Integer solutions to the equation $a^3+b^3+c^3=30$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find three integers $n_1, n_2, n_3$ such that $n_1^3 + n_2^3 + n_3^3 = 42$Are there any open mathematical puzzles?Solve $x^3 +y^3 + z^3 =57$Parametric solutions to $x^3+y^3+z^3 = N$?Integer solutions to the equation $a^3+b^3+c^3=colorred6$Conjecture: $a^3 + b^3 + c^3 = p^3 Rightarrow x^3 + y^3 + z^3 = big(fraca + b + c2big)^3$Solutions of the Diophantine equation $x^2(x^2+10)=3y^2(y^2+10)$Finding the integer solutions of $246x + 217y = 3$Number of Solutions to Diophantine Equationinteger solutions to bivariate polynomial of second degreeDescribe the integral solutions to this cubic equation.Solutions to the diophantine equation $6x^2 - 6x - y^2 + y=0$?Positive integer solutions to $fracxy+z+fracyx+z+fraczx+y=4$Determining number of integer solutions of Diophantine equationHow many positive integer solutions are there to the equation $𝑥^2 + 2𝑦^2 = 4𝑧^2$?Prove that there are no integer solutions to this equation
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Integer solutions to the equation $a^3+b^3+c^3=30$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find three integers $n_1, n_2, n_3$ such that $n_1^3 + n_2^3 + n_3^3 = 42$Are there any open mathematical puzzles?Solve $x^3 +y^3 + z^3 =57$Parametric solutions to $x^3+y^3+z^3 = N$?Integer solutions to the equation $a^3+b^3+c^3=colorred6$Conjecture: $a^3 + b^3 + c^3 = p^3 Rightarrow x^3 + y^3 + z^3 = big(fraca + b + c2big)^3$Solutions of the Diophantine equation $x^2(x^2+10)=3y^2(y^2+10)$Finding the integer solutions of $246x + 217y = 3$Number of Solutions to Diophantine Equationinteger solutions to bivariate polynomial of second degreeDescribe the integral solutions to this cubic equation.Solutions to the diophantine equation $6x^2 - 6x - y^2 + y=0$?Positive integer solutions to $fracxy+z+fracyx+z+fraczx+y=4$Determining number of integer solutions of Diophantine equationHow many positive integer solutions are there to the equation $𝑥^2 + 2𝑦^2 = 4𝑧^2$?Prove that there are no integer solutions to this equation
$begingroup$
The following problem was posed to me but I could not do much about it:
Determine if there are any integer solutions to the equation
$a^3+b^3+c^3=30$
I made a computer search that shows that there are no integers $a,b,c$ such that $a^3+b^3+c^3=30$ and $|a|,|b|,|c|<51$
Thank you a lot.
number-theory diophantine-equations computational-mathematics
edited Dec 8 '15 at 8:32
Tito Piezas III
27.9k369179
asked Aug 5 '15 at 23:33
AmrAmr
14.4k43296
$endgroup$
|
show 1 more comment
$begingroup$
The following problem was posed to me but I could not do much about it:
Determine if there are any integer solutions to the equation
$a^3+b^3+c^3=30$
I made a computer search that shows that there are no integers $a,b,c$ such that $a^3+b^3+c^3=30$ and $|a|,|b|,|c|<51$
Thank you a lot.
number-theory diophantine-equations computational-mathematics
edited Dec 8 '15 at 8:32
Tito Piezas III
27.9k369179
asked Aug 5 '15 at 23:33
AmrAmr
14.4k43296
$endgroup$
5
$begingroup$
ams.org/journals/mcom/2007-76-259/S0025-5718-07-01947-3/…
$endgroup$
– Will Jagy
Aug 5 '15 at 23:42
7
$begingroup$
From Wolfram: "... all numbers $N<1000$ and not of the form $9n pm 4$ are known to be expressible as the sum $N=A^3+B^3+C^3$ of three (positive or negative) cubes with the exception of $N=33, 42, 74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921$, and $975$ (Miller and Woollett 1955; Gardiner et al. 1964; Guy 1994, p. 151; Mishima; Elsenhaus and Jahnel 2007). Examples include $30 = (-283059965)^3+(-2218888517)^3+2220422932^3$ ..."
$endgroup$
– 727
Aug 5 '15 at 23:42
$begingroup$
@WillJagy - a great find!
$endgroup$
– hypergeometric
Aug 7 '15 at 16:16
$begingroup$
Table 2 in the paper cited by Jagy above needs an update.
$endgroup$
– Tito Piezas III
Dec 8 '15 at 6:25
$begingroup$
As of 2016, three solutions are now known for $N=30$, $$2220422932^3 - 283059965^3 - 2218888517^3 =30\ 3982933876681^3 - 636600549515^3 - 3977505554546^3 = 30\ -662037799708799^3 + 190809268841284^3 + 656711689254565^3 = 30$$ with the last found by S. Huisman.
$endgroup$
– Tito Piezas III
Dec 28 '17 at 7:00
|
show 1 more comment
$begingroup$
The following problem was posed to me but I could not do much about it:
Determine if there are any integer solutions to the equation
$a^3+b^3+c^3=30$
I made a computer search that shows that there are no integers $a,b,c$ such that $a^3+b^3+c^3=30$ and $|a|,|b|,|c|<51$
Thank you a lot.
number-theory diophantine-equations computational-mathematics
edited Dec 8 '15 at 8:32
Tito Piezas III
27.9k369179
asked Aug 5 '15 at 23:33
AmrAmr
14.4k43296
$endgroup$
The following problem was posed to me but I could not do much about it:
Determine if there are any integer solutions to the equation
$a^3+b^3+c^3=30$
I made a computer search that shows that there are no integers $a,b,c$ such that $a^3+b^3+c^3=30$ and $|a|,|b|,|c|<51$
Thank you a lot.
number-theory diophantine-equations computational-mathematics
number-theory diophantine-equations computational-mathematics
edited Dec 8 '15 at 8:32
Tito Piezas III
27.9k369179
asked Aug 5 '15 at 23:33
AmrAmr
14.4k43296
edited Dec 8 '15 at 8:32
Tito Piezas III
27.9k369179
asked Aug 5 '15 at 23:33
AmrAmr
14.4k43296
edited Dec 8 '15 at 8:32
Tito Piezas III
27.9k369179
edited Dec 8 '15 at 8:32
Tito Piezas III
27.9k369179
edited Dec 8 '15 at 8:32
Tito Piezas III
27.9k369179
27.9k369179
asked Aug 5 '15 at 23:33
AmrAmr
14.4k43296
asked Aug 5 '15 at 23:33
AmrAmr
14.4k43296
asked Aug 5 '15 at 23:33
AmrAmr
14.4k43296
14.4k43296
5
$begingroup$
ams.org/journals/mcom/2007-76-259/S0025-5718-07-01947-3/…
$endgroup$
– Will Jagy
Aug 5 '15 at 23:42
7
$begingroup$
From Wolfram: "... all numbers $N<1000$ and not of the form $9n pm 4$ are known to be expressible as the sum $N=A^3+B^3+C^3$ of three (positive or negative) cubes with the exception of $N=33, 42, 74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921$, and $975$ (Miller and Woollett 1955; Gardiner et al. 1964; Guy 1994, p. 151; Mishima; Elsenhaus and Jahnel 2007). Examples include $30 = (-283059965)^3+(-2218888517)^3+2220422932^3$ ..."
$endgroup$
– 727
Aug 5 '15 at 23:42
$begingroup$
@WillJagy - a great find!
$endgroup$
– hypergeometric
Aug 7 '15 at 16:16
$begingroup$
Table 2 in the paper cited by Jagy above needs an update.
$endgroup$
– Tito Piezas III
Dec 8 '15 at 6:25
$begingroup$
As of 2016, three solutions are now known for $N=30$, $$2220422932^3 - 283059965^3 - 2218888517^3 =30\ 3982933876681^3 - 636600549515^3 - 3977505554546^3 = 30\ -662037799708799^3 + 190809268841284^3 + 656711689254565^3 = 30$$ with the last found by S. Huisman.
$endgroup$
– Tito Piezas III
Dec 28 '17 at 7:00
|
show 1 more comment
5
$begingroup$
ams.org/journals/mcom/2007-76-259/S0025-5718-07-01947-3/…
$endgroup$
– Will Jagy
Aug 5 '15 at 23:42
7
$begingroup$
From Wolfram: "... all numbers $N<1000$ and not of the form $9n pm 4$ are known to be expressible as the sum $N=A^3+B^3+C^3$ of three (positive or negative) cubes with the exception of $N=33, 42, 74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921$, and $975$ (Miller and Woollett 1955; Gardiner et al. 1964; Guy 1994, p. 151; Mishima; Elsenhaus and Jahnel 2007). Examples include $30 = (-283059965)^3+(-2218888517)^3+2220422932^3$ ..."
$endgroup$
– 727
Aug 5 '15 at 23:42
$begingroup$
@WillJagy - a great find!
$endgroup$
– hypergeometric
Aug 7 '15 at 16:16
$begingroup$
Table 2 in the paper cited by Jagy above needs an update.
$endgroup$
– Tito Piezas III
Dec 8 '15 at 6:25
$begingroup$
As of 2016, three solutions are now known for $N=30$, $$2220422932^3 - 283059965^3 - 2218888517^3 =30\ 3982933876681^3 - 636600549515^3 - 3977505554546^3 = 30\ -662037799708799^3 + 190809268841284^3 + 656711689254565^3 = 30$$ with the last found by S. Huisman.
$endgroup$
– Tito Piezas III
Dec 28 '17 at 7:00
5
5
$begingroup$
ams.org/journals/mcom/2007-76-259/S0025-5718-07-01947-3/…
$endgroup$
– Will Jagy
Aug 5 '15 at 23:42
$begingroup$
ams.org/journals/mcom/2007-76-259/S0025-5718-07-01947-3/…
$endgroup$
– Will Jagy
Aug 5 '15 at 23:42
7
7
$begingroup$
From Wolfram: "... all numbers $N<1000$ and not of the form $9n pm 4$ are known to be expressible as the sum $N=A^3+B^3+C^3$ of three (positive or negative) cubes with the exception of $N=33, 42, 74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921$, and $975$ (Miller and Woollett 1955; Gardiner et al. 1964; Guy 1994, p. 151; Mishima; Elsenhaus and Jahnel 2007). Examples include $30 = (-283059965)^3+(-2218888517)^3+2220422932^3$ ..."
$endgroup$
– 727
Aug 5 '15 at 23:42
$begingroup$
From Wolfram: "... all numbers $N<1000$ and not of the form $9n pm 4$ are known to be expressible as the sum $N=A^3+B^3+C^3$ of three (positive or negative) cubes with the exception of $N=33, 42, 74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921$, and $975$ (Miller and Woollett 1955; Gardiner et al. 1964; Guy 1994, p. 151; Mishima; Elsenhaus and Jahnel 2007). Examples include $30 = (-283059965)^3+(-2218888517)^3+2220422932^3$ ..."
$endgroup$
– 727
Aug 5 '15 at 23:42
$begingroup$
@WillJagy - a great find!
$endgroup$
– hypergeometric
Aug 7 '15 at 16:16
$begingroup$
@WillJagy - a great find!
$endgroup$
– hypergeometric
Aug 7 '15 at 16:16
$begingroup$
Table 2 in the paper cited by Jagy above needs an update.
$endgroup$
– Tito Piezas III
Dec 8 '15 at 6:25
$begingroup$
Table 2 in the paper cited by Jagy above needs an update.
$endgroup$
– Tito Piezas III
Dec 8 '15 at 6:25
$begingroup$
As of 2016, three solutions are now known for $N=30$, $$2220422932^3 - 283059965^3 - 2218888517^3 =30\ 3982933876681^3 - 636600549515^3 - 3977505554546^3 = 30\ -662037799708799^3 + 190809268841284^3 + 656711689254565^3 = 30$$ with the last found by S. Huisman.
$endgroup$
– Tito Piezas III
Dec 28 '17 at 7:00
$begingroup$
As of 2016, three solutions are now known for $N=30$, $$2220422932^3 - 283059965^3 - 2218888517^3 =30\ 3982933876681^3 - 636600549515^3 - 3977505554546^3 = 30\ -662037799708799^3 + 190809268841284^3 + 656711689254565^3 = 30$$ with the last found by S. Huisman.
$endgroup$
– Tito Piezas III
Dec 28 '17 at 7:00
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
($colorredUpdate:$ March 2019. The case $N=33$ has been found.)
The equation
$$x^3+y^3+z^3 = N$$
has been oft-discussed in both MSE and MO. For example, see this, this, and this.
Searching a low range $|x,y,z|$ just won't do. It's quite interesting to see how search ranges have increased over the years using ever more clever algorithms.
I. 1993
$$134476^3+ 117367^3 -159380^3 = 39$$
D. Heath-brown, W. Lioen, and H. Te Riele, "On Solving the Diophantine Equation $x^3+y^3+z^3=k$ on a Vector Computer".
II. 1994
$$40500964^3+ 22894759^3-42805979^3 = 84$$
B. Conn and L. Vaserstein, "On sums of three integral cubes".
III. 1999
$$2220422932^3 -283059965^3 -2218888517^3 = 30$$
Michael Beck, Eric Pine, Wayne Tarrant, and Kim Yarbrough (see p.18 of Noam Elkies, "Rational points near curves and small non-zero $|x^3-y^2|$ via lattice reduction").
IV. 2001
$$25585441403^3 + 47272468418^3 - 49649244505^3 = 834$$
with search bound $10^11$ found by D.J. Bernstein.
V. 2009
$$2322626411251^3 + 19868127639556^3 - 19878702430997^3 =894$$
with search bound $10^14$ found by A. Elsenhans and J. Jahnel.
VI. 2016
$$66229832190556^3 + 283450105697727^3 −284650292555885^3 = 74$$
with search bound $10^15$ found by S. Huisman.
VII. 2019
$$8866128975287528^3 -8778405442862239^3 -2736111468807040^3 = 33$$
with search bound $10^16$ found by Andrew Booker.
Papers
In "New integer representations as the sum of three cubes" (2007) by Beck, Pine, Tarrant, and Yarbrough-Jensen they give a list of 28 $N<1000$ with no $x,y,z$ decomposition.
In "New sums of three cubes" by A. Elsenhans and J. Jahnel (2009) this has been reduced to just 14 unsolved $N$ (also quoted in Mathworld) namely,
$$N = colorred33, 42, colorred74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, 975$$
Update: Numbers in red have been solved, so there are now just 12 unsolved $N$. Hopefully, over the years, we can slowly complete this list.
Note: Relevant data are also given by Leonid Durman (inc. $x_1^4+x_2^4+x_3^4 = z^4$), by Mishima, while other solutions can be found in Elsenhans and Jahnel's site.
answered Dec 8 '15 at 6:21
Tito Piezas IIITito Piezas III
27.9k369179
$endgroup$
2
$begingroup$
You might want to update your table as $N=74$ is solved. $$ (-284650292555885)^3 + 66229832190556^3 + 283450105697727^3 = 74 $$
$endgroup$
– Yong Hao Ng
Dec 28 '17 at 5:58
3
$begingroup$
33 is solved as well! $33=8866128975287528^3+(-8778405442862239)^3+(-2736111468807040)^3$
$endgroup$
– mkocabas
Mar 10 at 6:22
1
$begingroup$
According to Gil Kalai, credit for finding that solution for 33 is due to Andrew Booker. Booker documented the search in his paper Cracking the Problem with 33.
$endgroup$
– Rosie F
Mar 11 at 16:42
1
$begingroup$
@RosieF. I've looked again at Heath-Brown et al's 1993 paper and they explicitly say they found the first solution for $N=39$. And I looked at Lazarus et al's 1964 paper. Their range was only about 66,000, so they couldn't have found $N=39$ back then. So the possibility is that your copy of Gardner's Knotted Doughnuts is not the 1986 first printing.
$endgroup$
– Tito Piezas III
Mar 12 at 1:39
1
$begingroup$
@RosieF: Gardner in p. 229 of Knotted Doughnuts remarks "Some were not easy to come by, notably the expression for 87 in which each cube has four digits." I find it strange that he makes this remark, while ignoring 39 in the same table with solutions of six digits. It's almost as if 39 was added later.
$endgroup$
– Tito Piezas III
Mar 12 at 2:36
|
show 6 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
($colorredUpdate:$ March 2019. The case $N=33$ has been found.)
The equation
$$x^3+y^3+z^3 = N$$
has been oft-discussed in both MSE and MO. For example, see this, this, and this.
Searching a low range $|x,y,z|$ just won't do. It's quite interesting to see how search ranges have increased over the years using ever more clever algorithms.
I. 1993
$$134476^3+ 117367^3 -159380^3 = 39$$
D. Heath-brown, W. Lioen, and H. Te Riele, "On Solving the Diophantine Equation $x^3+y^3+z^3=k$ on a Vector Computer".
II. 1994
$$40500964^3+ 22894759^3-42805979^3 = 84$$
B. Conn and L. Vaserstein, "On sums of three integral cubes".
III. 1999
$$2220422932^3 -283059965^3 -2218888517^3 = 30$$
Michael Beck, Eric Pine, Wayne Tarrant, and Kim Yarbrough (see p.18 of Noam Elkies, "Rational points near curves and small non-zero $|x^3-y^2|$ via lattice reduction").
IV. 2001
$$25585441403^3 + 47272468418^3 - 49649244505^3 = 834$$
with search bound $10^11$ found by D.J. Bernstein.
V. 2009
$$2322626411251^3 + 19868127639556^3 - 19878702430997^3 =894$$
with search bound $10^14$ found by A. Elsenhans and J. Jahnel.
VI. 2016
$$66229832190556^3 + 283450105697727^3 −284650292555885^3 = 74$$
with search bound $10^15$ found by S. Huisman.
VII. 2019
$$8866128975287528^3 -8778405442862239^3 -2736111468807040^3 = 33$$
with search bound $10^16$ found by Andrew Booker.
Papers
In "New integer representations as the sum of three cubes" (2007) by Beck, Pine, Tarrant, and Yarbrough-Jensen they give a list of 28 $N<1000$ with no $x,y,z$ decomposition.
In "New sums of three cubes" by A. Elsenhans and J. Jahnel (2009) this has been reduced to just 14 unsolved $N$ (also quoted in Mathworld) namely,
$$N = colorred33, 42, colorred74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, 975$$
Update: Numbers in red have been solved, so there are now just 12 unsolved $N$. Hopefully, over the years, we can slowly complete this list.
Note: Relevant data are also given by Leonid Durman (inc. $x_1^4+x_2^4+x_3^4 = z^4$), by Mishima, while other solutions can be found in Elsenhans and Jahnel's site.
answered Dec 8 '15 at 6:21
Tito Piezas IIITito Piezas III
27.9k369179
$endgroup$
2
$begingroup$
You might want to update your table as $N=74$ is solved. $$ (-284650292555885)^3 + 66229832190556^3 + 283450105697727^3 = 74 $$
$endgroup$
– Yong Hao Ng
Dec 28 '17 at 5:58
3
$begingroup$
33 is solved as well! $33=8866128975287528^3+(-8778405442862239)^3+(-2736111468807040)^3$
$endgroup$
– mkocabas
Mar 10 at 6:22
1
$begingroup$
According to Gil Kalai, credit for finding that solution for 33 is due to Andrew Booker. Booker documented the search in his paper Cracking the Problem with 33.
$endgroup$
– Rosie F
Mar 11 at 16:42
1
$begingroup$
@RosieF. I've looked again at Heath-Brown et al's 1993 paper and they explicitly say they found the first solution for $N=39$. And I looked at Lazarus et al's 1964 paper. Their range was only about 66,000, so they couldn't have found $N=39$ back then. So the possibility is that your copy of Gardner's Knotted Doughnuts is not the 1986 first printing.
$endgroup$
– Tito Piezas III
Mar 12 at 1:39
1
$begingroup$
@RosieF: Gardner in p. 229 of Knotted Doughnuts remarks "Some were not easy to come by, notably the expression for 87 in which each cube has four digits." I find it strange that he makes this remark, while ignoring 39 in the same table with solutions of six digits. It's almost as if 39 was added later.
$endgroup$
– Tito Piezas III
Mar 12 at 2:36
|
show 6 more comments
$begingroup$
($colorredUpdate:$ March 2019. The case $N=33$ has been found.)
The equation
$$x^3+y^3+z^3 = N$$
has been oft-discussed in both MSE and MO. For example, see this, this, and this.
Searching a low range $|x,y,z|$ just won't do. It's quite interesting to see how search ranges have increased over the years using ever more clever algorithms.
I. 1993
$$134476^3+ 117367^3 -159380^3 = 39$$
D. Heath-brown, W. Lioen, and H. Te Riele, "On Solving the Diophantine Equation $x^3+y^3+z^3=k$ on a Vector Computer".
II. 1994
$$40500964^3+ 22894759^3-42805979^3 = 84$$
B. Conn and L. Vaserstein, "On sums of three integral cubes".
III. 1999
$$2220422932^3 -283059965^3 -2218888517^3 = 30$$
Michael Beck, Eric Pine, Wayne Tarrant, and Kim Yarbrough (see p.18 of Noam Elkies, "Rational points near curves and small non-zero $|x^3-y^2|$ via lattice reduction").
IV. 2001
$$25585441403^3 + 47272468418^3 - 49649244505^3 = 834$$
with search bound $10^11$ found by D.J. Bernstein.
V. 2009
$$2322626411251^3 + 19868127639556^3 - 19878702430997^3 =894$$
with search bound $10^14$ found by A. Elsenhans and J. Jahnel.
VI. 2016
$$66229832190556^3 + 283450105697727^3 −284650292555885^3 = 74$$
with search bound $10^15$ found by S. Huisman.
VII. 2019
$$8866128975287528^3 -8778405442862239^3 -2736111468807040^3 = 33$$
with search bound $10^16$ found by Andrew Booker.
Papers
In "New integer representations as the sum of three cubes" (2007) by Beck, Pine, Tarrant, and Yarbrough-Jensen they give a list of 28 $N<1000$ with no $x,y,z$ decomposition.
In "New sums of three cubes" by A. Elsenhans and J. Jahnel (2009) this has been reduced to just 14 unsolved $N$ (also quoted in Mathworld) namely,
$$N = colorred33, 42, colorred74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, 975$$
Update: Numbers in red have been solved, so there are now just 12 unsolved $N$. Hopefully, over the years, we can slowly complete this list.
Note: Relevant data are also given by Leonid Durman (inc. $x_1^4+x_2^4+x_3^4 = z^4$), by Mishima, while other solutions can be found in Elsenhans and Jahnel's site.
answered Dec 8 '15 at 6:21
Tito Piezas IIITito Piezas III
27.9k369179
$endgroup$
2
$begingroup$
You might want to update your table as $N=74$ is solved. $$ (-284650292555885)^3 + 66229832190556^3 + 283450105697727^3 = 74 $$
$endgroup$
– Yong Hao Ng
Dec 28 '17 at 5:58
3
$begingroup$
33 is solved as well! $33=8866128975287528^3+(-8778405442862239)^3+(-2736111468807040)^3$
$endgroup$
– mkocabas
Mar 10 at 6:22
1
$begingroup$
According to Gil Kalai, credit for finding that solution for 33 is due to Andrew Booker. Booker documented the search in his paper Cracking the Problem with 33.
$endgroup$
– Rosie F
Mar 11 at 16:42
1
$begingroup$
@RosieF. I've looked again at Heath-Brown et al's 1993 paper and they explicitly say they found the first solution for $N=39$. And I looked at Lazarus et al's 1964 paper. Their range was only about 66,000, so they couldn't have found $N=39$ back then. So the possibility is that your copy of Gardner's Knotted Doughnuts is not the 1986 first printing.
$endgroup$
– Tito Piezas III
Mar 12 at 1:39
1
$begingroup$
@RosieF: Gardner in p. 229 of Knotted Doughnuts remarks "Some were not easy to come by, notably the expression for 87 in which each cube has four digits." I find it strange that he makes this remark, while ignoring 39 in the same table with solutions of six digits. It's almost as if 39 was added later.
$endgroup$
– Tito Piezas III
Mar 12 at 2:36
|
show 6 more comments
$begingroup$
($colorredUpdate:$ March 2019. The case $N=33$ has been found.)
The equation
$$x^3+y^3+z^3 = N$$
has been oft-discussed in both MSE and MO. For example, see this, this, and this.
Searching a low range $|x,y,z|$ just won't do. It's quite interesting to see how search ranges have increased over the years using ever more clever algorithms.
I. 1993
$$134476^3+ 117367^3 -159380^3 = 39$$
D. Heath-brown, W. Lioen, and H. Te Riele, "On Solving the Diophantine Equation $x^3+y^3+z^3=k$ on a Vector Computer".
II. 1994
$$40500964^3+ 22894759^3-42805979^3 = 84$$
B. Conn and L. Vaserstein, "On sums of three integral cubes".
III. 1999
$$2220422932^3 -283059965^3 -2218888517^3 = 30$$
Michael Beck, Eric Pine, Wayne Tarrant, and Kim Yarbrough (see p.18 of Noam Elkies, "Rational points near curves and small non-zero $|x^3-y^2|$ via lattice reduction").
IV. 2001
$$25585441403^3 + 47272468418^3 - 49649244505^3 = 834$$
with search bound $10^11$ found by D.J. Bernstein.
V. 2009
$$2322626411251^3 + 19868127639556^3 - 19878702430997^3 =894$$
with search bound $10^14$ found by A. Elsenhans and J. Jahnel.
VI. 2016
$$66229832190556^3 + 283450105697727^3 −284650292555885^3 = 74$$
with search bound $10^15$ found by S. Huisman.
VII. 2019
$$8866128975287528^3 -8778405442862239^3 -2736111468807040^3 = 33$$
with search bound $10^16$ found by Andrew Booker.
Papers
In "New integer representations as the sum of three cubes" (2007) by Beck, Pine, Tarrant, and Yarbrough-Jensen they give a list of 28 $N<1000$ with no $x,y,z$ decomposition.
In "New sums of three cubes" by A. Elsenhans and J. Jahnel (2009) this has been reduced to just 14 unsolved $N$ (also quoted in Mathworld) namely,
$$N = colorred33, 42, colorred74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, 975$$
Update: Numbers in red have been solved, so there are now just 12 unsolved $N$. Hopefully, over the years, we can slowly complete this list.
Note: Relevant data are also given by Leonid Durman (inc. $x_1^4+x_2^4+x_3^4 = z^4$), by Mishima, while other solutions can be found in Elsenhans and Jahnel's site.
answered Dec 8 '15 at 6:21
Tito Piezas IIITito Piezas III
27.9k369179
$endgroup$
($colorredUpdate:$ March 2019. The case $N=33$ has been found.)
The equation
$$x^3+y^3+z^3 = N$$
has been oft-discussed in both MSE and MO. For example, see this, this, and this.
Searching a low range $|x,y,z|$ just won't do. It's quite interesting to see how search ranges have increased over the years using ever more clever algorithms.
I. 1993
$$134476^3+ 117367^3 -159380^3 = 39$$
D. Heath-brown, W. Lioen, and H. Te Riele, "On Solving the Diophantine Equation $x^3+y^3+z^3=k$ on a Vector Computer".
II. 1994
$$40500964^3+ 22894759^3-42805979^3 = 84$$
B. Conn and L. Vaserstein, "On sums of three integral cubes".
III. 1999
$$2220422932^3 -283059965^3 -2218888517^3 = 30$$
Michael Beck, Eric Pine, Wayne Tarrant, and Kim Yarbrough (see p.18 of Noam Elkies, "Rational points near curves and small non-zero $|x^3-y^2|$ via lattice reduction").
IV. 2001
$$25585441403^3 + 47272468418^3 - 49649244505^3 = 834$$
with search bound $10^11$ found by D.J. Bernstein.
V. 2009
$$2322626411251^3 + 19868127639556^3 - 19878702430997^3 =894$$
with search bound $10^14$ found by A. Elsenhans and J. Jahnel.
VI. 2016
$$66229832190556^3 + 283450105697727^3 −284650292555885^3 = 74$$
with search bound $10^15$ found by S. Huisman.
VII. 2019
$$8866128975287528^3 -8778405442862239^3 -2736111468807040^3 = 33$$
with search bound $10^16$ found by Andrew Booker.
Papers
In "New integer representations as the sum of three cubes" (2007) by Beck, Pine, Tarrant, and Yarbrough-Jensen they give a list of 28 $N<1000$ with no $x,y,z$ decomposition.
In "New sums of three cubes" by A. Elsenhans and J. Jahnel (2009) this has been reduced to just 14 unsolved $N$ (also quoted in Mathworld) namely,
$$N = colorred33, 42, colorred74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, 975$$
Update: Numbers in red have been solved, so there are now just 12 unsolved $N$. Hopefully, over the years, we can slowly complete this list.
Note: Relevant data are also given by Leonid Durman (inc. $x_1^4+x_2^4+x_3^4 = z^4$), by Mishima, while other solutions can be found in Elsenhans and Jahnel's site.
answered Dec 8 '15 at 6:21
Tito Piezas IIITito Piezas III
27.9k369179
edited Mar 12 at 1:27
answered Dec 8 '15 at 6:21
Tito Piezas IIITito Piezas III
27.9k369179
answered Dec 8 '15 at 6:21
Tito Piezas IIITito Piezas III
27.9k369179
answered Dec 8 '15 at 6:21
Tito Piezas IIITito Piezas III
27.9k369179
27.9k369179
2
$begingroup$
You might want to update your table as $N=74$ is solved. $$ (-284650292555885)^3 + 66229832190556^3 + 283450105697727^3 = 74 $$
$endgroup$
– Yong Hao Ng
Dec 28 '17 at 5:58
3
$begingroup$
33 is solved as well! $33=8866128975287528^3+(-8778405442862239)^3+(-2736111468807040)^3$
$endgroup$
– mkocabas
Mar 10 at 6:22
1
$begingroup$
According to Gil Kalai, credit for finding that solution for 33 is due to Andrew Booker. Booker documented the search in his paper Cracking the Problem with 33.
$endgroup$
– Rosie F
Mar 11 at 16:42
1
$begingroup$
@RosieF. I've looked again at Heath-Brown et al's 1993 paper and they explicitly say they found the first solution for $N=39$. And I looked at Lazarus et al's 1964 paper. Their range was only about 66,000, so they couldn't have found $N=39$ back then. So the possibility is that your copy of Gardner's Knotted Doughnuts is not the 1986 first printing.
$endgroup$
– Tito Piezas III
Mar 12 at 1:39
1
$begingroup$
@RosieF: Gardner in p. 229 of Knotted Doughnuts remarks "Some were not easy to come by, notably the expression for 87 in which each cube has four digits." I find it strange that he makes this remark, while ignoring 39 in the same table with solutions of six digits. It's almost as if 39 was added later.
$endgroup$
– Tito Piezas III
Mar 12 at 2:36
|
show 6 more comments
2
$begingroup$
You might want to update your table as $N=74$ is solved. $$ (-284650292555885)^3 + 66229832190556^3 + 283450105697727^3 = 74 $$
$endgroup$
– Yong Hao Ng
Dec 28 '17 at 5:58
3
$begingroup$
33 is solved as well! $33=8866128975287528^3+(-8778405442862239)^3+(-2736111468807040)^3$
$endgroup$
– mkocabas
Mar 10 at 6:22
1
$begingroup$
According to Gil Kalai, credit for finding that solution for 33 is due to Andrew Booker. Booker documented the search in his paper Cracking the Problem with 33.
$endgroup$
– Rosie F
Mar 11 at 16:42
1
$begingroup$
@RosieF. I've looked again at Heath-Brown et al's 1993 paper and they explicitly say they found the first solution for $N=39$. And I looked at Lazarus et al's 1964 paper. Their range was only about 66,000, so they couldn't have found $N=39$ back then. So the possibility is that your copy of Gardner's Knotted Doughnuts is not the 1986 first printing.
$endgroup$
– Tito Piezas III
Mar 12 at 1:39
1
$begingroup$
@RosieF: Gardner in p. 229 of Knotted Doughnuts remarks "Some were not easy to come by, notably the expression for 87 in which each cube has four digits." I find it strange that he makes this remark, while ignoring 39 in the same table with solutions of six digits. It's almost as if 39 was added later.
$endgroup$
– Tito Piezas III
Mar 12 at 2:36
2
2
$begingroup$
You might want to update your table as $N=74$ is solved. $$ (-284650292555885)^3 + 66229832190556^3 + 283450105697727^3 = 74 $$
$endgroup$
– Yong Hao Ng
Dec 28 '17 at 5:58
$begingroup$
You might want to update your table as $N=74$ is solved. $$ (-284650292555885)^3 + 66229832190556^3 + 283450105697727^3 = 74 $$
$endgroup$
– Yong Hao Ng
Dec 28 '17 at 5:58
3
3
$begingroup$
33 is solved as well! $33=8866128975287528^3+(-8778405442862239)^3+(-2736111468807040)^3$
$endgroup$
– mkocabas
Mar 10 at 6:22
$begingroup$
33 is solved as well! $33=8866128975287528^3+(-8778405442862239)^3+(-2736111468807040)^3$
$endgroup$
– mkocabas
Mar 10 at 6:22
1
1
$begingroup$
According to Gil Kalai, credit for finding that solution for 33 is due to Andrew Booker. Booker documented the search in his paper Cracking the Problem with 33.
$endgroup$
– Rosie F
Mar 11 at 16:42
$begingroup$
According to Gil Kalai, credit for finding that solution for 33 is due to Andrew Booker. Booker documented the search in his paper Cracking the Problem with 33.
$endgroup$
– Rosie F
Mar 11 at 16:42
1
1
$begingroup$
@RosieF. I've looked again at Heath-Brown et al's 1993 paper and they explicitly say they found the first solution for $N=39$. And I looked at Lazarus et al's 1964 paper. Their range was only about 66,000, so they couldn't have found $N=39$ back then. So the possibility is that your copy of Gardner's Knotted Doughnuts is not the 1986 first printing.
$endgroup$
– Tito Piezas III
Mar 12 at 1:39
$begingroup$
@RosieF. I've looked again at Heath-Brown et al's 1993 paper and they explicitly say they found the first solution for $N=39$. And I looked at Lazarus et al's 1964 paper. Their range was only about 66,000, so they couldn't have found $N=39$ back then. So the possibility is that your copy of Gardner's Knotted Doughnuts is not the 1986 first printing.
$endgroup$
– Tito Piezas III
Mar 12 at 1:39
1
1
$begingroup$
@RosieF: Gardner in p. 229 of Knotted Doughnuts remarks "Some were not easy to come by, notably the expression for 87 in which each cube has four digits." I find it strange that he makes this remark, while ignoring 39 in the same table with solutions of six digits. It's almost as if 39 was added later.
$endgroup$
– Tito Piezas III
Mar 12 at 2:36
$begingroup$
@RosieF: Gardner in p. 229 of Knotted Doughnuts remarks "Some were not easy to come by, notably the expression for 87 in which each cube has four digits." I find it strange that he makes this remark, while ignoring 39 in the same table with solutions of six digits. It's almost as if 39 was added later.
$endgroup$
– Tito Piezas III
Mar 12 at 2:36
|
show 6 more comments
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5
$begingroup$
ams.org/journals/mcom/2007-76-259/S0025-5718-07-01947-3/…
$endgroup$
– Will Jagy
Aug 5 '15 at 23:42
7
$begingroup$
From Wolfram: "... all numbers $N<1000$ and not of the form $9n pm 4$ are known to be expressible as the sum $N=A^3+B^3+C^3$ of three (positive or negative) cubes with the exception of $N=33, 42, 74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921$, and $975$ (Miller and Woollett 1955; Gardiner et al. 1964; Guy 1994, p. 151; Mishima; Elsenhaus and Jahnel 2007). Examples include $30 = (-283059965)^3+(-2218888517)^3+2220422932^3$ ..."
$endgroup$
– 727
Aug 5 '15 at 23:42
$begingroup$
@WillJagy - a great find!
$endgroup$
– hypergeometric
Aug 7 '15 at 16:16
$begingroup$
Table 2 in the paper cited by Jagy above needs an update.
$endgroup$
– Tito Piezas III
Dec 8 '15 at 6:25
$begingroup$
As of 2016, three solutions are now known for $N=30$, $$2220422932^3 - 283059965^3 - 2218888517^3 =30\ 3982933876681^3 - 636600549515^3 - 3977505554546^3 = 30\ -662037799708799^3 + 190809268841284^3 + 656711689254565^3 = 30$$ with the last found by S. Huisman.
$endgroup$
– Tito Piezas III
Dec 28 '17 at 7:00