Task B-4.4. Determine all natural numbers $a$ for which the number $a^{3}+1$ is a power of 3.
## First solution.

If $a^{3}+1$ is divisible by 3, the number $a$ is of the form $a=3 p-1, p \in \mathbb{N}$. Then,

$$
a^{3}+1=(3 p-1)^{3}+1=27 p^{3}-27 p^{2}+9 p-1+1=9 p\left(3 p^{2}-3 p+1\right)
$$

The expression $3 p^{2}-3 p+1$ is not divisible by 3, and since $a^{3}+1$ is a power of 3, it can only be equal to $3^{0}$.

Then from

$$
3 p^{2}-3 p+1=1
$$

it follows that $p=1$, and $a^{3}+1=9$ so $a=2$.