Proving that for $a\in\Bbb{Z}$, $a^2\mid a$ iff $a\in\{-1, 0, 1\}$. My first argument seems shaky, my second argument seems overly simple. Suppose $a\in\Bbb{Z}$. Then $a^2\mid a$ iff $a\in\{-1, 0, 1\}$.
Showing $P\Rightarrow Q$.
For the sake of contradiction, assume $a^2\mid a$ and $a\not\in\{-1, 0, 1\}$. Then $a=a^2n, n\in\Bbb{Z}, n\not=0$. 
Dividing both sides by $a$ gives us $an=1$ and therefore $a=\frac{1}{n}$. 
If $n=1$ or $n=-1$ then $a\in\{-1, 0, 1\}$, a contradiction.
If $n\not=1$ or $n\not=-1$ then $a\not\in\Bbb{Z}$, a contradiction.
Now, showing $Q\Rightarrow P$.
Assume $a\in\{-1, 0, 1\}$.
If $a=-1$, we have $1\mid -1$, obviously true. If $a=0$, we have $0\mid 0$, obviously true. If $a=1$, we have $1\mid 1$, obviously true.
$\blacksquare$
I'm certain there is a simpler way to prove this but I want to know if the arguments I used are valid. I'm worried there is something wrong with my assumption that $n\not=0$ and I'm concerned by how trivial the second argument seems. 

(I expand here on my comment)
Yes the second argument is trivial, that's OK ! Three cases to check, you check them, you're good.
The first argument is almost perfect: you are right that it's not correct to assume $≠0$. But it comes automatically however since otherwise $=0$... 
Finally, you can simplify this first argument by removing the contradiction, and working by case separation. Assume  $a^2 \mid a$. Then there is $n\in \Bbb Z$ such that $a = a^2n$. Either $a=0$ and the proof is over, or $a \neq 0$ and you can divide, giving $1 = an$, so $a \mid 1$ so $a=\pm 1$, and the proof is over too.
