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Number Theory Succession 5 numbers risen up to "3"

mente oscura

Well-known member
Nov 29, 2013
172
Hello.

I contribute a small study:

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

Family 1ª)

[tex]a=3n^6+3n^3+1[/tex]

[tex]b=3n^6+3n^3[/tex]

[tex]c=3n^4+2n[/tex]

[tex]d=n[/tex]

[tex]e=1^3[/tex]

Example:

[tex]For \ n=1 \ then: 7^3=6^3+5^3+1^3+1^3[/tex]

[tex]For \ n=2 \ then: 217^3=216^3+52^3+2^3+1^3[/tex]

[tex]For \ n=3 \ then: 2269^3=2268^3+249^3+3^3+1^3[/tex]

[tex]For \ n=4 \ then: 12481^3=12480^3+776^3+4^3+1^3[/tex]

...

Regards.



[tex][/tex]
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
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
Nov 29, 2013
172
Hello.:)

Another:

[tex]a=192n^6-576n^5+720n^4-288n^3-108n^2+108n+43[/tex]

[tex]b=192n^6-576n^5+720n^4-288n^3-108n^2+108n-21[/tex]

[tex]c=192n^4-384n^3+288n^2+32n-52[/tex]

[tex]d=64n-32[/tex]

[tex]e=64[/tex]

Result:

[tex]For \ n=1 \ then \

91^3=27^3+76^3+32^3+64^3[/tex]

[tex]For \ n=2 \ then \

2899^3=2835^3+1164^3+96^3+64^3[/tex]

[tex]For \ n=3 \ then \

49939^3=49875^3+7820^3+160^3+64^3[/tex]

...

...

Regards.
 

mente oscura

Well-known member
Nov 29, 2013
172
Hello.:)

Another:

[tex]a=6n^3+1[/tex]

[tex]b=6n^3-1[/tex]

[tex]c=6n^2[/tex]

[tex]d=1[/tex]

[tex]e=1[/tex]

Example:

[tex]For \ n=1 \ then \

7^3=5^3+6^3+1^3+1^3[/tex]

[tex]For \ n=2 \ then \

49^3=47^3+24^3+1^3+1^3[/tex]

[tex]For \ n=3 \ then \

163^3=161^3+54^3+1^3+1^3[/tex]

...

...



And, another:

[tex]a=192n^6-576n^5+720n^4-672n^3+468n^2+180n+91[/tex]

[tex]b=192n^6-576n^5+720n^4-672n^3+468n^2-180n+27[/tex]

[tex]c=192n^4-384n^3+288n^2-224n+76[/tex]

[tex]d=-64n+32[/tex]

[tex]e=64[/tex]

Result:

[tex]For \ n=1 \ then \

43^3=-21^3-52^3-32^3+64^3[/tex]

[tex]For \ n=2 \ then \

1603^3=1539^3+780^3-96^3+64^3[/tex]

[tex]For \ n=3 \ then \

43939^3=43875^3+7180^3-160^3+64^3[/tex]

...

...

Regards.
 

mente oscura

Well-known member
Nov 29, 2013
172
Hello.

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

It is referred to 4 elements.



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

Succession family 1ª)

[tex]a=3n^3+3pn^2+2p^2n+p^3[/tex]

[tex]b=3n^3+3pn^2+2p^2n[/tex]

[tex]c=3pn^2+2p^2n+p^3[/tex]

[tex]d=p^2n[/tex]

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

Regards.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
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
Nov 29, 2013
172
Hello.

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

Succession family 2ª)

[tex]a=9n^4[/tex]

[tex]b=9n^4-3p^3n[/tex]

[tex]c=9pn^3-p^4[/tex]

[tex]d=p^4[/tex]

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

Regards.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
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
Nov 29, 2013
172
Hello.

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

Succession family 3ª)

[tex]a=9n^3-6pn^2+3p^2n[/tex]

[tex]b=9n^3-6pn^2+3p^2n-p^3[/tex]

[tex]c=6pn^2-3p^2n+p^3[/tex]

[tex]d=3pn^2[/tex]

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

Regards.
 

mente oscura

Well-known member
Nov 29, 2013
172
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:

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

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

[tex](9, 8, 6, 1)[/tex], [tex](172, 138, 135,1)[/tex], [tex](1010, 812, 791, 1)[/tex], [tex](505, 426, 372, 1)[/tex]

[tex](577, 486, 426, 1)[/tex], [tex](144, 138, 71, 1)[/tex], [tex](1210, 1207, 236, 1)[/tex], [tex](729, 720, 242, 1)[/tex]

[tex](2304, 2292, 575, 1)[/tex], [tex](5625, 5610, 1124, 1)[/tex], [tex](11664, 11646, 1943, 1)[/tex], [tex](21609, 21588, 3086, 1)[/tex]

[tex](904, 823, 566, 1)[/tex], [tex](8703, 8675, 1851, 1)[/tex], [tex](6756, 6702, 1943, 1)[/tex], [tex](3097, 2820, 1938, 1)[/tex] ...

Regards.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
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
Nov 29, 2013
172
Hello.

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

Succession family 4ª)

[tex]a=9n^3+6pn^2+3p^2n+p^3[/tex]

[tex]b=9n^3+6pn^2+3p^2n[/tex]

[tex]c=6pn^2+3p^2n+p^3[/tex]

[tex]d=3pn^2[/tex]

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

Regards.