# 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?

-
I think you mean $a=4$, not $a=2$: $1^3 + 2^3 + 3^3 = 36$ is not a cube, but $3^3 + 4^3 + 5^3 = 6^3$ –  Robert Israel Mar 14 '12 at 22:07
@RobertIsrael Yup, I changed that :) –  itdoesntwork Mar 14 '12 at 22:10
Have you read this article? It seems that that is the only three consecutive integers whose cubes sum to a cube. –  Jack Mar 14 '12 at 22:11
It is also known as a Plato's number en.wikipedia.org/wiki/Plato%27s_number –  Jeremy Carlos Mar 14 '12 at 22:12
@Jack is it difficult to prove? I think I've read that but still tried to find one. –  itdoesntwork Mar 14 '12 at 22:16

How about $$(-1)^3+0^3+1^3=0^3?$$

-

Here's a start. If you notice, $\gcd(a,a^2+2)=\gcd(a,2)$, so by the unique factorization theorem, \eqalign{ & 3a(a^2+2)=z^3 \\ & \text{is a perfect cube} } ~\iff~ \matrix{ a &=& a_1^3\cdot2^{a_2}\cdot3^{a_3} \\ a^2+2 &=& b_1^3\cdot2^{b_2}\cdot3^{b_3} } \quad \text{with} \quad \eqalign{ & 2,3\not\mid a_1,b_1 \\ & a_2+b_2\equiv0\pmod3 \\ & a_3+b_3\equiv2\pmod3, } with all $a_i,b_i$ nonnegative except for $a_1$, which has the same sign as $a$. This has known solutions $$a=0,\quad\text{for which}\quad \mathbb{a} =\left[\matrix{a_1\cr a_2\cr a_3}\right] =\left[\matrix{0\cr0\cr0}\right],\quad \mathbb{b} =\left[\matrix{b_1\cr b_2\cr b_3}\right] =\left[\matrix{1\cr1\cr0}\right], \tag{solution 0}$$ and $$a=\pm4,\quad\text{for which}\quad \mathbb{a}=\left[\matrix{\pm1\cr2\cr0}\right],\quad \mathbb{b}=\left[\matrix{1\cr1\cr2}\right]. \tag{solution 1}$$ But then for $a\ne0$, assume without loss of generality that $a>0$ (so $a_1,b_1\ge1$), and note that, looking at the powers of two and three in its prime factorization, \eqalign{ b_1^3\cdot2^{b_2}\cdot3^{b_3} & = a^2+2 = a_1^6\cdot2^{2a_2}\cdot3^{2a_3}+2 \\ & = 2 \left( a_1^6\cdot2^{2a_2-1}\cdot3^{2a_3} + 1\right) } implies that either $a_2\ge1=b_2$ (case 2.1) or $a_2=b_2=0$ (case 2.0), and by similar reasoning, either $a_3>0=b_3$ (case 3.1), or else $a_3=0$ and $b_3\equiv2\pmod3$, which I will call case 3.0. This leaves us with four combinations to check, and hopefully either rule out or simplify: $$\matrix{ \\\text{case }0=2.0\wedge3.0:&\quad a_2=a_3=b_2=0 ,\quad b_3\equiv2\pmod3 \\\text{case }1=2.1\wedge3.0:&\quad a_2\ge1=b_2,\quad a_3=0,\quad b_3\equiv2\pmod3 \\\text{case }2=2.0\wedge3.1:&\quad a_3>0=a_2=b_2=b_3 \\\text{case }3=2.1\wedge3.1:&\quad a_2,a_3\ge1=b_2,\quad b_3=0. }$$ These four cases comprise all nontrivial (i.e. other than solution 0), (WLOG) postivie solutions, and we hope to deduce that only case 1 with solution 1 is viable. Case 0 simplifies to $$b_1^3 \cdot 3^{b_3} = a_1^6+2 \qquad\text{for}\qquad a_1,b_1\not\equiv0\pmod{2,3} \quad\text{and}\quad b_3\equiv2\pmod3.$$ Since $b_3\ge2$, we have $a_1^6\equiv-2\pmod9$, which has no solutions $a_1$ (sixth powers are always $0$ or $1$ modulo $9$), ruling this case out. Case 2 simplifies to $$b_1^3 = a_1^6 \cdot 3^{2a_3} + 2 \qquad\text{for}\qquad a_1,b_1\not\equiv0\pmod{2,3} \quad\text{and}\quad a_3 > 0,$$ so that $b_1^3\equiv2\pmod9$, which we can also rule out, since cubes are always $0$ or $\pm1$ modulo $9$. Case 3 simplifies to $$2b_1^3 = 2^{2a_2 } 3^{2a_3} a_1^6 + 2 = 2\left( 2^{2a_2-1} 3^{2a_3} a_1^6 + 1 \right) \qquad\text{for}\qquad a_1,b_1\not\equiv0\pmod{2,3} \quad\text{and}\quad a_2,a_3 > 1,$$ so that $b_1^3\equiv1\pmod9\implies b_1$ is even, a contradiction. Lastly, case 1 simplifies to $$3^{b_3} \cdot b_1^3 = 2^{2a_2-1} a_1^6 + 1 \qquad\text{for}\qquad a_1,b_1\not\equiv0\pmod{2,3}, \quad a_2 \ge 1 \quad\text{and}\quad b_3\equiv2\pmod3.$$ so that $b_3\ge2\implies$ $2^{2a_2-1} a_1^6\equiv-1\pmod9$ $\implies a_1^6 \equiv-2^{1-2a_2} \equiv 2^{-2(a_2+1)} \pmod 9$ since $2$ is a primitive root modulo $9$, i.e. $\text{ord}_9(2)=6=\phi(9)$, and $2^{\pm3}\equiv-1\pmod9$. But then $a_1\equiv2^{a_0}\pmod9$ for some $a_0$ with $-2(a_2+1)\equiv6a_0\equiv0\pmod6$, so that $a_2+1\equiv3~\implies~a_2\equiv2\pmod6$. In particular, we see that $a_2\ge2$.

Well anyway, that's a start. If we could show that $a_2,b_3\le2$, then we could reduce case 1 to $9y^3-8x^6=1$, and our goal would be to show that its only positive solution is $(1,1)$, and a start at that might be to note that $9y^3\equiv8x^6+1\equiv9\pmod{72}$.

-

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 $P_1-P_2$: $$\left(-\frac{10}{11}\right)^3 + \left(\frac{1}{11}\right)^3 + \left(\frac{12}{11}\right)^3 = \left(\frac{9}{11}\right)^3$$

-
Shouldn't it be $6a$ instead of $6a^2$? –  joriki Mar 15 '12 at 16:01
About that last solution: That's nice; it contains the cubes for the Hardy–Ramanujan number :-) –  joriki Mar 15 '12 at 16:04
Thanks! Fixed it. Luckily, the projective equation $3Z^3+6ZY^2=X^3$ was correctly written, so the rest is OK. –  Álvaro Lozano-Robledo Mar 15 '12 at 16:04