Let $n$ be a natural number. We denote $1=d_{1}<d_{2}<d_{3}<\cdots<d_{k}=n$ its divisors. We notice that $n=d_{2}^{2}+d_{3}^{3}$. Determine all possible values of $n$.

## - Statements intended for all students -
If $d_{2} \neq 2$, then $n$ is odd, so $d_{3}$ is odd. However, $d_{2}^{2}+d_{3}^{2}$ is even, which contradicts the fact that $n$ is odd. Therefore, $d_{2}=2$ and $n$ is even.

Since $d_{3}^{3}=n-d_{2}^{2}$ is even, we can write $d_{3}=2 m$. Then $m$ is a divisor of $n$ such that $1<m<d_{3}$, so $m=2$, which implies $d_{3}=4$ and $n=2^{2}+4^{3}=68$.