## Task A-1.3.

Determine all natural numbers $n>2$ for which there exists a divisor $d$ of the number $n$ such that

$$
n=a^{3}+d^{3}
$$

where $a$ is the smallest divisor of the number $n$ greater than 1.
## First solution.

If we assume that $n$ is odd, then its divisors and their cubes are also odd, which means that $a^{3}+d^{3}$ is an even number. On the other hand, since $n=a^{3}+d^{3}$, this is impossible. Therefore, $n$ must be even.

3 points

For even numbers, the smallest positive divisor greater than 1 is 2, i.e., $a=2$.

2 points

If $d$ is a divisor of $n$, then it is also a divisor of $n-d^{3}$.

2 points

From $n-d^{3}=a^{3}=8$ it follows that $d \mid 8$, so the only possible candidates are $d=1,2,4,8$.

1 point

For $d=1$, we get $n=9$, which is not a solution because $n$ must be an even number.

1 point

From the remaining possibilities $d=2,4,8$, we get the solutions $n=16, n=72$, and $n=520$ respectively.

1 point