Let $n \in \mathbb {Z}$. If $n^2$ is even, then $n$ is even. I think that a proof by contrapositive can do, but I don't know how to complete it. 
Assume that $n$ is odd. Then, $n=2k+1$ for some integer $k$. Therefore, $n^2= (2k+1)^2 = 4k^2+4k+1$. 
I know that now I should prove that $n^2$ is odd. I also know that even + even + odd = odd, but I was wondering if there is another way to prove it. Thanks in advance for any help.
 A: You can write: $n^2= (2k+1)^2 = 4k^2+4k+1=2(2k^2+2k)+1=2m+1$, where $m=2k^2+2k.$ Since $k \in \mathbb {Z}$, we have $m \in \mathbb {Z}$. Thus, $n^2= 2m+1$ for some integer $m$. Hence, $n^2$ is an odd integer. 

You can also use a proof by contradiction. 

 Assume, for the sake of contradiction, that $n^2$ is even and $n$ is odd.  Then, $n=2k+1$ for some integer $k$. Thus, $n^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1=2m+1$, where $m=2k^2+2k$. Since $k \in \mathbb {Z}$, we have $m \in \mathbb {Z}$. Thus, $n^2= 2m+1$ for some integer $m$. Hence, $n^2$ is an odd integer. Contradiction. Therefore, we deduce that if $n^2$ is even, then $n$ is even. 

A: Note that $n^2-n=n(n-1)$. Either $n$ or $n-1$ is even, and hence $n(n-1)$ is even. So, $n^2-n$ is even. But it's given that $n^2$ is even. So, $n$ must be even :).
A: We can prove something more generic $$\text{odd}\cdot\text{odd}=\text{odd}\text{ and }\text{even}\cdot\text{(any integer)}=\text{even}$$
for if $\displaystyle a\equiv1\pmod2,b\equiv1\pmod2\implies a\cdot b\equiv1\cdot1\pmod2$
and if $a\equiv0\pmod2,b$ is any integer; $a\cdot b\equiv0\cdot b\pmod2\equiv0$
A: I think the proof you give is the most standard one but I do have another. It relies on the fact that for $p,a,b \in \mathbf{Z}$ with $p$ prime, $$p|ab \implies p|a \text{ or } p|b$$
Setting $p=2, \, a=b=n$ gives $$2|n^2=n \cdot n \implies 2|n \text{ or } 2|n \implies 2|n$$
The 'fact' I've given above is established formally in first-year undergraduate number theory courses and relies on a number of other theorems. This might raise concern about circularity, but I don't think that's an issue here.
