For what positive integer $k$ can the product of the first $k$ prime numbers be the sum of two positive cubes?
Solution. Let's find the positive integer $k$ for which

$$
2 \cdot 3 \cdot 5 \cdot \ldots \cdot p_{k}=a^{3}+b^{3}
$$

where $p_{k}$ denotes the $k$-th prime in the increasing sequence of prime numbers, and $a$ and $b$ are positive integers. If $k \geq 2$, then the primes include 3, but the entire product is only divisible by 3, not by 9. On the other hand, $a^{3}+b^{3}=(a+b)^{3}-3 a b (a+b)$, the right-hand side can only be divisible by 3 if $(a+b)^{3}$ is also divisible by 3, which only holds if $a+b$ is divisible by 3. This sum appears as a factor in the second term as well, so the right-hand side is also divisible by 9, while the left-hand side is not; thus, there is no solution for $k \geq 2$.

If, however, $k=1$, then $2=1^{3}+1^{3}$; this is a solution. There are no other possibilities, so the only solution to the problem is $k=1$.