Problem 1637

Let's break down the problem into two parts and provide a detailed solution for each.

### Part (a): Prove that $k = a^3 + b^3 + c^3$ is not a prime number.

1. Given: $a + b + c = 2pq(p^{30} + q^{30})$ where $p > q$ are positive integers.
2. Observation: Since $a, b, c$ are natural numbers, $a + b + c$ is a natural number.
3. Sum of Cubes: We know that $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.
4. Given Condition: $a + b + c = 2pq(p^{30} + q^{30})$.
5. Evenness: Since $a + b + c$ is even, $k = a^3 + b^3 + c^3$ must also be even.
6. Prime Check: The smallest even prime number is 2. For $k$ to be 2, we would need $a^3 + b^3 + c^3 = 2$. However, this is impossible because $a, b, c$ are natural numbers and the smallest sum of cubes of natural numbers is $1^3 + 1^3 + 1^3 = 3$.
7. Conclusion: Since $k$ is even and greater than 2, $k$ cannot be a prime number.

### Part (b): Prove that if $a \cdot b \cdot c$ is maximum, then $1984$ divides $k$.

1. Given: $a + b + c = 2pq(p^{30} + q^{30})$.
2. Divisibility by 3: Notice that $a + b + c$ is divisible by 3 because if $3 \nmid p, q$, then $3 \mid p^2 - q^2 \mid p^{30} - q^{30}$. Hence, $a + b + c$ is divisible by 3.
3. AM-GM Inequality: To maximize $a \cdot b \cdot c$, by the Arithmetic Mean-Geometric Mean Inequality (AM-GM), $a = b = c$.
4. **Expression for $a$**: Let $a = b = c = \frac{2pq(p^{30} + q^{30})}{3}$.
5. **Expression for $k$**:
$$k = 3a^3 = 3 \left( \frac{2pq(p^{30} + q^{30})}{3} \right)^3 = \frac{8p^3q^3(p^{30} + q^{30})^3}{9}$$
6. Divisibility by 64: If at least one of $p$ or $q$ is even, $k$ is visibly divisible by $64$. If both are odd, $(p^{30} - q^{30})^3$ has an extra factor of $8$ to complete the factor of $64$.
7. Divisibility by 31: If at least one of $p$ or $q$ is divisible by 31, so is $k$. If not, $(31, pq) = 1$ and by Fermat's Little Theorem, $p^{30} \equiv q^{30} \equiv 1 \pmod{31}$, so $p^{30} - q^{30} \equiv 0 \pmod{31}$. Thus, $k$ is divisible by 31.
8. Conclusion: Combining these, $k$ is divisible by $64 \cdot 31 = 1984$.

The final answer is $\boxed{ k }$ is divisible by $1984$.