Cube Hunting 2

Relevant wiki: Fermat's Last Theorem

Note first that $n=0$ is not a solution. If $m,n$ are integers such that $n^6 + 24n^3 + 192 = m^3$ , then $\begin{aligned} n^9 + 24n^6 + 192n^3 & = \; (mn)^3 \\ (n^3 + 8)^3 & = \; (mn)^3 + 8^3 \end{aligned}$ It is well-known that the only solutions in integers of the equation $x^3 + y^3 = z^3$ occur with one or more of $x,y,z$ being equal to $0$ . If $mn = 0$ then (since $n \neq 0$ ), we deduce that $m=0$ , and hence $(n^3 + 8)^3 = 8^3$ , so that $n^3 + 8 = 8$ , and hence $n=0$ . This case is not possible, Thus we must have $n^3 + 8 = 0$ , so that $n=-2$ and hence $m=4$ .

There is only one solution: $n=-2$ . Thus the answer is $1 + (-2) = \boxed{-1}$ .

Honestly I was, mistakenly, looking for all solutions for which this expression is not a perfect cube but rather a perfect square . Interesting enough, it gave the same result. Rewritten as $(n^3+12)^2+48=M^2$ for some integer $M$ , we deduce that the only possibilities are $M=(n^3+12)^2+2,\ M=(n^3+12)^2+4,\ M=(n^3+12)^2+6,$ , for which $(n^3+12)=11,4,1$ respectively. The first two give $n=-1,\ -2$ , the last one is invalid as -11 isn't a perfect cube (while -1, -8 are).

Thus $N=2$ , $\Sigma_{j=1}^2n_j=-3$ , and the final result is $-1$ , too. Weird, isn't it?

Also, following the correct solution for the original problem together with this result, it follows that for $n=-2$ the expression is a perfect sixth power , and is the only one. And indeed, it equals $64=2^6$ .