Sum of three consecutive cubes equals a perfect square 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?

What you can do (but not ask from seventh graders) is to turn the equation into an elliptic curve in Weierstrass normal form, by multiplying both sides by $9$ and putting $9S=Y^2$ and $X=3x$. Then we get
$$Y^2=X^3+18X$$
which indeed denotes an elliptic curve and we can use for instance SageMath to return its integral points: $(0, 0), (3, 9), (6, 18), (72, 612)$.
Then we find $x=0,1,2,24$.
In fact, something similar can be done if we have an odd number, say $2t+1$ of consecutive cubes whose sum must be a perfect square. If we let $n^3$ be the central element of the cubes, the sum is $(n-t)^3+\cdots +n^3+\cdots +(n+t)^3=(2t+1)n^3+6n(1+4+\cdots +t^2)=(2t+1)n^3+t(t+1)(2t+1)n$
We know this is a square when $n=t,t+1$ (since the sum of the first $k$ cubes is a perfect square), and less obvious, but easy to verify, is that $n=4t(t+1)(2t+1)$ (which is $24$ times the sum of the first $t$ squares) is also a solution.
