Belarusian Mathematical Olympiad

Answer: $n=39$. (Solution of A. Semchankau, A. Zhuk.) Let $$2013n=a^3-b^3.(1)$$ Then $$(1)\iff 61\cdot 11\cdot 3n=2013n=(a-b{)}^3+3ab(a-b)\Rightarrow (a-b)\ge 3,$$ i.e. $(a-b)=3k,k\in \mathbb{N}$. So $61\cdot 11\cdot 3n=3^3{k}^3+3^{2}abk$, whence $n\ge 3$, i.e. $n=3m,m\in \mathbb{N}$. Then (1) can be rewritten as $$3k^3+abk=61\cdot 11m.(2)$$ Note that if $k=11,b=10,a=43$, then $m=13$, i.e. $n=39$. Show that $n=39$ is the smallest possible value of $n$, i.e. $m=13$ is the smallest possible value of $m$. Indeed, from (2) it follows that $k(3k^2+3kb+b^2)\ge 11$, and it is easy to show that $k\ge 11$. But for $k\ge 22$ we have $m>13$.