1. Determine all natural numbers $n>2$ such that $n=a^{3}+b^{3}$, where $a$ is the smallest natural divisor of $n$ greater than 1 and $b$ is any natural divisor of $n$.
Solution. First, let's note that if $n$ is odd, then both divisors $a$ and $b$ must be odd. However, their sum is even, so $n$ must be even, which is a contradiction. Therefore, $n$ is even and $a=2$. Then $b$ must also be even. It also holds that $b \mid (n - b^3) = a^3 = 8$. From this, we get that $b \in \{2, 4, 8\}$. Therefore, the desired numbers $n$ are $n=16$, $n=72$, and $n=520$.