All-Soviet-Union Mathematical Olympiad

Solution:

We have $b^3 = (a^2 - c)(a^2 + c)$, so one possibility is that $a^2 \pm c$ are both cubes. So we want two cubes whose sum is twice a square. Looking at the small cubes, we soon find $8 + 64 = 2 \cdot 36$ giving $6^4 = 28^2 + 8^3$. Multiplying through by $k^{12}$ gives an infinite family of solutions. Note that the question does not ask for all solutions.