https://www.artofproblemsolving.com/Forum/viewtopic.php?p=1524235&sid=f9990e38c47f0c6b633568f39856a3b5#p1524235

My solution : shivangjindal wrote: mathVNpro wrote: Let $ a,b$ be $ 2$ distinct positive interger number such that $ (a^2+ab+b^2)|ab(a+b)$. Prove that: $ |a-b|>\sqrt [3] {ab}$. Let $d=(a,b)$ so let $a=dx$ and $b=dy$ where $(x,y)=1$ So now we have we have $ \frac{dxy(x+y)}{x^2+y^2+xy} \in Z $ now note that $(x^2+y^2+xy,x) =1 , (x^2+y^2+xy,y)=1$ and similarly since $(x+y,y)=1$ we have $(x^2+y^2+xy , x+y)=(y^2,x+y) = 1 $ . since it is a integer , we have $ x^2+xy+y^2 | d $ so $d \ge x^2+xy+y^2 $ so $(|a-b|)^3 = (|dx-dy|)^3 = d^3(|x-y|)^3 \ge d^2 \cdot d \ge d^2(x^2+y^2+xy) > d^2(xy) = (dy)(dy) = ab$ So we are done $\Box$

What is wrong here? Clearly $ab|ab(a+b)|a^2+ab+b^2$ so $\frac{a^2+b^2}{ab}$ is An integer. But it is easy to prove that if it is integer so it is $+-2$. it can be True if $a=+-b$ so it is easy. please Everyone help me.

It should be $a^2 + ab + b^2$ divides $ab(a+b)$ and not the other way round.

The problem is not written correctly!