Proof for elementary divisibility problem

Question

Not sure if my thinking is correct. For the problem "$a$ divides $b$ if and only if $a$ divides $b^2$." So far my proof goes: since $a$ divides $b$ there exists an integer $n$ such that $b=an$. Then $b^2=a^2*n^2=a(an^2)$. Hence $a$ divides $b^2$. I am having a problem proving the converse.

Answer 1

It is incorrect. $16$ divides $16=4^2$, but $16$ doesn't divide $4$.

Answer 2

The converse isn't true. let's say $$an=b^2$$ for some $n$. Now let's say $a>b$. Then $n<b$ and thus: $$a>b$$. which means $a$ doesn't divide $b$, but $a$ does divide $b^2$