Balkan Mathematical Olympiads

(a) Every perfect cube $k^3$ of a positive integer is special because we can write $$k^3=k^3\frac{a^3+2b^3}{a^3+2b^3}=\frac{(ka{)}^3+2(kb{)}^3}{a^3+2b^3},$$ for some positive integers $a$, $b$.

(b) Observe that $2014=2\cdot 19\cdot 53$. If $2014$ is special, then we have, $$x^3+2y^3=2014(u^3+2v^3)$$ for some positive integers $x$, $y$, $u$, $v$. We may assume that $x^3+2y^3$ is minimal with this property. Now, we will use the fact that if $19$ divides $x^3+2y^3$, then it divides both $x$ and $y$. Indeed, if $19$ does not divide $x$ then it does not divide $y$ too. The relation $x^3\equiv -2y^3(\mathrm{mod}19)$ implies $(x^3{)}^6\equiv (-2y^3{)}^6(\mathrm{mod}19)$. The latter congruence is equivalent to $x^{18}\equiv 2^6{y}^{18}(\mathrm{mod}19)$. Now, according to Fermat's Little Theorem, we obtain $1\equiv 2^6(\mathrm{mod}19)$, that is $19$ divides $63$, not possible. It follows $x=19x_1$, $y=19y_1$, for some positive integers $x_1$ and $y_1$. Replacing in the previous equation we get $$19^2(x_1^3+2y_1^3)=2\cdot 53(u^3+2v^3)$$ i.e. $19$ divides $u^3+2v^3$. It follows $u=19u_1$ and $v=19v_1$, and replacing in the previous equation we get $$x_1^3+2y_1^3=2014(u_1^3+2v_1^3).$$ Clearly, $x_1^3+2y_1^3\le x^3+2y^3$, contradicting the minimality of $x^3+2y^3$.