Divisible By Itself Once More?

It works for $A=B$ and $A=-B$ .

Let,

$~~~~~~~A=Bx$

and $B=Ay$

So, $A=Axy ~\Longrightarrow~ xy=1$ .

$\because x,y\in\mathbb{Z}$ as $A$ and $B$ are both integers.

$\therefore (x,y) = (1,1)$ or $(x,y)=(-1,-1)$

Which gives us $A=B$ or $A=-B$ .

Hence, $A$ can be equal to $B$ but it is not the only case where the statement of this problem holds true.