Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$. 
Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.

If $ab \mid (a+b)^²$, then $ab\mid a^2+2ab+b^2 \Longrightarrow ab\mid a^2, ab\mid 2ab$ and $ab\mid b^2$ right?
So since $(a-b)^2 = a^2-2ab+b^2$ from the assumption we have that $ab \mid a^2$ and $ab \mid b^2$. Now only remains to show that $ab \mid -2ab$ which is clearly true.
Is this valid? I'm not sure about the implication that $ab$ would divide all the terms in $a^2+2ab+b^2$.

As others have said $x \mid (y+z) \nRightarrow (x \mid y) \land (x \mid z)$
What you can say is $ab \mid (a+b)^2 \iff (a+b)^2 = kab, k \in \mathbb{Z}$
This implies $a^2 + b^2 +2ab = kab \implies a^2 + b^2 = (k-2)ab \implies a^2 + b^2 - 2ab = (k-4)ab \implies (a-b)^2 = (k-4)ab \iff ab \mid (a-b)^2$, as required.
