THE DANUBE MATHEMATICAL COMPETITION

First, we prove that $n$ is square-free. If $d^2$ divides $n$ for a positive integer $d>1$, then $(d^2{)}^2+d^2+1$ would be a prime number. But $d^4+d^2+1=(d^2-d+1)(d^2+d+1)$, with both factors larger than $1$, which is a contradiction.

Thus, $n=p_1\cdot p_2\cdot \cdots \cdot p_s$, where $s\in \mathbb{N}$ and $p_1<p_2<\cdots <p_s$ are prime numbers. Let $p>5$ be a prime number. Then $p\equiv 1(\mathrm{mod}6)$ or $p\equiv 5(\mathrm{mod}6)$. If $p\equiv 1(\mathrm{mod}6)$, then $p^2+p+1\equiv 3(\mathrm{mod}6)$, and $p^2+p+1>3$ is composite.

If $p\equiv 5(\mathrm{mod}6)$, then $p^2-p+1\equiv 3(\mathrm{mod}6)$, and $p^2-p+1>3$ is composite.

In conclusion, the only prime factors of $n$ can be $2$ and $3$, so $n\in \{2,3,6\}$. It is easy to check that all these three numbers fulfill the given condition.