Prove that for any integer $n$, if $b^2$ divides $n$, then $b$ divides $n$. 
Prove that for any integer $n$, if $b^2$ divides $n$, then $b$ divides $n$.

Trying to figure out this proof. The proof I'm looking at is written as
$n$ = any integer, if $25|n \implies 5|n$. 
I've been trying to figure this for days and have been running around in circles. Would appreciate a general proof for this.
$n$ = any integer, if $(b^2)|n \implies b|n$.

Since an answer has been given ...
The $\mid$ relation is transitive; if $x \mid y$ and $y\mid z$, then $x\mid z$. Let $x=b$, $y=b^2$, and $z=n$. Done.
