BMO 2010 Shortlist

Suppose first that $a+b+c=0$. Then we have $$\begin{array}{rl}|a^{3}b+b^{3}c+c^{3}a|& =|a^{3}b+b^3(-a-b)+(-a-b{)}^{3}a|\\ & =|-b^4-2b^{3}a-3a^2{b}^2-2a^{3}b-a^4|\\ & =(a^2+ab+b^2{)}^2.\end{array}$$ So it suffices to partition the set of all integers into triples of zero sum. One way to do this is the following:

a) Begin with the triple $(-1,0,1)$.

b) Then, let on every next step $p$ and $q$ be the least two positive integers not paired yet.

c) Form the triplets $(p,q,-p-q)$ and $(-p,-q,p+q)$ and repeat.

It is easy to see that this procedure works. $\square$