MHB Succession 5 numbers risen up to "3"

[mente oscura]

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]

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
.

 

[mente oscura]

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 \<br /> <br /> 91^3=27^3+76^3+32^3+64^3

For \ n=2 \ then \<br /> <br /> 2899^3=2835^3+1164^3+96^3+64^3

For \ n=3 \ then \<br /> <br /> 49939^3=49875^3+7820^3+160^3+64^3

...

...

Regards.

 

[mente oscura]

Hello.:)

Another:

a=6n^3+1

b=6n^3-1

c=6n^2

d=1

e=1

Example:

For \ n=1 \ then \<br /> <br /> 7^3=5^3+6^3+1^3+1^3

For \ n=2 \ then \<br /> <br /> 49^3=47^3+24^3+1^3+1^3

For \ n=3 \ then \<br /> <br /> 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 \<br /> <br /> 43^3=-21^3-52^3-32^3+64^3

For \ n=2 \ then \<br /> <br /> 1603^3=1539^3+780^3-96^3+64^3

For \ n=3 \ then \<br /> <br /> 43939^3=43875^3+7180^3-160^3+64^3

...

...

Regards.

 

[mente oscura]

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]

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.

 

[mente oscura]

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]

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]

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]

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]

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]

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.