Final Round of National Olympiad

Answer: No.

Suppose that $2013=x^3-y^3$ where $x$ and $y$ are integers. Note that $$x^3-y^3=(x-y{)}^3+3x^{2}y-3x{y}^2=(x-y{)}^3+3xy(x-y).$$ As $2013$ is divisible by $3$ and so is $3xy(x-y)$, the difference $(x-y{)}^3$ must be divisible by $3$. Thus also $x-y$ is divisible by $3$ as $3$ is prime. Consequently, $3xy(x-y)$ is divisible by $3^2$ and $(x-y{)}^3$ is divisible by $3^3$, whence the sum $x^3-y^3$ is divisible by $3^2$. But $2013$ is not divisible by higher powers of $3$. The contradiction shows that $2013$ cannot be represented as the difference of two cubes of integers.

---

Alternative solution.

Suppose that $2013=x^3-y^3$ where $x$ and $y$ are integers. Then $x^3\equiv y^3(\mathrm{mod}3)$. As integers are congruent to their cubes modulo $3$, this gives $x\equiv y(\mathrm{mod}3)$. Now write $2013=(x-y)(x^2+xy+y^2)$. By the congruence obtained above, the first factor in the r.h.s. is divisible by $3$ and the terms $x^2$, $xy$ and $y^2$ in the second factor are all congruent modulo $3$ whence $x^2+xy+y^2$ is divisible by $3$. Altogether, the product $(x-y)(x^2+xy+y^2)$ must be divisible by $9$, but $2013$ is not.