Prove the following: Let $n$ be an integer. If $2|n^2$, then $4|n^2$. I came up with the following:

$2|n^2$ implies that $2|n*n$. We proved in class that if $q|b*p$, then $q|b$ or $q|p$. Therefore, if $2|n^2$, then $2|n$ or $2|n$.
So, $2|n$ implies $n=2k$, for some $k ∈ Z$. 
So, $n^2|(2k)^2$,
    $n^2 = 4k^2$, where $k^2 ∈ Z$.

So, if $2|n^2$, then $4|n^2$ as desired.

What do you all think?
If $2|n^2$, $n^2$ is even. Note that  $n^2$ is even if and only if  $n$ is even (in $\Bbb Z$). So  $n=2m $ for some  $m \in \Bbb Z$. Hence  $n^2=4m^2$.
An import ant part of writing any proof is to determine from the outset what you can assume to be true.

If you can assume the fundamental theorem of arithmetic then you have the unique prime factorisation of $n=p_1^{i_2}\cdot p_2^{i_2}\cdot p_3^{i_3}\ldots$

Then the prime factorisation of $n^2=p_1^{2i_2}\cdot p_2^{2i_2}\cdot p_3^{2i_3}\ldots$

Then since every exponent is even, the power of every prime factor of $n^2$ is either $1$ or a power of $p^2$, which generalises the result you need to every prime.

(In $2^3=8$, 2 is the prime, 3 is the exponent and 8 is the power).