I have found this problem in an old German textbook: Find all sets of three consecutive integers such that the sum of their cubes is a perfect square.
We can write $$S = (x-1)^3 + x^3 + (x+1)^3 = (x-1+x+1)((x-1)^2 - (x-1)(x-1) + (x+1)^2) + x^3$$ which reduces to $$S = 3x(x^2 + 2).$$
If we set $x^2 + 2 = 3x$, we get $$x^2 - 3x + 2 = 0 \iff (x-1)(x-2) = 0$$ and we thus obtain the solutions $(0,1,2)$ and $(1,2,3)$.
At first I conjectured that these are the only solutions, but I couldn't prove this. However, I was wrong: $(23,24,25)$ also satisfies the relationship.
It is worth noting that these are the only solutions for $x \leq 100000$. Can anyboy help me prove that these are the only ones? Or otherwise, help me find all other triples?