# Number TheorySuccession 5 numbers risen up to "3"

#### mente oscura

##### Well-known member
Hello.

I contribute a small study:

$$For : a, b, c, d, e, n \in{N} / a^3=b^3+c^3+d^3+e^3$$

Family 1ª)

$$a=3n^6+3n^3+1$$

$$b=3n^6+3n^3$$

$$c=3n^4+2n$$

$$d=n$$

$$e=1^3$$

Example:

$$For \ n=1 \ then: 7^3=6^3+5^3+1^3+1^3$$

$$For \ n=2 \ then: 217^3=216^3+52^3+2^3+1^3$$

$$For \ n=3 \ then: 2269^3=2268^3+249^3+3^3+1^3$$

$$For \ n=4 \ then: 12481^3=12480^3+776^3+4^3+1^3$$

...

Regards.



#### mathbalarka

##### Well-known member
MHB Math Helper
Nice! Also, if I may add, the Boutin's identity :

$$(-a+b+c)^3 + (a-b+c)^3 + (a+b-c)^3 + d^3 = (a+b+c)^3$$

where $24abc = d^3$. Another found by Piezas :

$$\left (11x^2+xy+14y^2 \right )^3 + \left (12x^2-3xy+13y^2 \right )^3 + \left (13x^2+3xy+12y^2 \right )^3 + \left (14x^2-xy+11y^2 \right )^3 = 20^3 \left (x^2+y^2 \right )^3$$

Balarka
.

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#### mente oscura

##### Well-known member
Hello.

Another:

$$a=192n^6-576n^5+720n^4-288n^3-108n^2+108n+43$$

$$b=192n^6-576n^5+720n^4-288n^3-108n^2+108n-21$$

$$c=192n^4-384n^3+288n^2+32n-52$$

$$d=64n-32$$

$$e=64$$

Result:

$$For \ n=1 \ then \ 91^3=27^3+76^3+32^3+64^3$$

$$For \ n=2 \ then \ 2899^3=2835^3+1164^3+96^3+64^3$$

$$For \ n=3 \ then \ 49939^3=49875^3+7820^3+160^3+64^3$$

...

...

Regards.

#### mente oscura

##### Well-known member
Hello.

Another:

$$a=6n^3+1$$

$$b=6n^3-1$$

$$c=6n^2$$

$$d=1$$

$$e=1$$

Example:

$$For \ n=1 \ then \ 7^3=5^3+6^3+1^3+1^3$$

$$For \ n=2 \ then \ 49^3=47^3+24^3+1^3+1^3$$

$$For \ n=3 \ then \ 163^3=161^3+54^3+1^3+1^3$$

...

...

And, another:

$$a=192n^6-576n^5+720n^4-672n^3+468n^2+180n+91$$

$$b=192n^6-576n^5+720n^4-672n^3+468n^2-180n+27$$

$$c=192n^4-384n^3+288n^2-224n+76$$

$$d=-64n+32$$

$$e=64$$

Result:

$$For \ n=1 \ then \ 43^3=-21^3-52^3-32^3+64^3$$

$$For \ n=2 \ then \ 1603^3=1539^3+780^3-96^3+64^3$$

$$For \ n=3 \ then \ 43939^3=43875^3+7180^3-160^3+64^3$$

...

...

Regards.

#### mente oscura

##### Well-known member
Hello.

Another small study, that I already brought into other forums. I share it with you.

It is referred to 4 elements.

$$Let \ a, \ b, \ c, \ d, \ n, \ p \in{Z}/ a^3=b^3+c^3+d^3$$

Succession family 1ª)

$$a=3n^3+3pn^2+2p^2n+p^3$$

$$b=3n^3+3pn^2+2p^2n$$

$$c=3pn^2+2p^2n+p^3$$

$$d=p^2n$$

Being "n" number of the sequence, and "p" the number that you want.

Regards.

#### mathbalarka

##### Well-known member
MHB Math Helper
Ramanujan:

$$\left (3x^2+5xy-5y^2 \right)^3 + \left(4x^2-4xy+6y^2\right)^3 + \left(5x^2-5xy-3y^2\right)^3 = \left(6x^2-4xy+4y^2\right)^3$$

A well-known is Euler's:

$$(p+q)^3 + (p-q)^3 + (s-r)^3 = (r+s)^3$$

with $(p, q, r, s)$ being

$$\left ( 3(bc-ad)\left(c^2+3d^2\right), \left(a^2+3b^2\right)^2 - (ac+3bd)\left(c^2+3d^2\right), 3(bc-ad)\left(a^2+3b^2\right),-\left(c^2+3d^2\right)^2 + (ac+3bd)\left(a^2+3b^2\right) \right )$$

Another interesting phenomena is sum-of-consecutive-cubes being a cube, and a couple of years back I proved that $(0, 4)$ are the only solutions to $(a-1)^3+a^3+(a+1)^3=b^3$ using hard elliptic curve analysis.

Also, it is conjectured that the only $n$-tuples with the property that the $k$-th power of first $n-1$ is equal to $k$-th power of the other for some integer $k$ are $(3, 4, 5)$ with $(3, 4, 5, 6)$ for $k=3$, that is, $3^3+4^3+5^3=6^3$ and no other. All cases for $k \leq 5$ is proved, although I don't have a reference for that. See Erdos-Moser topic I have created in this forum.

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#### mente oscura

##### Well-known member
Hello.

$$Let \ a, \ b, \ c, \ d, \ n, \ p \in{Z}/ a^3=b^3+c^3+d^3$$

Succession family 2ª)

$$a=9n^4$$

$$b=9n^4-3p^3n$$

$$c=9pn^3-p^4$$

$$d=p^4$$

Being "n" number of the sequence, and "p" the number that you want.

Regards.

#### mathbalarka

##### Well-known member
MHB Math Helper
I just got hold of my copy of Perelman today, and here's the one :

$$(x, y, z, u) = (a + k\alpha, b + k\beta, c + k\gamma, -d - k\delta)$$

$$k = - \frac{a^2\alpha + b^2\beta + c^2\gamma + d^2\delta}{a\alpha^2+b\beta^2+c\gamma^2+d\delta^2}$$

Where $(a, b, c, -d)$ and $(\alpha, \beta, \gamma, -\delta)$ both satisfy the diophantine equation and are used as an initializing 'seed' here.

This is quite superior, indeed, as it not just generates tuples over $\mathbb{Z}^4$, but also over $\mathbb{Q}^4$. Looking at it, I believe it is plausible to conjecture

This solution over $\mathbb{Z}^4$ is complete.

Although much more general $\mathbb{Q}^4$ seems likely to be false, although a proof is lacking. I think I should mail this one to Piezas, don't know whether he have already seen this before....

#### mente oscura

##### Well-known member
Hello.

$$Let \ a, \ b, \ c, \ d, \ n, \ p \in{Z}/ a^3=b^3+c^3+d^3$$

Succession family 3ª)

$$a=9n^3-6pn^2+3p^2n$$

$$b=9n^3-6pn^2+3p^2n-p^3$$

$$c=6pn^2-3p^2n+p^3$$

$$d=3pn^2$$

Being "n" number of the sequence, and "p" the number that you want.

Regards.

#### mente oscura

##### Well-known member
I just got hold of my copy of Perelman today, and here's the one :

$$(x, y, z, u) = (a + k\alpha, b + k\beta, c + k\gamma, -d - k\delta)$$

$$k = - \frac{a^2\alpha + b^2\beta + c^2\gamma + d^2\delta}{a\alpha^2+b\beta^2+c\gamma^2+d\delta^2}$$

Where $(a, b, c, -d)$ and $(\alpha, \beta, \gamma, -\delta)$ both satisfy the diophantine equation and are used as an initializing 'seed' here.

This is quite superior, indeed, as it not just generates tuples over $\mathbb{Z}^4$, but also over $\mathbb{Q}^4$. Looking at it, I believe it is plausible to conjecture

This solution over $\mathbb{Z}^4$ is complete.

Although much more general $\mathbb{Q}^4$ seems likely to be false, although a proof is lacking. I think I should mail this one to Piezas, don't know whether he have already seen this before....
Hello.

I have much difficulty to assimilate this formula to a case study.

For example:

$$Let (a,b,c,d), \ / \ a, \ b, \ c, \ d \in{Z}/ \ a^3=b^3+c^3+d^3$$

Some results, of the following, arise with any of my formulas:

$$(9, 8, 6, 1)$$, $$(172, 138, 135,1)$$, $$(1010, 812, 791, 1)$$, $$(505, 426, 372, 1)$$

$$(577, 486, 426, 1)$$, $$(144, 138, 71, 1)$$, $$(1210, 1207, 236, 1)$$, $$(729, 720, 242, 1)$$

$$(2304, 2292, 575, 1)$$, $$(5625, 5610, 1124, 1)$$, $$(11664, 11646, 1943, 1)$$, $$(21609, 21588, 3086, 1)$$

$$(904, 823, 566, 1)$$, $$(8703, 8675, 1851, 1)$$, $$(6756, 6702, 1943, 1)$$, $$(3097, 2820, 1938, 1)$$ ...

Regards.

#### mathbalarka

##### Well-known member
MHB Math Helper
mente oscura said:
I have much difficulty to assimilate this formula to a case study.
I don't see why. Just pick up arbitrary seeds $a, \alpha, b, \beta, c, \gamma, d, \delta$ satisfying the intended equation and then do some calculation to arrive at another multiset. This is simply a binary operation acting over the multisets.

And if you come to analyze this one, you'll see how ingenious it is. And fortunately you came up with the issue, I completely forgot to mail it to Piezas! Let's see what he thinks.

#### mente oscura

##### Well-known member
Hello.

$$Let \ a, \ b, \ c, \ d, \ n, \ p \in{Z}/ a^3=b^3+c^3+d^3$$

Succession family 4ª)

$$a=9n^3+6pn^2+3p^2n+p^3$$

$$b=9n^3+6pn^2+3p^2n$$

$$c=6pn^2+3p^2n+p^3$$

$$d=3pn^2$$

Being "n" number of the sequence, and "p" the number that you want.

Regards.