Macedonian Mathematical Olympiad

It is easy to show that for every whole number $a$ holds $3|a^3-a$, and that the square of a number modulo $3$ can be $0$ or $1$.

Now the given equation is equivalent to $x^3-x+2(y^3-y)-3(x-y)+z^2=2012$. From the above concluded and from the fact that $2012\equiv 2(\mathrm{mod}3)$ we get that $z^2\equiv 2(\mathrm{mod}3)$, which is impossible. Hence the given equation has no solution in the set of whole numbers.