Prove that if $ a^2$ divides $a$, then $a$ is in ${-1,0,1}$ Having a little trouble writing a proof of this one:
Prove that if $a^2$ divides $a$, then $a \in \{-1,0,1\}$
Thanks!
 A: If this is to hold then $ma^2=a$ for some $m\in\mathbb{Z}$. Thus, $ma^2-a=a(ma-1)=0$. Necessarily, $a=0$ or $ma=1$. Suppose $a\ne 0$. Then $m\ne 0$, and so $a=1/m\in\mathbb{Z}$. The only values of $m$ for which $1/m$ is integral are $m=\pm1$. By extension, $a=\pm 1$.
A: As $(a)^2$ is divisible by a hence we can write $K(a)^2=a$ where K is any natural number.
So $K(a^2)-a=0$
Or $a (Ka-1)=0$
So either $a=0$ or $Ka=1$
$Ka=1$ gives you two possibilities that either $K=a=1$ or $K=a=-1$ because $K$ is natural number.
Hence $a=0,1,-1$
A: If $|a|>1$, then $|a^2|>|a|$, so $a^2$ can't possibly divide $a$. This is because for any nonzero integer k, it holds that $|k|\geq1$, so $|ka^2|>|a| \Rightarrow ka^2\neq a$, and for $k=0$ the equality doesn't hold anyway. Basically, there doesn't exist a $k\in\mathbb{Z}$ such that $ka^2=a$, failing the definition of divisibility. The idea behind this is exactly the fact that you cannot divide by a number of greater absolute value.
The remaining cases, since $a$ is an integer, are $a=1$, $a=-1$ and $a=0$ whose proofs are trivial.
