Find all solutions in integers of (a, b)

[anemone]

Find all solutions in integers of the equation $a^3+(a+1)^3+(a+2)^3+\cdots+(a+7)^3=b^3$

 

[mente oscura]

Find all solutions in integers of the equation $a^3+(a+1)^3+(a+2)^3+\cdots+(a+7)^3=b^3$

Hello.

A little bit. to brute force.

Spoiler

My experience with cubic powers:

$$3^3+4^3+5^3=6^3$$

It no more, correlative of 4 elements.

You need 5 correlative elements, which are cancelled:

$$(-2)^3+(-1)^3+0^3+1^3+2^3=0$$

Solution:

1º) $$(-6)^3=(-5)^3+(-4)^3+ \cdots+1^3+2^3$$

$$a=-5 \ and \ b=-6$$

2º) $$6^3=5^3+4^3+ \dots +(-1)^3+(-2)^3$$

$$a=-2 \ and \ b=6$$

Regards.

 

[anemone]

Hello.

A little bit. to brute force.

Spoiler

My experience with cubic powers:

$$3^3+4^3+5^3=6^3$$

It no more, correlative of 4 elements.

You need 5 correlative elements, which are cancelled:

$$(-2)^3+(-1)^3+0^3+1^3+2^3=0$$

Solution:

1º) $$(-6)^3=(-5)^3+(-4)^3+ \cdots+1^3+2^3$$

$$a=-5 \ and \ b=-6$$

2º) $$6^3=5^3+4^3+ \dots +(-1)^3+(-2)^3$$

$$a=-2 \ and \ b=6$$

Regards.

Hi mente oscura, the two set solutions are correct, but there are a total of 4 answers, could you please also identify the other two?

 

[magneto1]

Hi mente oscura, the two set solutions are correct, but there are a total of 4 answers, could you please also identify the other two?

Spoiler

$(-4,-4)$: $-64 -27 -8 -1 +0 +1 +8 + 27 = -64 = (-4)^3$

$(-3,4)$: $-27-8-1+0+1+8+27+64=4^3$.

Edit: Oops, made a typo: $8 = 2^3 \neq 4$.

 

[mathbalarka]

Spoiler

$$a^3 + \cdots + (a+7)^3 = 8a^3 + 84a^2 + 420a + 784$$

So the cubic curve of interest is $8a^3 + 84a^2 + 420a + 784 = b^3$ over $\Bbb Q$. This obviously have a rational point, take for example $(a, b) = (-4, -4)$. Thus it's isomorphic to an elliptic curve.

In particular, use the parametrization

$$(x, y) = \left (\frac{1008b}{2a+7}, \frac{254016}{2a + 7} \right )$$

To reduce the above into the well-known Mordell form

$$\mathbf {E} : y^2 = x^3 -1024192512$$

I do hope this has rank $1$, though. (I believe BSD)

First, the trivial solutions over $\Bbb Q$ include $(-7/2, 0), (-4, -4), (-3, 4)$, the first being simply a real zero of the cubic. Now about computing the torsion, one can try a relatively easy hands-on derivation by either Nagell-Lutz or Mazur (I am not sure which one is easier here). Precisely, it's $\Bbb Z/2\Bbb Z$ so the group of rational points over the elliptic curve is really $\Bbb Z \times \Bbb Z/ 2\Bbb Z$.

The extra solutions may come from the torsion points. It seems that a generator for $\mathbf E$ is $(2016, 84672)$ and from this comes the corresponding integral point $(-2, 6)$.

I am missing something here.

EDIT : Darn, I forgot the negative $(2016, -84672)$. Of course, plugging this in gives $a = -5$ and $b = 6$, and thus all the integer points (and as well rational points, which are the multiplies of the generator precisely).

 

[mente oscura]

Hi mente oscura, the two set solutions are correct, but there are a total of 4 answers, could you please also identify the other two?

Hello.

Spoiler

You're right, I am missing identities:

$$(-3)^3+(-2)^3+(-1)^3+0+1^3+2^3+3^3=0$$

Therefore:

3º) $$a=4 \ and \ b=4$$

4º) $$a=-4 \ and \ b=-4$$

Regards.

 

[anemone]

Thank you all for participating!:)

Solution I saw somewhere online:

Spoiler

Let $f(a)=a^3+(a+1)^3+(a+2)^3+(a+3)^3+(a+4)^3+(a+5)^3+(a+6)^3+(a+7)^3=8a^3+84a^2+420a+784$Case I:
If $a>0$:

Note that $(2a+7)^3=8a^3+84a^2+294a+343$ and $(2a+10)^3=8a^3+120a^2+600a+1000$

So if $a>0$, we have

$(2a+7)^3<f(a)<(2a+10)^3$, i.e. $2a+7<b<2a+10$ which implies $b=2a+8$ or $b=2a+9$.

And neither of the equations $f(a)-(2a+8)^3=-12a^2+36a+272=0$ and $ f(a)-(2a+9)^3=-24a^2-66a+55=0$ have any integer roots, so we can conclude that there are no solutions with $x>0$.

Case II:
Note that $f$ satisfies $f(-a-7)=-f(a)$, so $(a,\,b)$ is a solution iff $(-a-7,\,-b)$. Therefore, there are no solutions with $a<-7$.

So for $(a,\,b)$ to be a solution, we must have $-6<a<-1$.

Checking each case we have

$f(-1)=440$ and this isn't a cube.

$f(-2)=216=6^3$

$f(-3)=64=4^3$

Therefore, $(-2,\,6)$ and $(-3,\,4)$ are the only solutions with $-3<a<-1$ and $(-4,\,-4)$ and $(-5,\,-6)$ are the only solutions with $-6<a<-4$.

 

[mathbalarka]

Darn it. I never think straight. :D