Is 42 a sum of three cubes?

Is 42 a sum of three cubes?

Postby Guest » Sat Apr 06, 2019 3:10 pm

"Dr. Andrew Booker, Reader of Pure Mathematics from the University's School of Mathematics, has now discovered the solution for number 33: (8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³."

Source:

https://phys.org/news/2019-04-bristol-mathematician-diophantine-puzzle.html
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sat Apr 06, 2019 5:04 pm

Guest wrote:"Dr. Andrew Booker, Reader of Pure Mathematics from the University's School of Mathematics, has now discovered the solution for number 33: (8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³."

Source:

https://phys.org/news/2019-04-bristol-mathematician-diophantine-puzzle.html


Since 33 = (8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³, we seek the solution of variables, a, b, and c such that

42 = (8,866,128,975,287,528 + a)³ + (–8,778,405,442,862,239 + b )³ + (–2,736,111,468,807,040 + c)³.

Use Newton Method and the link(modify the four variables to three variables):

https://www.math10.com/forum/viewtopic.php?f=63&t=7803&start=20
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sun Apr 07, 2019 8:53 am

Guest wrote:
Guest wrote:"Dr. Andrew Booker, Reader of Pure Mathematics from the University's School of Mathematics, has now discovered the solution for number 33: (8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³."

Source:

https://phys.org/news/2019-04-bristol-mathematician-diophantine-puzzle.html


Since 33 = (8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³, we seek the solution of integer variables, a, b, and c such that

42 = (8,866,128,975,287,528 + a)³ + (–8,778,405,442,862,239 + b )³ + (–2,736,111,468,807,040 + c)³.

Use Newton Method and the link(modify the four variables to three variables):

https://www.math10.com/forum/viewtopic.php?f=63&t=7803&start=20

...
You can also solve the Diophantine equation,

[tex]x^{3}[/tex]+ [tex]y^{3}[/tex]+ [tex]z^{3}[/tex] = 42.

Good luck!
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Mon Apr 08, 2019 3:49 pm

What does the graph of [tex]x^{3}+y^{3}+z^{3}=1[/tex] look like?

Source:

https://www.quora.com/What-does-the-graph-of-x-3+y-3+z-3-1-look-like
Guest
 


Re: Is 42 a sum of three cubes?

Postby Guest » Fri Apr 12, 2019 3:53 pm

Guest wrote:
Guest wrote:
Guest wrote:"Dr. Andrew Booker, Reader of Pure Mathematics from the University's School of Mathematics, has now discovered the solution for number 33: (8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³."

Source:

https://phys.org/news/2019-04-bristol-mathematician-diophantine-puzzle.html


Since 33 = (8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³, we seek the solution of integer variables, a, b, and c such that

42 = (8,866,128,975,287,528 + a)³ + (–8,778,405,442,862,239 + b )³ + (–2,736,111,468,807,040 + c)³.

Use Newton Method and the link(modify the four variables to three variables):

https://www.math10.com/forum/viewtopic.php?f=63&t=7803&start=20

...
You can also solve the Diophantine equation,

[tex]x^{3}[/tex]+ [tex]y^{3}[/tex]+ [tex]z^{3}[/tex] = 42.

Good luck!


If we decide to use the Newton Method to solve our system of three nonlinear equations, our initial vector could be,

[[tex]x_{0}, y_{0}, z_{0}[/tex]] = [8,866,128,975,287,528, –8,778,405,442,862,239, –2,736,111,468,807,040].

Therefore, we have,

[tex]n_{10} = y_{0} + z_{0}[/tex];

[tex]n_{20} = x_{0} + z_{0}[/tex];

[tex]n_{30} = x_{0} + y_{0}[/tex].

And [tex]N_{0 }[/tex]= [[tex]n_{10 }, n_{20}, n_{30}[/tex]].
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Fri Apr 12, 2019 3:59 pm

Update:

[tex]n_{10} = y_{0}^{3} + z_{0}^{3}[/tex];

[tex]n_{20} = x_{0}^{3}+ z_{0}^{3}[/tex];

[tex]n_{30} = x_{0}^{3}+ y_{0}^{3}[/tex].
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Fri Apr 12, 2019 11:42 pm

Guest wrote:Update:

[tex]n_{10} = y_{0}^{3} + z_{0}^{3}[/tex];

[tex]n_{20} = x_{0}^{3}+ z_{0}^{3}[/tex];

[tex]n_{30} = x_{0}^{3}+ y_{0}^{3}[/tex].


[tex]x_{k}= (42 - n_{1k})^{1/3}[/tex];

[tex]y_{k}= (42 - n_{2k})^{1/3}[/tex];

[tex]z_{k}= (42 - n_{3k})^{1/3}[/tex];

[tex]N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}[/tex]

where [tex]N_{k}[/tex] = [[tex]n_{1k}, n_{2k}, n_{3k}][/tex] (column vector)

and where [tex]J_{k}^{-1}[/tex] is inverse of the

Jacobian matrix (3 x 3), [tex]J_{k}[/tex]...
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sat Apr 13, 2019 1:08 am

Guest wrote:
Guest wrote:Update:

[tex]n_{10} = y_{0}^{3} + z_{0}^{3}[/tex];

[tex]n_{20} = x_{0}^{3}+ z_{0}^{3}[/tex];

[tex]n_{30} = x_{0}^{3}+ y_{0}^{3}[/tex].


[tex]x_{k}= (42 - n_{1k})^{1/3}[/tex];

[tex]y_{k}= (42 - n_{2k})^{1/3}[/tex];

[tex]z_{k}= (42 - n_{3k})^{1/3}[/tex];

[tex]N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}[/tex]

where [tex]N_{k}[/tex] = [[tex]n_{1k}, n_{2k}, n_{3k}][/tex] (column vector)

and where [tex]J_{k}^{-1}[/tex] is inverse of the

Jacobian matrix (3 x 3), [tex]J_{k}[/tex]...


[tex]J_{k} = \frac{\partial(n_{1k}, n_{2k}, n_{3k})}{\partial(x_{k}, y_{k}, z_{k})}

= \begin {bmatrix}
0 & 3(42-n_{2k})^{2/3}&3(42-n_{3k})^{2/3}\\
3(42-n_{1k})^{2/3} & 0 &3(42-n_{3k})^{2/3}\\
3(42-n_{1k})^{2/3} &3(42-n_{2k})^{2/3}& 0
\end{bmatrix}[/tex]...
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sat Apr 13, 2019 1:45 am

We recall our previous equations:

E. [tex]x^{3} + y^{3} + z^{3} = 42[/tex];

[tex]n_{1k} = y_{k}^{3} + z_{k}^{3}[/tex];

[tex]n_{2k} = x_{k}^{3}+ z_{k}^{3}[/tex];

[tex]n_{3k} = x_{k}^{3}+ y_{k}^{3}[/tex];

[tex]x_{k}= (42 - n_{1k})^{1/3}[/tex];

[tex]y_{k}= (42 - n_{2k})^{1/3}[/tex];

[tex]z_{k}= (42 - n_{3k})^{1/3}[/tex].

Therefore, we derive the following equations:

[tex]\frac{\partial n_{1k} }{\partial x_{k} }= 0[/tex];

[tex]\frac{\partial n_{1k} }{\partial y_{k} }= 3y_{k}^{2}= 3(42-n_{2k})^{2/3}[/tex];

[tex]\frac{\partial n_{1k} }{\partial z_{k} }= 3z_{k}^{2}= 3(42-n_{3k })^{2/3}[/tex]...
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sat Apr 13, 2019 2:22 pm

[tex]N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}[/tex]

where [tex]N_{k}[/tex] = [[tex]n_{1k}, n_{2k}, n_{3k}][/tex] (column vector)

and where [tex]J_{k}^{-1}[/tex] is inverse of the

Jacobian matrix (3 x 3), [tex]J_{k}[/tex].

[tex]J_{k}^{-1}

= \begin {bmatrix}
-(42-n_{1k})^{-2/3}/6 & (42-n_{1k})^{-2/3}/6&(42-n_{1k})^{-2/3}/6\\
(42-n_{2k})^{-2/3}/6&-(42-n_{2k})^{-2/3}/6&(42-n_{2k})^{-2/3}/6\\
(42-n_{3k})^{-2/3}/6&(42-n_{3k})^{-2/3}/6&(42-n_{3k})^{-2/3}/6 \end {bmatrix}[/tex].

Therefore,

[tex]N_{k+1}

= \begin {bmatrix}
n_{1k } + (42-n_{1k})^{-2/3}/6 * n_{1k } - (42-n_{1k})^{-2/3}/6 * n_{2k } - (42-n_{1k})^{-2/3}/6* n_{3k }\\
n_{2k } - (42-n_{2k})^{-2/3}/6 * n_{1k } + (42-n_{2k})^{-2/3}/6 * n_{2k } - (42-n_{2k})^{-2/3}/6* n_{3k }\\
n_{3k } - (42-n_{3k})^{-2/3}/6 * n_{1k } - (42-n_{3k})^{-2/3}/6 * n_{2k } + (42-n_{3k})^{-2/3}/6* n_{3k } \end {bmatrix}[/tex].
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sat Apr 13, 2019 2:49 pm

MSP12742361766bfahghhb400004f2bi2g70bab6hf2.gif
Please see above value for n_10:
MSP12742361766bfahghhb400004f2bi2g70bab6hf2.gif (2.86 KiB) Viewed 149 times
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sat Apr 13, 2019 2:52 pm

MSP310223defbda43b9g187000040fg7c30f9di6c0h.gif
Please see above value for n_20:
MSP310223defbda43b9g187000040fg7c30f9di6c0h.gif (2.86 KiB) Viewed 149 times
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sat Apr 13, 2019 2:55 pm

MSP243019hghe6224d4b25c00001625890ide000b25.gif
Please see above value for n_30:
MSP243019hghe6224d4b25c00001625890ide000b25.gif (2.85 KiB) Viewed 149 times
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sat Apr 13, 2019 3:10 pm

[tex]N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}[/tex]

where [tex]N_{k}[/tex] = [[tex]n_{1k}, n_{2k}, n_{3k}][/tex] (column vector)

and where [tex]J_{k}^{-1}[/tex] is inverse of the

Jacobian matrix (3 x 3), [tex]J_{k}[/tex].

[tex]J_{k}^{-1}

= \begin {bmatrix}
-(42-n_{1k})^{-2/3}/6 & (42-n_{1k})^{-2/3}/6&(42-n_{1k})^{-2/3}/6\\
(42-n_{2k})^{-2/3}/6&-(42-n_{2k})^{-2/3}/6&(42-n_{2k})^{-2/3}/6\\
(42-n_{3k})^{-2/3}/6&(42-n_{3k})^{-2/3}/6&(42-n_{3k})^{-2/3}/6 \end {bmatrix}[/tex].

Therefore,

[tex]N_{k+1}

= \begin {bmatrix}
n_{1k } + (42-n_{1k})^{-2/3}/6 * n_{1k } - (42-n_{1k})^{-2/3}/6 * n_{2k } - (42-n_{1k})^{-2/3}/6* n_{3k }\\
n_{2k } - (42-n_{2k})^{-2/3}/6 * n_{1k } + (42-n_{2k})^{-2/3}/6 * n_{2k } - (42-n_{2k})^{-2/3}/6* n_{3k }\\
n_{3k } - (42-n_{3k})^{-2/3}/6 * n_{1k } - (42-n_{3k})^{-2/3}/6 * n_{2k } + (42-n_{3k})^{-2/3}/6* n_{3k } \end {bmatrix}[/tex] = [[tex]n_{1k+1 }, n_{2k+1 },
n_{3k+1 }[/tex]] (column vector).

We recall the following equations:

[tex]x_{k+1}= (42 - n_{1k+1})^{1/3}[/tex];

[tex]y_{k+1}= (42 - n_{2k+1})^{1/3}[/tex];

[tex]z_{k+1}= (42 - n_{3k+1})^{1/3}[/tex].
Guest
 

Re: 42 a sum of three cubes?

Postby Guest » Sat Apr 13, 2019 5:29 pm

Guest wrote:[tex]N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}[/tex]

where [tex]N_{k}[/tex] = [[tex]n_{1k}, n_{2k}, n_{3k}][/tex] (column vector)

and where [tex]J_{k}^{-1}[/tex] is inverse of the

Jacobian matrix (3 x 3), [tex]J_{k}[/tex].

[tex]J_{k}^{-1}

= \begin {bmatrix}
-(42-n_{1k})^{-2/3}/6 & (42-n_{1k})^{-2/3}/6&(42-n_{1k})^{-2/3}/6\\
(42-n_{2k})^{-2/3}/6&-(42-n_{2k})^{-2/3}/6&(42-n_{2k})^{-2/3}/6\\
(42-n_{3k})^{-2/3}/6&(42-n_{3k})^{-2/3}/6&(42-n_{3k})^{-2/3}/6 \end {bmatrix}[/tex].

Therefore,

[tex]N_{k+1}

= \begin {bmatrix}
n_{1k } + (42-n_{1k})^{-2/3}/6 * n_{1k } - (42-n_{1k})^{-2/3}/6 * n_{2k } - (42-n_{1k})^{-2/3}/6* n_{3k }\\
n_{2k } - (42-n_{2k})^{-2/3}/6 * n_{1k } + (42-n_{2k})^{-2/3}/6 * n_{2k } - (42-n_{2k})^{-2/3}/6* n_{3k }\\
n_{3k } - (42-n_{3k})^{-2/3}/6 * n_{1k } - (42-n_{3k})^{-2/3}/6 * n_{2k } + (42-n_{3k})^{-2/3}/6* n_{3k } \end {bmatrix}[/tex] = [[tex]n_{1k+1 }, n_{2k+1 },
n_{3k+1 }[/tex]] (column vector).

We recall the following equations:

[tex]x_{k+1}= (42 - n_{1k+1})^{1/3}[/tex];

[tex]y_{k+1}= (42 - n_{2k+1})^{1/3}[/tex];

[tex]z_{k+1}= (42 - n_{3k+1})^{1/3}[/tex].


For what integers, [tex]k, x_{k }, y_{k }[/tex], and [tex]z_{k}[/tex],

does [tex]x_{k}^{3} + y_{k}^{3}+ z_{k}^{3} = 42[/tex]

when [tex]k \ge 0[/tex]?
Guest
 

Re: 42 a sum of three cubes?

Postby Guest » Sat Apr 13, 2019 8:03 pm

Guest wrote:
Guest wrote:[tex]N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}[/tex]

where [tex]N_{k}[/tex] = [[tex]n_{1k}, n_{2k}, n_{3k}][/tex] (column vector)

and where [tex]J_{k}^{-1}[/tex] is inverse of the

Jacobian matrix (3 x 3), [tex]J_{k}[/tex].

[tex]J_{k}^{-1}

= \begin {bmatrix}
-(42-n_{1k})^{-2/3}/6 & (42-n_{1k})^{-2/3}/6&(42-n_{1k})^{-2/3}/6\\
(42-n_{2k})^{-2/3}/6&-(42-n_{2k})^{-2/3}/6&(42-n_{2k})^{-2/3}/6\\
(42-n_{3k})^{-2/3}/6&(42-n_{3k})^{-2/3}/6&(42-n_{3k})^{-2/3}/6 \end {bmatrix}[/tex].

Therefore,

[tex]N_{k+1}

= \begin {bmatrix}
n_{1k } + (42-n_{1k})^{-2/3}/6 * n_{1k } - (42-n_{1k})^{-2/3}/6 * n_{2k } - (42-n_{1k})^{-2/3}/6* n_{3k }\\
n_{2k } - (42-n_{2k})^{-2/3}/6 * n_{1k } + (42-n_{2k})^{-2/3}/6 * n_{2k } - (42-n_{2k})^{-2/3}/6* n_{3k }\\
n_{3k } - (42-n_{3k})^{-2/3}/6 * n_{1k } - (42-n_{3k})^{-2/3}/6 * n_{2k } + (42-n_{3k})^{-2/3}/6* n_{3k } \end {bmatrix}[/tex] = [[tex]n_{1k+1 }, n_{2k+1 },
n_{3k+1 }[/tex]] (column vector).

We recall the following equations:

[tex]x_{k+1}= (42 - n_{1k+1})^{1/3}[/tex];

[tex]y_{k+1}= (42 - n_{2k+1})^{1/3}[/tex];

[tex]z_{k+1}= (42 - n_{3k+1})^{1/3}[/tex].


For what integers, [tex]k, x_{k }, y_{k }[/tex], and [tex]z_{k}[/tex],

does [tex]x_{k}^{3} + y_{k}^{3}+ z_{k}^{3} = 42[/tex]

when [tex]k \ge 0[/tex]?


Keywords:. David Hilbert's Tenth Problem

Hmm. We may need a supercomputer to solve our problem. And there could be no solutions (doubtful) or there could be a few solutions (or at least one solution) or there could be infinitely many solutions too. Hmm. ??? We need a sound theory to explain any result... Does our search have a happy ending (at least one solution) or does it become an endless search according to the MRDP Theorem?

https://en.m.wikipedia.org/wiki/Diophantine_set

Dave.
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sat Apr 13, 2019 11:41 pm

FYI: 'New sums of three cubes':

"Introduction. It is a long standing problem whether every rational integer
n ≡! 4, 5 (mod 9) may be written as a sum of three integral cubes. According to
the web page http://cr.yp.to/threecubes.html of Daniel Bernstein, the first attacks by computer were carried out as early as in 1955..."

Please see attachment for more details.
Attachments
elk_ants6c.pdf
(73.94 KiB) Downloaded 2 times
Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sun Apr 14, 2019 12:19 am

Guest
 

Re: Is 42 a sum of three cubes?

Postby Guest » Sun Apr 14, 2019 5:26 pm

For the given problem, how do we adapt the Newton Method to find integral solutions (assuming their existence) as fast as possible?
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