Sum of three consecutive cubes When I noticed that $3^3+4^3+5^3=6^3$, I wondered if there are any other times where $(a-1)^3+a^3+(a+1)^3$ equals another cube. That expression simplifies to $3a(a^2+2)$ and I'm still trying to find another value of $a$ that satisfies the condition (the only one found is $a=4$)
Is this impossible? (It doesn't happen for $3 \leq a \leq 10000$) Is it possible to prove?
 A: How about $$(-1)^3+0^3+1^3=0^3?$$
A: How about 
$$\left(-\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^3 + \left(\frac{3}{2}\right)^3 = \left(\frac{3}{2}\right)^3 ?$$
After all, the OP didn't specify where $a$ lives... (by the way, there are infinitely many distinct rational solutions of this form!).
Now for a more enlightened answer: no, there are no other integral solutions with $a\in\mathbb{Z}$, other than $a=0$ and $a=4$. Here is why (what follows is a sketch of the argument, several details would be too lengthy to write fully).
Suppose $(a-1)^3+a^3+(a+1)^3=b^3$. Then $3a^3+6a=b^3$. Hence $(a:b:1)$ is a point on the elliptic curve $E:3Z^3+6ZY^2=X^3$ with origin at $(0:1:0)$. In particular, a theorem of Siegel tells us that there are at most finitely many integral solutions of $3a^3+6a=b^3$ with $a,b\in\mathbb{Z}$. Now the hard part is to prove that there are exactly $2$ integral solutions.
With a change of variables $U=X/Z$ and $V=Y/Z$ followed by a change $U=x/6$ and $V=y/36$, we can look instead at the curve $E':y^2=x^3-648$. This curve has a trivial torsion subgroup and rank $2$, with generators $(18,72)$ and $(9,9)$. Moreover each point $(x,y)$ in $E'$ corresponds to a (projective) point $(x/6:y/36:1)$ on $E$, and a point $(X:Y:Z)$ on $E$ corresponds to a solution $a=Z/Y$ and $b=X/Y$. This means that $E$ is generated by $P_1=(3:2:1)$ and $P_2=(18:3:12)$ which correspond respectively to $(a,b)=(1/2,3/2)$ and $(4,6)$. The origin $(0:1:0)$ corresponds to $(a,b)=(0,0)$.
Now it is a matter of looking through all $\mathbb{Z}$-linear combinations of $P_1$ and $P_2$ to see if any gives another $(a,b)$ integral. However, this is a finite search, because of the way heights of points work, and one can calculate a bound on the height for a point $(a,b)$ to have both integral coordinates. Once this bound is found, a search among a few small linear combinations of $P_1$ and $P_2$ shows that $(0,0)$ and $(4,6)$ are actually the only two possible integral solutions.
Here is another rational solution, not as trivial as the first one I offered, that appears from $P_1-P_2$:
$$\left(-\frac{10}{11}\right)^3 + \left(\frac{1}{11}\right)^3 + \left(\frac{12}{11}\right)^3 = \left(\frac{9}{11}\right)^3 $$
