# Is 42 a sum of three cubes?

### Is 42 a sum of three cubes?

"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?

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?

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,

$$x^{3}$$+ $$y^{3}$$+ $$z^{3}$$ = 42.

Good luck!
Guest

### Re: Is 42 a sum of three cubes?

What does the graph of $$x^{3}+y^{3}+z^{3}=1$$ look like?

Source:

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

Guest

### Re: Is 42 a sum of three cubes?

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,

$$x^{3}$$+ $$y^{3}$$+ $$z^{3}$$ = 42.

Good luck!

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

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

Therefore, we have,

$$n_{10} = y_{0} + z_{0}$$;

$$n_{20} = x_{0} + z_{0}$$;

$$n_{30} = x_{0} + y_{0}$$.

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

### Re: Is 42 a sum of three cubes?

Update:

$$n_{10} = y_{0}^{3} + z_{0}^{3}$$;

$$n_{20} = x_{0}^{3}+ z_{0}^{3}$$;

$$n_{30} = x_{0}^{3}+ y_{0}^{3}$$.
Guest

### Re: Is 42 a sum of three cubes?

Guest wrote:Update:

$$n_{10} = y_{0}^{3} + z_{0}^{3}$$;

$$n_{20} = x_{0}^{3}+ z_{0}^{3}$$;

$$n_{30} = x_{0}^{3}+ y_{0}^{3}$$.

$$x_{k}= (42 - n_{1k})^{1/3}$$;

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

$$z_{k}= (42 - n_{3k})^{1/3}$$;

$$N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}$$

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

and where $$J_{k}^{-1}$$ is inverse of the

Jacobian matrix (3 x 3), $$J_{k}$$...
Guest

### Re: Is 42 a sum of three cubes?

Guest wrote:
Guest wrote:Update:

$$n_{10} = y_{0}^{3} + z_{0}^{3}$$;

$$n_{20} = x_{0}^{3}+ z_{0}^{3}$$;

$$n_{30} = x_{0}^{3}+ y_{0}^{3}$$.

$$x_{k}= (42 - n_{1k})^{1/3}$$;

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

$$z_{k}= (42 - n_{3k})^{1/3}$$;

$$N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}$$

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

and where $$J_{k}^{-1}$$ is inverse of the

Jacobian matrix (3 x 3), $$J_{k}$$...

$$J_{k} = \frac{\partial(n_{1k}, n_{2k}, n_{3k})}{\partial(x_{k}, y_{k}, z_{k})} = \begin {bmatrix} 0 & 3(42-n_{2k})^{2/3}&3(42-n_{3k})^{2/3}\\ 3(42-n_{1k})^{2/3} & 0 &3(42-n_{3k})^{2/3}\\ 3(42-n_{1k})^{2/3} &3(42-n_{2k})^{2/3}& 0 \end{bmatrix}$$...
Guest

### Re: Is 42 a sum of three cubes?

We recall our previous equations:

E. $$x^{3} + y^{3} + z^{3} = 42$$;

$$n_{1k} = y_{k}^{3} + z_{k}^{3}$$;

$$n_{2k} = x_{k}^{3}+ z_{k}^{3}$$;

$$n_{3k} = x_{k}^{3}+ y_{k}^{3}$$;

$$x_{k}= (42 - n_{1k})^{1/3}$$;

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

$$z_{k}= (42 - n_{3k})^{1/3}$$.

Therefore, we derive the following equations:

$$\frac{\partial n_{1k} }{\partial x_{k} }= 0$$;

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

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

### Re: Is 42 a sum of three cubes?

$$N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}$$

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

and where $$J_{k}^{-1}$$ is inverse of the

Jacobian matrix (3 x 3), $$J_{k}$$.

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

Therefore,

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

### Re: Is 42 a sum of three cubes?

Please see above value for n_10:
MSP12742361766bfahghhb400004f2bi2g70bab6hf2.gif (2.86 KiB) Viewed 2039 times
Guest

### Re: Is 42 a sum of three cubes?

Please see above value for n_20:
MSP310223defbda43b9g187000040fg7c30f9di6c0h.gif (2.86 KiB) Viewed 2039 times
Guest

### Re: Is 42 a sum of three cubes?

Please see above value for n_30:
MSP243019hghe6224d4b25c00001625890ide000b25.gif (2.85 KiB) Viewed 2039 times
Guest

### Re: Is 42 a sum of three cubes?

$$N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}$$

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

and where $$J_{k}^{-1}$$ is inverse of the

Jacobian matrix (3 x 3), $$J_{k}$$.

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

Therefore,

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

We recall the following equations:

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

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

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

### Re: 42 a sum of three cubes?

Guest wrote:$$N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}$$

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

and where $$J_{k}^{-1}$$ is inverse of the

Jacobian matrix (3 x 3), $$J_{k}$$.

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

Therefore,

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

We recall the following equations:

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

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

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

For what integers, $$k, x_{k }, y_{k }$$, and $$z_{k}$$,

does $$x_{k}^{3} + y_{k}^{3}+ z_{k}^{3} = 42$$

when $$k \ge 0$$?
Guest

### Re: 42 a sum of three cubes?

Guest wrote:
Guest wrote:$$N_{k+1} = N_{k} +J_{k}^{-1} * N_{k}$$

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

and where $$J_{k}^{-1}$$ is inverse of the

Jacobian matrix (3 x 3), $$J_{k}$$.

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

Therefore,

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

We recall the following equations:

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

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

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

For what integers, $$k, x_{k }, y_{k }$$, and $$z_{k}$$,

does $$x_{k}^{3} + y_{k}^{3}+ z_{k}^{3} = 42$$

when $$k \ge 0$$?

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?

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
Guest

Guest

### Re: Is 42 a sum of three cubes?

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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