Proof: $\;n^2\;$ is even if and only if $\;n\;$ is even. Please help how would you go about doing this? I'm studying for a final. This is on a study guide. I'm having a lot of trouble with this class.

Prove that $n^2$ is even if and only if $n$ is even.

 A: *

*The forward implication: is pretty easy: take $n=2k $ so $n^2=2\times 2k^2$ is even.

*The backward implication: we have $n^2-n=n(n-1)=\text{even}\times\text{odd}=2\times\alpha$ is even so $n=n^2-2\alpha$ is even.

A: First, we want to show that if $n$ is even then $n^2$ is even. So if $n$ is even what happens? Well, assume $n$ is even. Then what can we say about $n$?

 Then $n=2m$ for some integer $m$.

So then $n^2=$?

 $n^2=(2m)^2=4m^2=2(2m^2)$

So then $n^2$ is even because is of the form $2 \times \text{integer}$.
Now we need to show the other way around (to get the if and only if, sometimes written iff). So let's assume that $n^2$ is even. Notice if we try what we did before, we get irrational numbers and the whole thing is a mess (because of the $\sqrt{2m}$ we would get). So we do this part by contradiction. Suppose that $n^2$ is even but in fact, $n$ is not even. Then $n$ has to be odd because it is an integer, so it is either even or it is odd. Then by assumption, $n^2$ is even but $n$ is odd. Since $n$ is odd, what can we say about $n$?

 $n=2m+1$ for some integer $m$

But then what does $n^2$ have to be?

 $n^2=(2m+1)^2=4m^2+4m+1=2(2m^2+2m)+1$

But then $n^2$ is odd! But we just said it was even! That's a contradiction. That means that we must have been wrong in assuming that when $n^2$ is even that $n$ is odd. So it must be that when $n^2$ is even that $n$ is even. 
Notice we proved both directions, so we have our 'if and only if'. Hence, we have our proof and we're done. 
Big Q to the E to the D (Q.E.D.)
A: Start with the forward implication: $n$ even implies $n^2$ even. This can be done by recognizing that every even number is of the form $2k$ for some integer $k$. What happens when you square $2k$?
Next, start with the backward implication: $n^2$ even implies $n$ even. One way is by contradiction. Assume $n$ is odd. What would happen to $n^2$ if $n$ was odd? Could $n^2$ be even? You want to show that it would be impossible.
A: Let $p \in \mathbb{Z}$ s.t. $p$ is even. That is, $\exists m \in \mathbb{Z}$ s.t. $p=2m$. On squaring the equation we get, $p^2=4m^2$ $\implies$ $p^2 = 2(2m^2)$, where $2m^2 \in \mathbb{Z}$. Therefore,  $2 \mid p^2$ $\implies p^2$ is even.
Conversely, let $p^2 \in \mathbb{Z}^+$ s.t. $p^2$ is even. That is, $\exists n \in \mathbb{Z}$ s.t. $p^2=2n$, which implies that $2 \mid p^2$. We have to prove that $p$ is even. Let us assume on the contrary that $p$ is not even. That is, $\not\exists m \in \mathbb{Z}$ s.t. $p=2m$, or $2 \nmid p$; which implies that $2 \nmid p^2$. But $2 \mid p^2$ since $p^2$ is even. Therefore, we arrive at a contradiction and our assumption is wrong.
Therefore, $p$ is even. Hence Proved.
A: Hint:  If $n$ is even, you can write it as $2k$ for some $k$.  Now what is $n^2$?
A: An even number $N$ can be written as: $N=2k$. If $n$ is even you can write $n=2r$, so: $n^2=4r^2=2(2r^2)$. If you put $2r^2=k$ you get: $N=2k$
