# cube+cube+cube=cube

• May 14th 2011, 02:12 PM
Zeno
cube+cube+cube=cube
Hi everyone. (This is my first post here)

Where can I find an ordered list of primitive equations of the form:

$\displaystyle {a}^{3}+ {b}^{3}+{c}^{3}={d}^{3}$ where $\displaystyle a, b, c$ are consecutive numbers.

Or alternatively, can anyone give me a formula to produce a list of such equations?

Do these equations have a name? The similar situation in squares are called "Pythagorean Triplets". But I'm not familiar with any name associated with this similar form in cubes. If they have a specific name I could probably just type that into Google.

Thank you for your precious time.
• May 14th 2011, 02:23 PM
HallsofIvy
If a, b, and c are consecutive integers, then a= n, b= n+1, c= n+ 2 for some integer n and so $\displaystyle a^3+ b^3+ c^3= n^3+ n^3+ 3n^2+ 3n+ 1+ n^3+ 6n^2+ 12n+ 8= 3(n^3+ 3n^2+ 5n+ 3)= d^3$ so, first of all, $\displaystyle d^3$ must be divisible by 3. But 3 is prime so d must be divisible by 3, d= 3m and so $\displaystyle d^3= 27m^3$: $\displaystyle 3(n^3+ 3n^2+ 6n+ 3)= 27m^3$ and then $\displaystyle n^3+ 3n^2+ 6n+ 3= 9m^3$
• May 14th 2011, 04:12 PM
Zeno
Thank you very much!

I think you meant $\displaystyle n^3+ 3n^2+ 5n+ 3= 9m^3$ though.

You actually typed $\displaystyle n^3+ 3n^2+ 6n+ 3= 9m^3$

But I see the big picture now.

I thought that would be an easy calculation.

But now how do I go about finding m & n's that are whole number solutions to that equation?

I'm seeking a quick way to generate a list of these consecutive equations.
When I type this into spreadsheet and chose an arbitrary n I get solutions for m that aren't whole numbers, or vice versa.

I would settle for a list on the Internet of these consecutive equations if someone knows where I can find such a list.

Surely these must have been calculated by someone at some point in time.
• May 14th 2011, 04:47 PM
Jester
I'd like to ask a question (maybe a silly one at that). Why do you think that these numbers exist?
• May 14th 2011, 05:02 PM
Zeno
Quote:

Originally Posted by Danny
I'd like to ask a question (maybe a silly one at that). Why do you think that these numbers exist?

Well, I'm not sure that they do. And if they don't exist, I would like to know that too. And potentially see the proof of how we can be certain that they don't.

I do know that at least one such equation exists

Namely, $\displaystyle {3}^{3}+{4}^{3}+{5}^{3}={6}^{3}=27+64+125=216$

Is that the only one? I don't know?
• May 14th 2011, 05:02 PM
topsquark
Quote:

Originally Posted by Zeno
Thank you very much!

I think you meant $\displaystyle n^3+ 3n^2+ 5n+ 3= 9m^3$ though.

You actually typed $\displaystyle n^3+ 3n^2+ 6n+ 3= 9m^3$

But I see the big picture now.

I thought that would be an easy calculation.

But now how do I go about finding m & n's that are whole number solutions to that equation?

I'm seeking a quick way to generate a list of these consecutive equations.
When I type this into spreadsheet and chose an arbitrary n I get solutions for m that aren't whole numbers, or vice versa.

I would settle for a list on the Internet of these consecutive equations if someone knows where I can find such a list.

Surely these must have been calculated by someone at some point in time.

Let's see if we can whittle this down a bit more.
$\displaystyle n^3 + 3n^2 + 5n + 3 = 9m^3$

$\displaystyle n^3 + 3n^2 + 5n + 3 \equiv 0 ~\text{(mod 9)}$

has solutions n = 3, 4, and 8 (mod 9).

Let's take n = 9p + 3, for all p.

Then the equation becomes
$\displaystyle 729p^3 + 972p^2 + 450p + 72 = 9m^3$

$\displaystyle 81p^3 + 108p^2 + 50p + 8 = m^3$

One immediate solution of this is p = 0, m = 2. (a = 3, b = 4, c = 5, d = 6). p obviously can't be negative. I checked up to p = 25 and could not find any more solutions. I can't think of a way to get closer than this.

I'll let you work out n = 4, 8 (mod 9).

-Dan

Edit: Unless I programmed Excel wrong, there are no solutions up to p = 1000 in any of the three categories except for the one mentioned above. I suspect that's the only one.
• May 14th 2011, 05:42 PM
Zeno
Quote:

Originally Posted by topsquark
Edit: Unless I programmed Excel wrong, there are no solutions up to p = 1000 in any of the three categories except for the one mentioned above. I suspect that's the only one.

Thank you Dan, you're probably right.

I didn't know, and that's why I asked.

Thank you very much for your time and effort.
• May 14th 2011, 10:54 PM
CaptainBlack
Quote:

Originally Posted by Zeno
Thank you Dan, you're probably right.

I didn't know, and that's why I asked.

Thank you very much for your time and effort.

Wolfram-Mathworld states (without proof or reference) that 3,4,5 are the only three consecutive numbers for which the sum of their cubes is a cube.

CB
• May 14th 2011, 11:42 PM
topsquark
Quote:

Originally Posted by CaptainBlack
Wolfram-Mathworld states (without proof or reference) that 3,4,5 are the only three consecutive numbers for which the sum of their cubes is a cube.

CB

I thought about using that but couldn't figure out how to word the input. How did you do that?

-Dan
• May 15th 2011, 01:10 AM
CaptainBlack
Quote:

Originally Posted by topsquark
I thought about using that but couldn't figure out how to word the input. How did you do that?

-Dan

Something like "three consecutive cubes sum cube"

CB
• May 1st 2014, 03:43 AM
individ
Re: cube+cube+cube=cube
For Diophantine equations: $\displaystyle X^3+Y^3+Z^3=R^3$

Symmetric solutions can be written as: $\displaystyle \left|\frac{X+Y}{R-Z}\right|=\frac{b^2}{a^2}$

Solutions have the form:

$\displaystyle X=b(a^6-b^6)p^3+3ba^6p^2s+3ba^6ps^2+b(a^6-b^6)s^3$

$\displaystyle Y=(a^7-ab^6)p^3+3(a^7-a^4b^3-ab^6)p^2s+3(a^7-2a^4b^3)ps^2+$$\displaystyle (a^7-3a^4b^3+2ab^6)s^3 \displaystyle Z=(2ba^6+3a^3b^4+b^7)p^3+3(2ba^6+a^3b^4)p^2s+3(2ba ^6-a^3b^4)ps^2+$$\displaystyle (2ba^6-3a^3b^4+b^7)s^3$

$\displaystyle R=(a^7+3a^4b^3+2ab^6)p^3+3(a^7+2a^4b^3)p^2s+3(a^7+ a^4b^3-ab^6)ps^2+$$\displaystyle (a^7-ab^6)s^3 Solutions have the form: \displaystyle X=3(b^6-7a^6)as^2-9a^4ps-ap^2 \displaystyle Y=(6ab^6+21a^7-27b^3a^4)s^2-3(2ab^3-3a^4)ps+ap^2 \displaystyle Z=(3b^7+33ba^6-18a^3b^4)s^2+3(4ba^3-b^4)ps+bp^2 \displaystyle R=(3b^7+6ba^6-9a^3b^4)s^2+3(2ba^3-b^4)ps+bp^2 Solutions have the form: \displaystyle X=(b^6-7a^6)ap^2+9a^4ps-3as^2 \displaystyle Y=(2ab^6+9b^3a^4+7a^7)p^2-3(2ab^3+3a^4)ps+3as^2 \displaystyle Z=(3b^4a^3+2ba^6+b^7)p^2-3(2ba^3+b^4)ps+3bs^2 \displaystyle R=(6b^4a^3+11ba^6+b^7)p^2-3(4ba^3+b^4)ps+3bs^2 Solutions have the form: \displaystyle X=(aj-bt)^2sp^6-3t^2j^2p^2s^5+3(aj+bt)jtps^6-(a^2j^2+abtj+b^2t^2)s^7 \displaystyle Y=2(aj-bt)^2sp^6-6(aj-bt)jtp^4s^3+3(a^2j^2-b^2t^2)p^3s^4+3t^2j^2p^2s^5-$$\displaystyle 3(aj+bt)tjps^6+(a^2j^2+abtj+b^2t^2)s^7$

$\displaystyle Z=(aj-bt)^2p^7-3(aj-bt)tjp^5s^2+3(aj-bt)btp^4s^3+3t^2j^2p^3s^4-$$\displaystyle 6bjt^2p^2s^5-(a^2j^2-2abtj-2b^2t^2)ps^6 \displaystyle R=(aj-bt)^2p^7-3(aj-bt)tjp^5s^2+3(aj-bt)ajp^4s^3+3t^2j^2p^3s^4-$$\displaystyle 6atj^2p^2s^5-(b^2t^2-2abtj-2a^2j^2)ps^6$

$\displaystyle a,j,t,b,p,s$ - what some integers.