is even, then it can be written $n^2=2m$ (where $m$ is an integer) in the same way that $n=2k$ was written (where $k$ is an integer).
The "since $2k^2$ is an integer" bit is just reiterating that the "where $m$ is an integer" bit is proven already by the choice of $m=2k^2$ (because multiplying integers by integers gives integers making the right-hand-side an integer and thus making $m$ on the left an integer as well).
