Making Cubes

Define $f(a) = 2a^3 + 24a$ . Then we note the following sequence of equalities: $\begin{aligned} f(a) = 2a^3 + 24a & = 2(a^3 + 12a) \\ & = 2(a^3 + 12a) + (6a^2 - 6a^2) + (8 - 8) \\ & = (a^3 + 6a^2 + 12a + 8) + (a^3 - 6a^2 + 12a - 8) \\ & = (a + 2)^3 + (a - 2)^3 \end{aligned}$ Suppose that for some integer $x$ , $(a + 2)^3 + (a - 2)^3 = x^3$ . We analyze three exhaustive cases.

• Case 1 : $x > 0$ . Then, since $a$ is positive, $a + 2$ is positive. Fermat's Last Theorem rules out the possibility that $a - 2$ is positive, so we require $a - 2 \leq 0$ , or $a \leq 2$ . Since $a > 0$ , the only two values we consider are $a = 1,2$ . We have $f(1) = 2 + 24 = 26$ , which is not a perfect cube. However, $f(2) = 16 + 48 = 64 = 4^3$ , so $a = 2$ is a solution.
• Case 2 : $x = 0$ . Then, we are solving $2a^3 + 24a = 0$ . This factors as $2a(a^2 + 12) = 0$ , so the only solutions are $a = 0, \pm 2\sqrt{3}i$ , none of which are positive integers. Therefore, this case yields no solutions.
• Case 3 : $x < 0$ . Then, for some positive integer $y$ , $x = -y$ . Hence, we have $(a + 2)^3 + (a - 2)^3 = -y^3$ . Rearranging slightly gives $(a + 2)^3 + y^3 = (2 - a)^3$ Since $a + 2$ and $y$ are positive, the left side is necessarily positive. However, the right side is negative for all $a > 2$ , so since $a > 0$ , we once again only allow $a = 1,2$ . We have already computed $f(1)$ and $f(2)$ in case 1, and neither of these values are negative, so they cannot be equal to $-y^3$ . Therefore, this case yields no solutions.

Hence, there is exactly one solution, namely $a = 2$ as obtained in the first case, so the answer is $\boxed{1}$ .