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?

Here is an approachable (but not exhaustive) algebraic route to obtain the solution $23,24,25$.
OP obtains $S=3x(x^2+2)$. Plainly, $x^2+2$ cannot itself be a square (there are no two integer squares which differ by $2$), so OP makes the (limiting) assumption that to obtain a square, you must set $3x=x^2+2$. It is also possible, however, that $3x=a\cdot b^2$ and $x^2+2=a\cdot c^2$, yielding a product $a^2b^2c^2$. Furthermore, since we have one explicit factor of $3$ to account for, one of $x$ or $x^2+2$ must contain another factor of $3$.
Let's assume here that $x=3k$. Then $3x(x^2+2)=9k(9k^2+2)$; and if that is a square, then so is $k(9k^2+2)$. Now $gcd(k,(9k^2+2))=1,2$. If it is $1$, then $k=d^2$ and $(9k^2+2)=9d^4+2$. But this is impossible because $9d^4+2$ is $2$ greater than a square, and therefore cannot be a square.
So we conclude that $k=2n$ and $k(9k^2+2)=2n(36n^2+2)=4n(18n^2+1)$. Again, since this last expression is a square, so is $n(18n^2+1)$. Here, $gcd(n,(18n^2+1))=1$, so we arrive at $n=d^2$ and $18n^2+1=18d^4+1$.
$18d^4+1=m^2$ has the solution $d=2,m=17$, from which $n=4,k=8,x=24$, yielding the overall solution $23,24,25$.
